Sample 2022-23 Class 12 Maths Q32 Solution: Area of {(x, y): 0 ≤ y ≤ x², 0 ≤ y ≤ x, 0 ≤ x ≤ 2}

The question

Make a rough sketch of the region $\{(x, y): 0 \le y \le x^2, 0 \le y \le x, 0 \le x \le 2\}$ and find the area of the region using integration.

Exam answer

The region lies under both the parabola $y = x^2$ and the line $y = x$ ⁠, above the x-axis, from $x = 0$ to $x = 2$ (shaded).

Why?

A point is in the region when it satisfies all three conditions. $0 \le y \le x^2$ and $0 \le y \le x$ together mean the point is above the x-axis and below both graphs, so below whichever graph is lower. The condition $0 \le x \le 2$ gives the left and right ends.

The region (shaded): under

y = x^2

from

x = 0

to 1 and under

y = x

from

x = 1

to 2.

The points of intersection of the parabola $y = x^2$ and the line $y = x$ ⁠: $x^2 = x \Rightarrow x = 0, 1$ ⁠, i.e. $(0, 0)$ and $(1, 1)$ ⁠.

Why?

At a point of intersection both equations hold, so equate the two values of $y$ ⁠:

\begin{aligned} x^2 = x &\Rightarrow x^2 - x = 0 \\ &\Rightarrow x(x - 1) = 0 \end{aligned}

so $x = 0$ or $x = 1$ ⁠, and $y = x$ gives $y = 0$ or $y = 1$ ⁠.

For $0 \le x \le 1$ ⁠, $x^2 \le x$ ⁠, and for $1 \le x \le 2$ ⁠, $x \le x^2$ ⁠. So

Why?

Test one point in each part to see which graph is lower:

\begin{aligned} x = \frac12&:\ x^2 = \frac14 < \frac12 = x \\ x = \frac32&:\ x = \frac32 < \frac94 = x^2 \end{aligned}

So the top of the region is the parabola from 0 to 1, and the line from 1 to 2. The region must be split at $x = 1$ ⁠.

\begin{aligned} &\text{Required area} \\ &= \int_0^1 y_{\text{parabola}}\,dx + \int_1^2 y_{\text{line}}\,dx \\ &= \int_0^1 x^2\,dx + \int_1^2 x\,dx \end{aligned}

Why?

The area between a curve $y = f(x) \ge 0$ ⁠, the x-axis and the lines $x = a$ ⁠, $x = b$ is

\int_a^b y\,dx

Each piece uses the lower graph found above as its top boundary.

\begin{aligned} &= \left[\frac{x^3}{3}\right]_0^1 + \left[\frac{x^2}{2}\right]_1^2 \\ &= \frac13 + \left(2 - \frac12\right) = \frac13 + \frac32 \end{aligned}

Why?

By the power rule ( $n \ne -1$ ⁠),

\int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C

Then take upper limit minus lower limit:

\begin{aligned} \left[\frac{x^3}{3}\right]_0^1 &= \frac{1^3}{3} - \frac{0^3}{3} \\ &= \frac13 - 0 = \frac13 \\[4pt] \left[\frac{x^2}{2}\right]_1^2 &= \frac{2^2}{2} - \frac{1^2}{2} \\ &= \frac42 - \frac12 \\ &= 2 - \frac12 = \frac32 \end{aligned}

Finally $\frac13 + \frac32 = \frac{2}{6} + \frac{9}{6} = \frac{11}{6}$ ⁠.

\begin{aligned} \frac13 &= \frac{1 \times 2}{3 \times 2} = \frac26 \\ \frac32 &= \frac{3 \times 3}{2 \times 3} = \frac96 \\ \frac26 + \frac96 &= \frac{2 + 9}{6} = \frac{11}{6} \end{aligned}

An area is given in square units.

$\therefore$ Required area $= \dfrac{11}{6}$ sq. units

Where this comes from

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

Question type: A curve and a line

How to spot it: click to see

Type 7. A curve and a line: top minus bottom

Spot it: the region is shut in by a curve and a line, like a parabola and $y = x$ ⁠, sometimes written as a set with $\le$ signs. Your book has no worked question of this kind, but CBSE sample papers ask it.

Here: Read the set: under both $y = x^2$ and $y = x$ ⁠. They meet at $(0, 0)$ and $(1, 1)$ ⁠, so the lower of the two is the edge: the parabola from 0 to 1, the line from 1 to 2.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q32: Area of {(x,y):0≤y≤x2,0≤y≤x,0≤x≤2}\{(x, y): 0 \le y \le x^2, 0 \le y \le x, 0 \le x \le 2\}{(x,y):0≤y≤x2,0≤y≤x,0≤x≤2}

Q32: Area of $\{(x, y): 0 \le y \le x^2, 0 \le y \le x, 0 \le x \le 2\}$