Definite Integrals and Area for CUET Maths

Q: What does the fundamental theorem of calculus say?

It comes in two parts. The first: the area function A(x) = ∫ₐ^x f(t) dt has derivative A'(x) = f(x), so integrating and then differentiating gives back the function. The second: ∫ₐ^b f(x) dx = F(b) - F(a) for any antiderivative F of f, which is how every definite integral is evaluated.

Q: How do you change the limits in a substitution?

When you put t = g(x), change the limits too: the new lower limit is g(a) and the new upper limit is g(b). Then finish in t, with no need to go back to x.

Q: What is the integral of an odd function from −a to a?

It is 0. An odd function satisfies f(-x) = -f(x), so its graph is symmetric about the origin, and the area below the axis on one side exactly cancels the area above it on the other. So ∫₋₂^2 (x^3 - 5x) dx = 0 with no working at all.

Q: How do you find the area under a curve?

For a curve y = f(x) that stays above the x -axis between x = a and x = b, the area of the region between the curve, the axis and those two lines is ∫ₐ^b f(x) dx. If the curve is below the axis, the integral comes out negative and the area is its absolute value; if it crosses the axis, split at the crossing and add the sizes of the parts.

About 5 min to learn · 71 questions

Definite integrals and the area under a curve are part of Section A of CUET Maths, the section every candidate answers, whether you take Mathematics or Applied Mathematics. They turn an integral into a number, and a handful of properties turn long calculations into one-line answers. Learn each rule here, from the fundamental theorem and changing the limits to the odd, even and a − x properties and the area under a simple curve, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section A1

On this page · 10 sections

01The definite integral
02The fundamental theorem of calculus
03Substitution and by parts with limits
04Properties of definite integrals
05Area under a simple curve
06How CUET asks this chapter
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

The definite integral

The ball's velocity is $20 - 10t$ m/s. How far does it rise in its first 2 seconds? Its height is $20t - 5t^2$ plus a constant, so the rise is the height at $t = 2$ minus the height at $t = 0$ : $(40 - 20) - 0 = 20$ metres. The constant cancels. That subtraction is a definite integral.

What is a definite integral?

A definite integral is a number, not a function: the integral of f from a to b. For a function that stays above the x-axis, it is the area between the graph and the axis from a to b. It is worked out from any antiderivative F as F(b) minus F(a), and the constant of integration always cancels. It is written $\int_a^b f(x)\,dx$ .

The fundamental theorem of calculus

What does the fundamental theorem of calculus say?

It comes in two parts. The first: the area function $A(x) = \int_a^x f(t)\,dt$ has derivative $A'(x) = f(x)$ , so integrating and then differentiating gives back the function. The second: $\int_a^b f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$ of $f$ , which is how every definite integral is evaluated.

Example 1 · The second fundamental theorem

Evaluate $\int_1^3 (2x + 3x^2)\,dx$ .

An antiderivative is $F(x) = x^2 + x^3$ .
$F(3) - F(1) = (9 + 27) - (1 + 1)$ .

Answer

Example 2 · The first fundamental theorem

Find $\frac{d}{dx}\int_0^{x^3} e^t\,dt$ .

The integral is $e^{x^3} - 1$ , the area function evaluated at $x^3$ .
Differentiate with the chain rule: $e^{x^3} \cdot 3x^2$ .

Answer

$3x^2e^{x^3}$ : the integrand at the upper limit, times the derivative of the limit.

Watch out

Subtract in the right order, top minus bottom. Writing $F(a) - F(b)$ gives the right number with the wrong sign, and forgetting $F(a)$ altogether is just as common.

Substitution and by parts with limits

How do you change the limits in a substitution?

When you put $t = g(x)$ , change the limits too: the new lower limit is $g(a)$ and the new upper limit is $g(b)$ . Then finish in $t$ , with no need to go back to $x$ .

Example 3 · New variable, new limits

Evaluate $\int_0^1 \frac{x^2}{1 + x^3}\,dx$ .

Put $t = 1 + x^3$ , so $dt = 3x^2\,dx$ . At $x = 0$ , $t = 1$ ; at $x = 1$ , $t = 2$ .
The integral is $\frac{1}{3}\int_1^2 \frac{dt}{t} = \frac{1}{3}\left(\log 2 - \log 1\right)$ .

Answer

\frac{1}{3}\log 2

By parts works the same way with limits: $\int_a^b u\,dv = \left[uv\right]_a^b - \int_a^b v\,du$ .

Example 4 · By parts with limits

Evaluate $\int_0^{\pi} x\cos x\,dx$ .

$u = x$ , $dv = \cos x\,dx$ : an antiderivative is $x\sin x + \cos x$ .
At $\pi$ : $0 + (-1) = -1$ . At 0: $0 + 1 = 1$ .

Answer

-1 - 1 = -2

Properties of definite integrals

These properties often replace a long calculation with a single line.

The properties CUET uses

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx, \qquad \int_a^b f = \int_a^c f + \int_c^b f

\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx, \qquad \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \text{ if } f \text{ is even}, \qquad \int_{-a}^{a} f(x)\,dx = 0 \text{ if } f \text{ is odd}

What is the integral of an odd function from −a to a?

It is 0. An odd function satisfies $f(-x) = -f(x)$ , so its graph is symmetric about the origin, and the area below the axis on one side exactly cancels the area above it on the other. So $\int_{-2}^{2} (x^3 - 5x)\,dx = 0$ with no working at all.

A product is odd when one factor is odd and the other even: $x^3\cos x$ is odd, while $x\sin x$ is even.

Example 5 · Using a − x

Evaluate $\int_0^2 x(2 - x)^3\,dx$ .

Replace $x$ by $2 - x$ : $\int_0^2 (2 - x)x^3\,dx = \int_0^2 (2x^3 - x^4)\,dx$ .
$\left[\frac{x^4}{2} - \frac{x^5}{5}\right]_0^2 = 8 - \frac{32}{5}$ .

Answer

\frac{8}{5}

Key idea

The same idea settles a whole family at once. For $I = \int_0^{\frac{\pi}{2}} \frac{g(\sin x)}{g(\sin x) + g(\cos x)}\,dx$ , replacing $x$ by $\frac{\pi}{2} - x$ swaps $\sin$ and $\cos$ ; adding the two forms gives $2I = \int_0^{\frac{\pi}{2}} 1\,dx$ , so $I = \frac{\pi}{4}$ whatever $g$ is.

