CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q24: Area bounded by $y^2 = 8x$ and the line $x = 2$

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

The question

Find the area of the region bounded by the parabola $y^2 = 8x$ and the line $x = 2$ ⁠.

Exam answer

Putting $x = 2$ in $y^2 = 8x$ ⁠,

y^2 = 16 \Rightarrow y = \pm 4

Why?

The corners of the region are where the line meets the parabola, so substitute $x = 2$ ⁠: $y^2 = 8 \times 2 = 16$ ⁠. Both signs count, because the parabola has an upper half and a lower half.

\begin{aligned} y^2 &= 8(2) = 16 \\ y &= \pm\sqrt{16} = \pm 4 \end{aligned}

The line $x = 2$ meets $y^2 = 8x$ at $(2, 4)$ and $(2, -4)$ ⁠. The region is symmetric about the x-axis.

The region (shaded) between the parabola

y^2 = 8x

and the line

x = 2

⁠.

\begin{aligned} \text{Area} &= 2\int_0^2\sqrt{8x}\,dx \end{aligned}

Why?

The area between a curve and the x-axis, from $x = a$ to $x = b$ ⁠, is

\text{Area} = \int_a^b y\,dx

The upper half of the parabola is $y = \sqrt{8x}$ (taking $y \ge 0$ ⁠), from $x = 0$ at O to $x = 2$ at the line. That gives the part above the x-axis. The part below is its mirror image, so we double it.

\begin{aligned} \text{Area} &= 2 \times 2\sqrt2\int_0^2 x^{\frac12}\,dx \end{aligned}

Why?

Split the square root: $\sqrt{8x} = \sqrt8\,\sqrt x$ ⁠, and

\sqrt8 = \sqrt{4 \times 2} = 2\sqrt2

Writing $\sqrt x = x^{\frac12}$ gets it ready for the power rule.

\begin{aligned} \text{Area} &= 4\sqrt2\left[\frac23x^{\frac32}\right]_0^2 \\ &= \frac{8\sqrt2}{3}\left[2^{\frac32} - 0\right] \\ &= \frac{8\sqrt2}{3} \times 2\sqrt2 \end{aligned}

Why?

First, $2 \times 2\sqrt2 = 4\sqrt2$ in front.

Power rule with $n = \frac12$ ⁠: the new power is $\frac32$ ⁠, and dividing by $\frac32$ is multiplying by $\frac23$ ⁠.

\begin{aligned} \int x^{\frac12}\,dx &= \frac{x^{\frac12 + 1}}{\frac12 + 1} = \frac{x^{\frac32}}{\frac32} \\ &= \frac23x^{\frac32} \\[4pt] 4\sqrt2 \times \frac23 &= \frac{8\sqrt2}{3} \end{aligned}

At the upper limit, $2^{\frac32} = 2 \times 2^{\frac12} = 2\sqrt2$ ⁠. At the lower limit, $0^{\frac32} = 0$ ⁠.

Finally $\sqrt2 \times \sqrt2 = 2$ ⁠, so the product is $\frac{8 \times 2 \times 2}{3} = \frac{32}{3}$ ⁠.

\begin{aligned} \frac{8\sqrt2}{3} \times 2\sqrt2 &= \frac{8 \times 2 \times \sqrt2 \times \sqrt2}{3} \\ &= \frac{8 \times 2 \times 2}{3} \\ &= \frac{32}{3} \end{aligned}

$\therefore$ Required area $= \dfrac{32}{3}$ sq. units

Where this comes from

Chapter 8, Application of Integrals: Area bounded by a parabola and a line.

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: The line $x = 2$ meets $y^2 = 8x$ at $(2, \pm4)$ ⁠, and the region is symmetric about the $x$ -axis. Take the top half $y = \sqrt{8x}$ from 0 to 2, and double it.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Application of Integrals in past papers: question 16 of 17 · see them all

Q24: Area bounded by y2=8xy^2 = 8xy2=8x and the line x=2x = 2x=2

Q24: Area bounded by $y^2 = 8x$ and the line $x = 2$