CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q8: Area bounded by $y = x^2$ ⁠, the x-axis, $x = -1$ and $x = 1$

Sample 2020-21Ch 8 · Application of Integrals Ch 7 · Integrals

The question

Find the area bounded by $y = x^2$ ⁠, the x-axis and the lines $x = -1$ and $x = 1$ ⁠.

Exam answer

The curve $y = x^2$ lies above the x-axis and is symmetric about the y-axis.

Why?

$x^2 \ge 0$ for every $x$ ⁠, so the region lies above the x-axis and its area is simply $\int y\,dx$ ⁠, with no sign change.

$(-x)^2 = x^2$ ⁠, so the part from $-1$ to $0$ is the mirror image of the part from $0$ to $1$ ⁠.

\begin{aligned} A &= \int_{-1}^{1} x^2\,dx = 2\int_0^1 x^2\,dx \\ &= \frac{2}{3}\Big[x^3\Big]_0^1 = \frac{2}{3}(1 - 0) \end{aligned}

Why?

The symmetry gives the even-function property:

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx

Then $\int x^2\,dx = \frac{x^3}{3}$ ⁠, so the 2 and the $\frac{1}{3}$ combine to $\frac{2}{3}$ ⁠. At the limits, $1^3 - 0^3 = 1$ ⁠.

\begin{aligned} 2\int_0^1 x^2\,dx &= 2\Big[\frac{x^3}{3}\Big]_0^1 \\ &= \frac{2}{3}\Big[x^3\Big]_0^1 \\ &= \frac{2}{3}\big(1^3 - 0^3\big) \\ &= \frac{2}{3}(1 - 0) \\ &= \frac{2}{3}(1) = \frac{2}{3} \end{aligned}

$\therefore$ Required area $= \dfrac{2}{3}$ sq. units.

Where this comes from

Chapter 8, Application of Integrals: Area under a parabola.

Question type: Area under a curve

How to spot it: click to see

Type 1. Area under a curve: vertical strips

Spot it: a curve $y = f(x)$ above the $x$ -axis, and two lines such as $x = 1$ and $x = 2$ ⁠.

Here: $y = x^2$ stays above the axis between the lines $x = -1$ and $x = 1$ ⁠, so the area is $\int_{-1}^1 x^2\,dx$ ⁠; by symmetry, $2\int_0^1 x^2\,dx = \frac23$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q8: Area bounded by y=x2y = x^2y=x2⁠, the x-axis, x=−1x = -1x=−1 and x=1x = 1x=1

Q8: Area bounded by $y = x^2$ ⁠, the x-axis, $x = -1$ and $x = 1$