Example 6 · An absolute value

Evaluate $\int_{-1}^{2} \lvert x \rvert\,dx$ .

$\lvert x \rvert = -x$ for $x < 0$ and $x$ for $x \ge 0$ , so split at 0.
$\int_{-1}^{0} (-x)\,dx + \int_0^2 x\,dx = \frac{1}{2} + 2$ .

Answer

\frac{5}{2}

Area under a simple curve

How do you find the area under a curve?

For a curve $y = f(x)$ that stays above the $x$ -axis between $x = a$ and $x = b$ , the area of the region between the curve, the axis and those two lines is $\int_a^b f(x)\,dx$ . If the curve is below the axis, the integral comes out negative and the area is its absolute value; if it crosses the axis, split at the crossing and add the sizes of the parts.

The graph of y = x² + 1 with the region under it from x = 0 to x = 2 shaded, down to the x-axis. — The shaded area is the integral of x² + 1 from 0 to 2.

Example 7 · The shaded area

Find the area between $y = x^2 + 1$ , the $x$ -axis, $x = 0$ and $x = 2$ .

The curve is above the axis throughout, so the area is $\int_0^2 (x^2 + 1)\,dx$ .
$\left[\frac{x^3}{3} + x\right]_0^2 = \frac{8}{3} + 2$ .

Answer

$\frac{14}{3}$ square units

Watch out

An integral can be 0 while the area is not. $\int_0^{\pi} \cos x\,dx = 0$ , but the region between $y = \cos x$ and the axis on $[0, \pi]$ has area 2: the part below the axis counts with a plus sign in an area.

A region against the $y$ -axis, between $x = g(y)$ and the lines $y = c$ and $y = d$ , has area $\int_c^d g(y)\,dy$ . And integrating over a quarter and multiplying by 4 gives the two areas worth remembering: a circle of radius $r$ has area $\pi r^2$ , and the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has area $\pi ab$ .

How CUET asks this chapter

Evaluating: a definite integral of a standard function, or by substitution or by parts.
The first theorem: the derivative of an integral with a variable limit, or an unknown limit from a given value.
Properties: odd and even functions over $[-a, a]$ , the $a - x$ and $a + b - x$ tricks, an absolute value, or a value from given integrals, often as a statement question.
Area: under a simple curve, often drawn; a region below the axis or crossing it; a region against the $y$ -axis; a circle or an ellipse.

Exam shortcuts

Odd function over $[-a, a]$ : 0, at once.
$\int_0^{\frac{\pi}{2}} \frac{g(\sin x)}{g(\sin x) + g(\cos x)}\,dx = \frac{\pi}{4}$ .
$\int_a^b \frac{h(x)}{h(x) + h(a + b - x)}\,dx = \frac{b - a}{2}$ .
$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$ .
Ellipse area $\pi ab$ ; circle area $\pi r^2$ .

Common mistakes

Subtracting $F(b)$ from $F(a)$ , or forgetting $F(a)$ .
Keeping the old limits after a substitution.
Treating the integral as the area when the curve dips below the axis.
Calling a function odd or even without checking $f(-x)$ .
Forgetting the chain-rule factor $g'(x)$ when the upper limit is a function.

Quick revision

Evaluating

$\int_a^b f = F(b) - F(a)$ ; $C$ cancels
Substitution: change the limits with the variable

Properties

Swap limits: change sign; split at any $c$
$f(a + b - x)$ and $f(a - x)$ leave the value unchanged
Odd: 0; even: twice the half

Area

Above the axis: $\int_a^b f$ ; below: its absolute value
Crossing the axis: split and add the sizes

An equation that contains a derivative is solved by integrating: differential equations.

Next · 09Quick revision

Check your understanding

71 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

The fundamental theorem of calculusQ1 of 71Q1

The fundamental theorem of calculus

17 questions

Question 1

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Evaluate $\int_{0}^{2} \left(3 x^{2} + 2 x\right)\,dx$ .

-12

Correct answerYour answer

12

Correct answerYour answer

13

Correct answerYour answer

24

Correct answerYour answer

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AnswerB · $12$

An antiderivative is $F(x) = x^{2} \left(x + 1\right)$ .
By the fundamental theorem, the integral is $F(2) - F(0) = 12$ .

Worth noting

$-12$ subtracts the wrong way, $F(0) - F(2)$ .

Question 2

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Evaluate $\int_{1}^{e} \frac{1}{x}\,dx$ .

-1

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

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2

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AnswerC · $1$

An antiderivative is $F(x) = \log x$ .
By the fundamental theorem, the integral is $F(e) - F(1) = 1$ .

Worth noting

$-1$ subtracts the wrong way, $F(1) - F(e)$ .

Question 3

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Evaluate $\int_{0}^{\frac{\pi}{4}} \sec^{2} x\,dx$ .

-1

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

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2

Correct answerYour answer

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AnswerC · $1$

An antiderivative is $F(x) = \tan x$ .
By the fundamental theorem, the integral is $F(\frac{\pi}{4}) - F(0) = 1$ .

Worth noting

$-1$ subtracts the wrong way, $F(0) - F(\frac{\pi}{4})$ .

Question 4

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Evaluate $\int_{1}^{2} \left(4 x^{3} - 2 x\right)\,dx$ .

-12

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6

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11

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12

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AnswerD · $12$

An antiderivative is $F(x) = x^{4} - x^{2}$ .
By the fundamental theorem, the integral is $F(2) - F(1) = 12$ .

Worth noting

$-12$ subtracts the wrong way, $F(1) - F(2)$ .

Question 5

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Evaluate $\int_{0}^{\frac{\pi}{2}} \sin x\,dx$ .

-1

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0

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\frac{1}{2}

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1

Correct answerYour answer

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AnswerD · $1$

An antiderivative is $F(x) = - \cos x$ .
By the fundamental theorem, the integral is $F(\frac{\pi}{2}) - F(0) = 1$ .

Worth noting

$0$ is $F(\frac{\pi}{2})$ alone: $F(0)$ was not subtracted.

Question 6

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Evaluate $\int_{0}^{\log 3} e^{x}\,dx$ .

-2

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0

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1

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2

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AnswerD · $2$

An antiderivative is $F(x) = e^{x}$ .
By the fundamental theorem, the integral is $F(\log 3) - F(0) = 2$ .

Worth noting

$3$ is $F(\log 3)$ alone: $F(0)$ was not subtracted.

Question 7

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Evaluate $\int_{0}^{1} \frac{1}{x^{2} + 1}\,dx$ .

