Read the two statements and choose the correct option. I. For c > 0, the region bounded by y^2 = 4x and the line x = c has 2/3 of the area of the rectangle 0 ≤ x ≤ c, -2√(c) ≤ y ≤ 2√(c). II. The region bounded by y = x^2 and the line y = 1 has twice the area of the region bounded by y = x^2, the x-axis and the lines x = -1 and x = 1.

Question 74

MediumOriginal question

Read the two statements and choose the correct option.
I. For $c > 0$ , the region bounded by $y^{2} = 4x$ and the line $x = c$ has $\frac{2}{3}$ of the area of the rectangle $0 \le x \le c$ , $-2\sqrt{c} \le y \le 2\sqrt{c}$ .
II. The region bounded by $y = x^{2}$ and the line $y = 1$ has twice the area of the region bounded by $y = x^{2}$ , the $x$ -axis and the lines $x = -1$ and $x = 1$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

I: $2\int_0^{c} 2\sqrt{x}\,dx = \frac{8}{3}c^{3/2}$ , and the rectangle is $c \times 4\sqrt{c} = 4c^{3/2}$ ; the ratio is $\frac{2}{3}$ . True.
II: $\int_{-1}^{1} (1 - x^{2})\,dx = \frac{4}{3}$ and $\int_{-1}^{1} x^{2}\,dx = \frac{2}{3}$ . True.

Worth noting

Doubting II mixes up the two regions: the one inside the parabola is what is left of the $2 \times 1$ rectangle, $2 - \frac{2}{3}$ .

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