- \frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{2}

Correct answerYour answer

\frac{\pi}{4} + 1

Correct answerYour answer

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AnswerB · $\frac{\pi}{4}$

An antiderivative is $F(x) = \tan^{-1}{\left(x \right)}$ .
By the fundamental theorem, the integral is $F(1) - F(0) = \frac{\pi}{4}$ .

Worth noting

$- \frac{\pi}{4}$ subtracts the wrong way, $F(0) - F(1)$ .

Question 8

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If $\int_1^c 3x^2\,dx = 26$ and $c > 1$ , what is $c$ ?

\sqrt[3]{26}

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3

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9

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26

Correct answerYour answer

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AnswerB · $3$

$\int_1^c 3x^2\,dx = c^3 - 1$ .
$c^3 - 1 = 26$ gives $c^3 = 27$ , so $c = 3$ .

Worth noting

$\sqrt[3]{26}$ forgets to subtract the value at the lower limit, $1^3 = 1$ .

Question 9

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If $A(x) = \int_1^x (t^2 + 1)\,dt$ , what is $A'(2)$ ?

0

Correct answerYour answer

\frac{10}{3}

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4

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5

Correct answerYour answer

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AnswerD · $5$

By the first fundamental theorem, $A'(x) = x^2 + 1$ : the area function's derivative is the integrand at the upper limit.
$A'(2) = 4 + 1 = 5$ .

Worth noting

$\frac{10}{3}$ is $A(2)$ , the integral itself, not its derivative.

Question 10

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Which of the following statements are true?

I. A definite integral $\int_a^b f(x)\,dx$ is a number, not a function of $x$ .

II. $\int_a^a f(x)\,dx = 0$ for any $f$ defined at $a$ .

III. The constant $C$ of the antiderivative changes the value of $F(b) - F(a)$ .

IV. $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ .

I, II and IV onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerA · I, II and IV only

The limits are numbers, so the result is a number (I); with equal limits $F(a) - F(a) = 0$ (II); and the letter used for the variable does not matter (IV).
III is false: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$ , so $C$ cancels.

Question 11

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Evaluate $\int_{1}^{9} \frac{1}{\sqrt{x}}\,dx$ .

-4

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4

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6

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8

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AnswerB · $4$

An antiderivative is $F(x) = 2 \sqrt{x}$ .
By the fundamental theorem, the integral is $F(9) - F(1) = 4$ .

Worth noting

$6$ is $F(9)$ alone: $F(1)$ was not subtracted.

Question 12

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What is $\frac{d}{dx}\int_x^5 (t^3 + 1)\,dt$ ?

x^{3} + 1

Correct answerYour answer

- x^{3} - 1

Correct answerYour answer

- 3 x^{2}

Correct answerYour answer

\frac{x^{4}}{4} + x

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AnswerB · $- x^{3} - 1$

$\int_x^5 f(t)\,dt = -\int_5^x f(t)\,dt$ , and $\frac{d}{dx}\int_5^x f(t)\,dt = f(x)$ .
So the derivative is $-(x^3 + 1)$ .

Worth noting

$x^3 + 1$ forgets that $x$ is the lower limit, which turns the sign.

Question 13

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Evaluate $\int_{0}^{3} \frac{1}{x + 1}\,dx$ .

- \log 4

Correct answerYour answer

-1 + \log 4

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\frac{\log 4}{2}

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\log 4

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AnswerD · $\log 4$

An antiderivative is $F(x) = \log{\left(x + 1 \right)}$ .
By the fundamental theorem, the integral is $F(3) - F(0) = \log 4$ .

Worth noting

$- \log 4$ subtracts the wrong way, $F(0) - F(3)$ .

Question 14

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Evaluate $\int_{0}^{\log 2} e^{2 x}\,dx$ .

- \frac{3}{2}

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\frac{3}{4}

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\frac{3}{2}

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2

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AnswerC · $\frac{3}{2}$

An antiderivative is $F(x) = \frac{e^{2 x}}{2}$ .
By the fundamental theorem, the integral is $F(\log 2) - F(0) = \frac{3}{2}$ .

Worth noting

$2$ is $F(\log 2)$ alone: $F(0)$ was not subtracted.

Question 15

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If $\int_0^k (2x + 1)\,dx = 12$ and $k > 0$ , what is $k$ ?

3

Correct answerYour answer

2 \sqrt{3}

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4

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6

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AnswerA · $3$

$\int_0^k (2x + 1)\,dx = k^2 + k$ .
$k^2 + k - 12 = 0$ , so $(k + 4)(k - 3) = 0$ ; with $k > 0$ , $k = 3$ .

Worth noting

$2\sqrt{3}$ drops the $+1$ and solves $k^2 = 12$ .

Question 16

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What is $\frac{d}{dx}\int_0^{x^2} \cos t\,dt$ ?

2 x \cos{\left(x^{2} \right)}

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\cos{\left(x^{2} \right)}

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\sin{\left(x^{2} \right)}

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2 x \sin{\left(x^{2} \right)}

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AnswerA · $2 x \cos{\left(x^{2} \right)}$

The integral is $\sin(x^2)$ , since $\int_0^u \cos t\,dt = \sin u$ .
By the chain rule its derivative is $\cos(x^2) \cdot 2x$ .

Worth noting

$\cos(x^2)$ puts the upper limit into the integrand but forgets the derivative of the limit, $2x$ .

Question 17

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What is $\frac{d}{dx}\int_x^{x^2} t\,dt$ ?

x^{2} - x

Correct answerYour answer

2 x^{3} + x

Correct answerYour answer

2 x - 1

Correct answerYour answer

2 x^{3} - x

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AnswerD · $2 x^{3} - x$

$\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v)v' - f(u)u'$ .
$= x^2 \cdot 2x - x \cdot 1 = 2x^3 - x$ .

Worth noting

$x^2 - x$ puts the limits into $f(t) = t$ but forgets to multiply by their derivatives.

Substitution and by parts with limits

16 questions

Question 18

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Evaluate $\int_0^{\frac{\pi}{4}} \tan x\sec^2 x\,dx$ .

- \frac{1}{2}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

Correct answerYour answer

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AnswerC · $\frac{1}{2}$

Put $t = \tan x$ , $dt = \sec^2 x\,dx$ ; the limits become 0 and 1.
$\int_0^1 t\,dt = \frac{1}{2}$ .

Worth noting

1 forgets to halve $t^2$ when integrating $t$ .

Question 19

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Evaluate $\int_{0}^{1} x e^{x^{2}}\,dx$ .

\frac{1}{2} - \frac{e}{2}

Correct answerYour answer

- \frac{1}{4} + \frac{e}{4}

Correct answerYour answer

- \frac{1}{2} + \frac{e}{2}

Correct answerYour answer

\frac{e}{2}

Correct answerYour answer

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AnswerC · $- \frac{1}{2} + \frac{e}{2}$

An antiderivative is $F(x) = \frac{e^{x^{2}}}{2}$ .
By the fundamental theorem, the integral is $F(1) - F(0) = - \frac{1}{2} + \frac{e}{2}$ .

Worth noting

$\frac{e}{2}$ is $F(1)$ alone: $F(0)$ was not subtracted.

Question 20

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Evaluate $\int_{0}^{\frac{\pi}{2}} \sin^{2} x \cos x\,dx$ .

- \frac{2}{3}

Correct answerYour answer

- \frac{1}{3}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

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AnswerD · $\frac{1}{3}$

An antiderivative is $F(x) = \frac{\sin^{3} x}{3}$ .
By the fundamental theorem, the integral is $F(\frac{\pi}{2}) - F(0) = \frac{1}{3}$ .

Worth noting

$- \frac{1}{3}$ subtracts the wrong way, $F(0) - F(\frac{\pi}{2})$ .

Question 21

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Evaluate $\int_{0}^{1} \frac{2 x}{x^{2} + 1}\,dx$ .

- \log 2

Correct answerYour answer

\log 2

Correct answerYour answer

2 \log 2

Correct answerYour answer

\log 2 + 1

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AnswerB · $\log 2$

An antiderivative is $F(x) = \log{\left(x^{2} + 1 \right)}$ .
By the fundamental theorem, the integral is $F(1) - F(0) = \log 2$ .

Worth noting

$- \log 2$ subtracts the wrong way, $F(0) - F(1)$ .

Question 22

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Evaluate $\int_{0}^{4} \sqrt{2 x + 1}\,dx$ .

- \frac{26}{3}

Correct answerYour answer

\frac{13}{3}

Correct answerYour answer

\frac{26}{3}

Correct answerYour answer

9

Correct answerYour answer

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AnswerC · $\frac{26}{3}$

An antiderivative is $F(x) = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{3}$ .
By the fundamental theorem, the integral is $F(4) - F(0) = \frac{26}{3}$ .

Worth noting

$9$ is $F(4)$ alone: $F(0)$ was not subtracted.

Question 23

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With the substitution $t = x^2$ , what does $\int_0^2 xe^{x^2}\,dx$ become?

\frac{1}{2}\int_0^4 e^t\,dt

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\frac{1}{2}\int_0^2 e^t\,dt

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2\int_0^4 e^t\,dt

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\int_0^2 e^t\,dt

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AnswerA · $\frac{1}{2}\int_0^4 e^t\,dt$

$dt = 2x\,dx$ , so $x\,dx = \frac{dt}{2}$ .
The limits change too: $x = 0$ gives $t = 0$ and $x = 2$ gives $t = 4$ . So the integral is $\frac{1}{2}\int_0^4 e^t\,dt$ .

Worth noting

Keeping the old limits 0 and 2 for $t$ is the usual slip: they belong to $x$ .

Question 24

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Evaluate $\int_1^e \log x\,dx$ .

0

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1

Correct answerYour answer

-1 + e

Correct answerYour answer

e

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AnswerB · $1$

By parts, $\int \log x\,dx = x\log x - x$ .
$[x\log x - x]_1^e = (e - e) - (0 - 1) = 1$ .

Worth noting

$e - 1$ integrates $\log x$ as if it were $e^{x}$ or forgets the $-x$ .

Question 25

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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x}\,dx$ .

\frac{\log 2}{2}

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- \log 2 + \log 3

Correct answerYour answer

\log 2

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1

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AnswerC · $\log 2$

The top is the derivative of the bottom, so an antiderivative is $\log(1 + \sin x)$ .
$\log 2 - \log 1 = \log 2$ .

Worth noting

1 treats the integral as $\int_0^{\pi/2}\cos x\,dx$ , ignoring the denominator.

Question 26

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Evaluate $\int_0^1 x\left(x^2 + 1\right)^3 dx$ .

\frac{15}{8}

Correct answerYour answer

\frac{23}{8}

Correct answerYour answer

\frac{15}{4}

Correct answerYour answer

4

Correct answerYour answer

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AnswerA · $\frac{15}{8}$

Put $t = x^2 + 1$ , $dt = 2x\,dx$ ; $x = 0 \to t = 1$ , $x = 1 \to t = 2$ .
$\frac{1}{2}\int_1^2 t^3\,dt = \frac{1}{2} \cdot \frac{16 - 1}{4} = \frac{15}{8}$ .

Worth noting

$\frac{15}{4}$ forgets the $\frac{1}{2}$ from $x\,dx = \frac{dt}{2}$ .

Question 27

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Evaluate $\int_0^{\pi} \frac{\sin x}{2 - \cos x}\,dx$ .

- \log 3

Correct answerYour answer

-1 + \log 3

Correct answerYour answer

\log 2

Correct answerYour answer

\log 3

Correct answerYour answer

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AnswerD · $\log 3$

Put $t = 2 - \cos x$ , $dt = \sin x\,dx$ ; $x = 0 \to t = 1$ , $x = \pi \to t = 3$ .
$\int_1^3 \frac{dt}{t} = \log 3$ .

Worth noting

$-\log 3$ takes $dt = -\sin x\,dx$ ; the derivative of $-\cos x$ is $+\sin x$ .

Question 28

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Evaluate $\int_0^{\sqrt{3}} \frac{x}{\sqrt{1 + x^2}}\,dx$ .

\frac{1}{2}

Correct answerYour answer

-1 + \sqrt{3}

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerC · $1$

$\frac{d}{dx}\sqrt{1 + x^2} = \frac{x}{\sqrt{1 + x^2}}$ , so an antiderivative is $\sqrt{1 + x^2}$ .
$\sqrt{4} - \sqrt{1} = 1$ .

Worth noting

2 is the antiderivative at the upper limit alone; $\sqrt{1} = 1$ must be subtracted.

Question 29

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Evaluate $\int_1^2 \frac{\log x}{x}\,dx$ .

\frac{\left(\log 2\right)^{2}}{2}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\log 2

Correct answerYour answer

2 \log 2

Correct answerYour answer

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AnswerA · $\frac{\left(\log 2\right)^{2}}{2}$

Put $t = \log x$ , $dt = \frac{dx}{x}$ ; the limits become 0 and $\log 2$ .
$\int_0^{\log 2} t\,dt = \frac{(\log 2)^2}{2}$ .

Worth noting

$\log 2$ is the new upper limit, not the value of the integral.

Question 30

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Evaluate $\int_{0}^{1} x e^{x}\,dx$ .

-1

Correct answerYour answer

0

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

Correct answerYour answer

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AnswerD · $1$

An antiderivative is $F(x) = \left(x - 1\right) e^{x}$ .
By the fundamental theorem, the integral is $F(1) - F(0) = 1$ .

Worth noting

$0$ is $F(1)$ alone: $F(0)$ was not subtracted.

Question 31

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Evaluate $\int_{0}^{\frac{\pi}{2}} x \cos x\,dx$ .

1 - \frac{\pi}{2}

Correct answerYour answer

-1 + \frac{\pi}{2}

Correct answerYour answer

-2 + \pi

Correct answerYour answer

\frac{\pi}{2}

Correct answerYour answer

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AnswerB · $-1 + \frac{\pi}{2}$

An antiderivative is $F(x) = x \sin x + \cos x$ .
By the fundamental theorem, the integral is $F(\frac{\pi}{2}) - F(0) = -1 + \frac{\pi}{2}$ .

Worth noting

$\frac{\pi}{2}$ is $F(\frac{\pi}{2})$ alone: $F(0)$ was not subtracted.

Question 32

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Evaluate $\int_0^1 xe^{-x}\,dx$ .

1 - \frac{2}{e}

Correct answerYour answer

e^{-1}

Correct answerYour answer

1 - e^{-1}

Correct answerYour answer

2 - \frac{2}{e}

Correct answerYour answer

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AnswerA · $1 - \frac{2}{e}$

By parts with $u = x$ , $dv = e^{-x}\,dx$ : $\int xe^{-x}\,dx = -xe^{-x} - e^{-x}$ .
$[-xe^{-x} - e^{-x}]_0^1 = \left(-\frac{2}{e}\right) - (-1) = 1 - \frac{2}{e}$ .

Worth noting

$1 - \frac{1}{e}$ is $\int_0^1 e^{-x}\,dx$ : the term $-xe^{-x}$ was dropped.

Question 33

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Evaluate $\int_0^1 \tan^{-1} x\,dx$ .

- \log 2 + \frac{\pi}{4}

Correct answerYour answer

- \frac{\log 2}{2} + \frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

\frac{\log 2}{2} + \frac{\pi}{4}

Correct answerYour answer

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AnswerB · $- \frac{\log 2}{2} + \frac{\pi}{4}$

By parts with $u = \tan^{-1} x$ , $dv = dx$ : $\int \tan^{-1} x\,dx = x\tan^{-1} x - \frac{1}{2}\log(1 + x^2)$ .
From 0 to 1: $\frac{\pi}{4} - \frac{1}{2}\log 2$ .

Worth noting

$\frac{\pi}{4}$ stops at $uv$ and leaves out $-\int v\,du$ .

Properties of definite integrals

20 questions

Question 34

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If $\int_1^4 f(x)\,dx = 5$ , what is $\int_4^1 f(x)\,dx$ ?

-5

Correct answerYour answer

0

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

5

Correct answerYour answer

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AnswerA · $-5$

Swapping the limits changes the sign: $\int_b^a f = -\int_a^b f$ .
So the answer is $-5$ .

Question 35

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$\int_0^5 f(x)\,dx = 4$ and $\int_0^2 f(x)\,dx = -2$ . What is $\int_2^5 f(x)\,dx$ ?

-8

Correct answerYour answer

-6

Correct answerYour answer

2

Correct answerYour answer

6

Correct answerYour answer

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AnswerD · $6$

$\int_0^5 f = \int_0^2 f + \int_2^5 f$ .
So $\int_2^5 f = 4 - (-2) = 6$ .

Worth noting

2 adds the two given values instead of subtracting one from the other.

Question 36

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Evaluate $\int_{-1}^{1} \left(x^{5} + 3 x\right)\,dx$ .

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

\frac{10}{3}

Correct answerYour answer

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AnswerA · $0$

$f(-x) = -f(x)$ for $f(x) = x^{5} + 3 x$ : the function is odd.
The integral of an odd function over $[-a, a]$ is 0: the area below the axis cancels the area above it.

Worth noting

Doubling $\int_0^a$ is the rule for an even function.

Question 37

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If $\int_0^3 f(x)\,dx = 5$ and $\int_0^3 g(x)\,dx = 2$ , what is $\int_0^3 \left(3f(x) - 2g(x)\right)dx$ ?

3

Correct answerYour answer

10

Correct answerYour answer

11

Correct answerYour answer

19

Correct answerYour answer

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AnswerC · $11$

The integral is linear: $3\int_0^3 f\,dx - 2\int_0^3 g\,dx$ .
$3 \times 5 - 2 \times 2 = 11$ .

Worth noting

3 is $5 - 2$ , leaving out the multipliers 3 and 2.

Question 38

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Evaluate $\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{7} x\,dx$ .

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

\pi

Correct answerYour answer

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AnswerA · $0$

$\sin^7(-x) = -\sin^7 x$ : the integrand is odd.
An odd function over $[-a, a]$ integrates to 0.

Worth noting

2 doubles the integral from 0, which is the rule for an even function.

Question 39

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If $\int_0^2 f(x)\,dx = 3$ , what is $\int_0^2 \left(2f(x) + 1\right)dx$ ?

6

Correct answerYour answer

7

Correct answerYour answer

8

Correct answerYour answer

9

Correct answerYour answer

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AnswerC · $8$

Split: $2\int_0^2 f(x)\,dx + \int_0^2 1\,dx = 2 \times 3 + 2$ .
That is 8.

Worth noting

7 takes $\int_0^2 1\,dx$ as 1; it is the length of the interval, 2.

Question 40

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Evaluate $\int_{-1}^{1} \left(x^{4} + x^{2}\right)\,dx$ .

0

Correct answerYour answer

\frac{8}{15}

Correct answerYour answer

\frac{16}{15}

Correct answerYour answer

\frac{32}{15}

Correct answerYour answer

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AnswerC · $\frac{16}{15}$

$x^4 + x^2$ is even: $f(-x) = f(x)$ , so $\int_{-1}^{1} f = 2\int_0^1 f$ .
$\int_0^1 (x^4 + x^2)\,dx = \frac{1}{5} + \frac{1}{3} = \frac{8}{15}$ , so the integral is $\frac{16}{15}$ .

Worth noting

0 is the rule for odd functions; an even function's two halves add.

Question 41

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Evaluate $\int_0^4 \lvert x - 2 \rvert\,dx$ .

0

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

8

Correct answerYour answer

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AnswerB · $4$

$\lvert x - 2 \rvert = 2 - x$ for $x < 2$ and $x - 2$ for $x \ge 2$ , so split at $x = 2$ .
$\int_0^2 (2 - x)\,dx + \int_2^4 (x - 2)\,dx = 2 + 2 = 4$ .

Worth noting

0 integrates $x - 2$ without the absolute value, so the part below the axis cancels the part above.

Question 42

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Which of the following statements are true?

I. $\int_{-a}^{a} f(x)\,dx = 0$ whenever $f$ is odd.

II. $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ whenever $f$ is even.

III. $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ .

IV. $\int_a^b f(x)g(x)\,dx = \int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx$ .

All four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answer

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AnswerD · I, II and III only

I, II and III are properties of definite integrals: the odd and even rules, and the reflection $x \to a + b - x$ , which only reverses the direction of travel across $[a, b]$ .
IV is false: on $[0, 1]$ with $f = g = x$ , $\int_0^1 x^2\,dx = \frac{1}{3}$ , but $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ .

Question 43

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Evaluate $\int_{-2}^{2} \left(x^{3} + x^{2}\right)\,dx$ .

\frac{8}{3}

Correct answerYour answer

\frac{13}{3}

Correct answerYour answer

\frac{16}{3}

Correct answerYour answer

\frac{32}{3}

Correct answerYour answer

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AnswerC · $\frac{16}{3}$

$x^3$ is odd, so its part over $[-2, 2]$ is 0; $x^2$ is even.
$2\int_0^2 x^2\,dx = 2 \cdot \frac{8}{3} = \frac{16}{3}$ .

Worth noting

$\frac{8}{3}$ is $\int_0^2 x^2\,dx$ ; the even part must be doubled over $[-2, 2]$ .

Question 44

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Assertion (A): $\int_{-1}^{1} x\lvert x \rvert\,dx = 0$ .

Reason (R): $x\lvert x \rvert$ is an odd function.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R: $(-x)\lvert -x \rvert = -x\lvert x \rvert$ , so the function is odd. True.
An odd function over $[-1, 1]$ integrates to 0, so A is true because of R.

Question 45

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Evaluate $\int_{-1}^{1} e^{\lvert x \rvert}\,dx$ .

0

Correct answerYour answer

-1 + e

Correct answerYour answer

e - e^{-1}

Correct answerYour answer

-2 + 2 e

Correct answerYour answer

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AnswerD · $-2 + 2 e$

$e^{\lvert x \rvert}$ is even, so the integral is $2\int_0^1 e^{x}\,dx$ .
$2(e - 1)$ .

Worth noting

$e - \frac{1}{e}$ is $\int_{-1}^{1} e^{x}\,dx$ , which ignores the absolute value.

Question 46

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Evaluate $\int_0^3 \lvert x - 1 \rvert\,dx$ .

\frac{1}{2}

Correct answerYour answer

\frac{3}{2}

Correct answerYour answer

2

Correct answerYour answer

\frac{5}{2}

Correct answerYour answer

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AnswerD · $\frac{5}{2}$

Split at $x = 1$ : $\int_0^1 (1 - x)\,dx + \int_1^3 (x - 1)\,dx$ .
$\frac{1}{2} + 2 = \frac{5}{2}$ .

Worth noting

$\frac{3}{2}$ is $\int_0^3 (x - 1)\,dx$ , which lets the part below the axis cancel.

Question 47

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Which of the following statements are true?

I. $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ .

II. $\int_a^a f(x)\,dx = 0$ .

III. $\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ for every $f$ .

IV. $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ .

All four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerII and IV onlyCorrect answerYour answerI, II and IV onlyCorrect answerYour answer

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AnswerD · I, II and IV only

I: the name of the variable does not matter. II: an interval of length 0 has no area. IV: reversing the limits changes the sign.
III holds only when $f(2a - x) = f(x)$ ; for $f(x) = x$ on $[0, 2]$ , $\int_0^2 x\,dx = 2$ but $2\int_0^1 x\,dx = 1$ .

Question 48

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Evaluate $\int_{-1}^{1} x^{3} \cos^{2} x\,dx$ .

0

Correct answerYour answer

- \frac{\sin 2}{4} + \frac{3 \cos 2}{8} + \frac{5}{8}

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerA · $0$

$f(-x) = -f(x)$ for $f(x) = x^{3} \cos^{2} x$ : the function is odd, since $\cos^2(-x) = \cos^2 x$ and $(-x)^3 = -x^3$ .
The integral of an odd function over $[-a, a]$ is 0: the area below the axis cancels the area above it.

Worth noting

Doubling $\int_0^a$ is the rule for an even function.

Question 49

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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\,dx$ .

0

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

1

Correct answerYour answer

\frac{\pi}{2}

Correct answerYour answer

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AnswerB · $\frac{\pi}{4}$

Call the integral $I$ . By $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ , replacing $x$ by $\frac{\pi}{2} - x$ swaps $\sin$ and $\cos$ : $I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x}\,dx$ .
Adding the two forms, the integrand becomes 1, so $2I = \frac{\pi}{2}$ and $I = \frac{\pi}{4}$ .

Worth noting

$\frac{\pi}{2}$ is $2I$ ; the last step halves it.

Question 50

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Using $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ , evaluate $\int_0^1 x(1 - x)^5\,dx$ .

\frac{1}{42}

Correct answerYour answer

\frac{1}{30}

Correct answerYour answer

\frac{1}{7}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

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AnswerA · $\frac{1}{42}$

Replacing $x$ by $1 - x$ : $\int_0^1 (1 - x)x^5\,dx = \int_0^1 (x^5 - x^6)\,dx$ .
$= \frac{1}{6} - \frac{1}{7} = \frac{1}{42}$ .

Worth noting

$\frac{1}{6}$ integrates only the $x^5$ term and drops $-x^6$ .

Question 51

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Evaluate $\int_2^5 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{7 - x}}\,dx$ .

\frac{3}{2}

Correct answerYour answer

\frac{5}{2}

Correct answerYour answer

3

Correct answerYour answer

\frac{7}{2}

Correct answerYour answer

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AnswerA · $\frac{3}{2}$

By $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ with $a + b = 7$ , $I = \int_2^5 \frac{\sqrt{7 - x}}{\sqrt{7 - x} + \sqrt{x}}\,dx$ .
Adding, $2I = \int_2^5 1\,dx = 3$ , so $I = \frac{3}{2}$ .

Worth noting

3 is $2I$ , the length of the interval; the integral is half of it.

Question 52

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Evaluate $\int_0^{\frac{\pi}{2}} \log(\tan x)\,dx$ .

- \frac{\pi \log 2}{4}

Correct answerYour answer

0

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{2}

Correct answerYour answer

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AnswerB · $0$

With $x \to \frac{\pi}{2} - x$ , $I = \int_0^{\frac{\pi}{2}} \log(\cot x)\,dx$ .
Adding: $2I = \int_0^{\frac{\pi}{2}} \log(\tan x\cot x)\,dx = \int_0^{\frac{\pi}{2}} \log 1\,dx = 0$ , so $I = 0$ .

Worth noting

$\frac{\pi}{4}$ is the answer for $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$ , a different use of the same property.

Question 53

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Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$ .

\frac{\pi}{24}

Correct answerYour answer

\frac{\pi}{12}

Correct answerYour answer

\frac{\pi}{6}

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

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AnswerB · $\frac{\pi}{12}$

With $x \to \frac{\pi}{6} + \frac{\pi}{3} - x$ , $\tan x$ becomes $\cot x$ , and $\frac{1}{1 + \sqrt{\cot x}} = \frac{\sqrt{\tan x}}{\sqrt{\tan x} + 1}$ .
Adding the two forms: $2I = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{6}$ , so $I = \frac{\pi}{12}$ .

Worth noting

$\frac{\pi}{6}$ is $2I$ , the length of the interval; the integral is half of it.

Area under a simple curve

18 questions

Question 54

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The figure shows the region bounded by $y = x^{2}$ , the $x$ -axis and the lines $x = 0$ and $x = 3$ . What is its area?

$The graph of y = x^{2} with the region under it between x = 0 and x = 3 shaded, down to the x-axis.$

\frac{9}{2}

Correct answerYour answer

8

Correct answerYour answer

9

Correct answerYour answer

18

Correct answerYour answer

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AnswerC · $9$

The curve stays above the $x$ -axis on $[0, 3]$ , so the area is $\int_{0}^{3} x^{2}\,dx$ .
That is $9$ square units.

Question 55

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The figure shows the region bounded by $y = \sqrt{x}$ , the $x$ -axis and the lines $x = 0$ and $x = 4$ . What is its area?

$The graph of y = \sqrt{x} with the region under it between x = 0 and x = 4 shaded, down to the x-axis.$

\frac{8}{3}

Correct answerYour answer

\frac{16}{3}

Correct answerYour answer

\frac{19}{3}

Correct answerYour answer

\frac{32}{3}

Correct answerYour answer

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AnswerB · $\frac{16}{3}$

The curve stays above the $x$ -axis on $[0, 4]$ , so the area is $\int_{0}^{4} \sqrt{x}\,dx$ .
That is $\frac{16}{3}$ square units.

Question 56

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The figure shows the region bounded by $y = \sin x$ , the $x$ -axis and the lines $x = 0$ and $x = \pi$ . What is its area?

$The graph of y = \sin x with the region under it between x = 0 and x = \pi shaded, down to the x-axis.$

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

4

Correct answerYour answer

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AnswerC · $2$

The curve stays above the $x$ -axis on $[0, \pi]$ , so the area is $\int_{0}^{\pi} \sin x\,dx$ .
That is $2$ square units.

Question 57

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The figure shows the region bounded by $y = x^{3}$ , the $x$ -axis and the lines $x = 0$ and $x = 2$ . What is its area?

$The graph of y = x^{3} with the region under it between x = 0 and x = 2 shaded, down to the x-axis.$

2

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

8

Correct answerYour answer

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AnswerB · $4$

The curve stays above the $x$ -axis on $[0, 2]$ , so the area is $\int_{0}^{2} x^{3}\,dx$ .
That is $4$ square units.

Question 58

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The figure shows the region bounded by $y = \cos x$ , the $x$ -axis and the lines $x = 0$ and $x = \frac{\pi}{2}$ . What is its area?

$The graph of y = \cos x with the region under it between x = 0 and x = \frac{\pi}{2} shaded, down to the x-axis.$

0

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerC · $1$

The curve stays above the $x$ -axis on $[0, \frac{\pi}{2}]$ , so the area is $\int_{0}^{\frac{\pi}{2}} \cos x\,dx$ .
That is $1$ square units.

Question 59

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The figure shows the region bounded by $y = e^{x}$ , the $x$ -axis and the lines $x = 0$ and $x = 1$ . What is its area?

$The graph of y = e^{x} with the region under it between x = 0 and x = 1 shaded, down to the x-axis.$

- \frac{1}{2} + \frac{e}{2}

Correct answerYour answer

-1 + e

Correct answerYour answer

e

Correct answerYour answer

-2 + 2 e

Correct answerYour answer

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AnswerB · $-1 + e$

The curve stays above the $x$ -axis on $[0, 1]$ , so the area is $\int_{0}^{1} e^{x}\,dx$ .
That is $-1 + e$ square units.

Question 60

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The figure shows the region bounded by $y = 4 - x^{2}$ , the $x$ -axis and the lines $x = -2$ and $x = 2$ . What is its area?

$The graph of y = 4 - x^{2} with the region under it between x = -2 and x = 2 shaded, down to the x-axis.$

0

Correct answerYour answer

\frac{16}{3}

Correct answerYour answer

\frac{32}{3}

Correct answerYour answer

\frac{64}{3}

Correct answerYour answer

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AnswerC · $\frac{32}{3}$

The curve stays above the $x$ -axis on $[-2, 2]$ , so the area is $\int_{-2}^{2} 4 - x^{2}\,dx$ .
That is $\frac{32}{3}$ square units.

Question 61

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What is the area bounded by $y = x^{2} - 4$ , the $x$ -axis and the lines $x = 0$ and $x = 2$ ?

- \frac{16}{3}

Correct answerYour answer

\frac{16}{3}

Correct answerYour answer

\frac{19}{3}

Correct answerYour answer

\frac{32}{3}

Correct answerYour answer

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AnswerB · $\frac{16}{3}$

On $[0, 2]$ the curve lies below the $x$ -axis, so the integral comes out negative: $- \frac{16}{3}$ .
An area is never negative: it is $\left|- \frac{16}{3}\right| = \frac{16}{3}$ square units.

Worth noting

$- \frac{16}{3}$ is the value of the integral, not the area.

Question 62

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What is the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ ?

6 \pi

Correct answerYour answer

1 + 6 \pi

Correct answerYour answer

13 \pi

Correct answerYour answer

36 \pi

Correct answerYour answer

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AnswerA · $6 \pi$

By symmetry the area is 4 times the part in the first quadrant: $4\int_0^3 \frac{2}{3}\sqrt{9 - x^2}\,dx$ .
$\int_0^3 \sqrt{9 - x^2}\,dx$ is a quarter circle of radius 3, $\frac{9\pi}{4}$ , so the area is $4 \cdot \frac{2}{3} \cdot \frac{9\pi}{4} = 6\pi$ , that is $\pi ab$ with $a = 3$ , $b = 2$ .

Worth noting

$36\pi$ uses $a^2$ and $b^2$ , the denominators, instead of $a = 3$ and $b = 2$ .

Question 63

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Assertion (A): $\int_{-1}^{1} x\,dx = 0$ , so the line $y = x$ encloses no area with the $x$ -axis between $x = -1$ and $x = 1$ .

Reason (R): A definite integral counts area below the $x$ -axis as negative.

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AnswerD · A is false, but R is true.

The two triangles each have area $\frac{1}{2}$ , so the region's area is 1, not 0: A is false.
R is true, and it is exactly why the integral is 0 while the area is not.

Question 64

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What is the area bounded by $y = \cos x$ , the $x$ -axis and the lines $x = \frac{\pi}{2}$ and $x = \pi$ ?

-1

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0

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1

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2

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AnswerC · $1$

On $[\frac{\pi}{2}, \pi]$ the curve lies below the $x$ -axis, so the integral comes out negative: $-1$ .
An area is never negative: it is $\left|-1\right| = 1$ square units.

Worth noting

$-1$ is the value of the integral, not the area.

Question 65

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The figure shows the region bounded by $y = \frac{1}{x}$ , the $x$ -axis and the lines $x = 1$ and $x = e^{2}$ . What is its area?

$The graph of y = \frac{1}{x} with the region under it between x = 1 and x = e^{2} shaded, down to the x-axis.$

1 - e^{-2}

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1

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2

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4

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AnswerC · $2$

The curve stays above the $x$ -axis on $[1, e^{2}]$ , so the area is $\int_{1}^{e^{2}} \frac{1}{x}\,dx$ .
That is $2$ square units.

Question 66

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What is the total area between the line $y = x$ and the $x$ -axis from $x = -1$ to $x = 2$ ?

\frac{3}{2}

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\frac{5}{2}

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3

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\frac{7}{2}

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AnswerB · $\frac{5}{2}$

The line is below the axis on $[-1, 0]$ and above it on $[0, 2]$ : find each area separately.
$\left|\int_{-1}^{0} x\,dx\right| + \int_0^2 x\,dx = \frac{1}{2} + 2 = \frac{5}{2}$ .

Worth noting

$\frac{3}{2}$ is $\int_{-1}^{2} x\,dx$ , where the part below the axis cancels some of the rest.

Question 67

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Using integration, what is the area enclosed by the circle $x^2 + y^2 = 16$ ?

4 \pi

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8 \pi

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-1 + 16 \pi

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16 \pi

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AnswerD · $16 \pi$

By symmetry the area is $4\int_0^4 \sqrt{16 - x^2}\,dx$ , four quarter-circles.
$\int_0^4 \sqrt{16 - x^2}\,dx = \left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}\right]_0^4 = 4\pi$ , so the area is $16\pi$ .

Worth noting

$4\pi$ is one quarter: the integral from 0 to 4 covers only the first quadrant.

Question 68

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What is the area of the region bounded by the parabola $y^2 = 4x$ and the line $x = 1$ ?

\frac{4}{3}

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2

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\frac{8}{3}

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4

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AnswerC · $\frac{8}{3}$

The region is symmetric about the $x$ -axis; the upper half lies under $y = 2\sqrt{x}$ from 0 to 1.
$2\int_0^1 2\sqrt{x}\,dx = 2 \cdot \frac{4}{3} = \frac{8}{3}$ .

Worth noting

$\frac{4}{3}$ is only the half above the $x$ -axis.

Question 69

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What is the area bounded by $y = x^2 - 1$ and the $x$ -axis?

- \frac{4}{3}

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\frac{1}{3}

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\frac{2}{3}

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\frac{4}{3}

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AnswerD · $\frac{4}{3}$

The curve meets the $x$ -axis at $x = \pm 1$ and lies below it in between.
$\int_{-1}^{1} (x^2 - 1)\,dx = - \frac{4}{3}$ , so the area is $\frac{4}{3}$ square units.

Worth noting

$- \frac{4}{3}$ is the integral; an area is its absolute value.

Question 70

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What is the total area between $y = \sin x$ and the $x$ -axis from $x = 0$ to $x = 2\pi$ ?

0

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2

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3

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4

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AnswerD · $4$

The curve crosses the axis at $x = \pi$ : above it on $(0, \pi)$ , below it on $(\pi, 2\pi)$ . Take the two parts separately.
$\int_0^{\pi} \sin x\,dx = 2$ and $\left|\int_{\pi}^{2\pi} \sin x\,dx\right| = 2$ , so the total area is 4 square units.

Worth noting

0 is the single integral from 0 to $2\pi$ , in which the two parts cancel.

Question 71

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What is the area of the region bounded by the curve $x = y^2$ , the $y$ -axis and the lines $y = 1$ and $y = 2$ ?

\frac{7}{3}

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3

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4 - \frac{2 \sqrt{2}}{3}

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4

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AnswerA · $\frac{7}{3}$

The region is between the curve and the $y$ -axis, so integrate along $y$ : Area $= \int_1^2 x\,dy = \int_1^2 y^2\,dy$ .
$= \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ square units.

Worth noting

3 multiplies the width 1 by the height $2^2 - 1^2$ ; the curve is not a straight line, so the integral is needed.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.