CBSE Board Examination 2022 (Term 2) · Section C · 4 marks

Q11: Area of $\{4x^2 + 9y^2 \le 36,\ 2x + 3y \ge 6\}$ ⁠, or under $x - y + 1 = 0$ from $x = -2$ to $x = 3$

Board 2022 T2Ch 8 · Application of Integrals Ch 7 · Integrals

The question

(a) Using integration, find the area of the region $\{(x, y) : 4x^2 + 9y^2 \le 36,\ 2x + 3y \ge 6\}$ ⁠.

(b) Using integration, find the area of the region bounded by lines $x - y + 1 = 0$ ⁠, $x = -2$ ⁠, $x = 3$ and $x$ -axis.

Exam answer

(a)

$4x^2 + 9y^2 = 36$ is the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ ⁠, and $2x + 3y = 6$ is the line $\frac{x}{3} + \frac{y}{2} = 1$ ⁠. The region is the shaded part of the ellipse on and above the line.

Why?

Dividing $4x^2 + 9y^2 = 36$ by 36 gives the standard form, with semi-axes $a = 3$ along $x$ and $b = 2$ along $y$ ⁠. Dividing $2x + 3y = 6$ by 6 shows the line cuts the axes at $x = 3$ and $y = 2$ ⁠.

$4x^2 + 9y^2 \le 36$ means inside the ellipse. For $2x + 3y \ge 6$ ⁠, test the origin: $0 \ge 6$ is false, so the region is on the side of the line away from O.

The ellipse

4x^2 + 9y^2 = 36

and the line

2x + 3y = 6

⁠; the region is shaded.

The points of intersection are $(3, 0)$ and $(0, 2)$ ⁠.

Why?

From the line, $3y = 6 - 2x$ ⁠, so $9y^2 = (6 - 2x)^2$ ⁠. Put this into the ellipse:

\begin{aligned} 4x^2 + (6 - 2x)^2 &= 36 \\ 8x^2 - 24x &= 0 \\ 8x(x - 3) &= 0 \end{aligned}

In more detail, expand the square first:

\begin{aligned} (6 - 2x)^2 &= 6^2 - 2(6)(2x) + (2x)^2 \\ &= 36 - 24x + 4x^2 \end{aligned}

\begin{aligned} 4x^2 + 36 - 24x + 4x^2 &= 36 \\ 8x^2 - 24x + 36 &= 36 \\ 8x^2 - 24x &= 0 \\ 8x(x - 3) &= 0 \end{aligned}

So $x = 0$ (then $y = 2$ ⁠) or $x = 3$ (then $y = 0$ ⁠).

\begin{aligned} x = 0&:\ 3y = 6 - 0 = 6,\ y = 2 \\ x = 3&:\ 3y = 6 - 6 = 0,\ y = 0 \end{aligned}

On the ellipse (upper half), $y = \frac23\sqrt{9 - x^2}$ ⁠; on the line, $y = \frac23(3 - x)$ ⁠.

\begin{aligned} \text{Area} &= \frac23\int_0^3 \sqrt{9 - x^2}\,dx \\ &\quad - \frac23\int_0^3 (3 - x)\,dx \end{aligned}

Why?

Solve each curve for $y$ ⁠. Ellipse: $9y^2 = 36 - 4x^2 = 4(9 - x^2)$ ⁠, so $y = \frac23\sqrt{9 - x^2}$ ⁠, taking the $+$ root for the upper half. Line: $3y = 6 - 2x = 2(3 - x)$ ⁠.

\begin{aligned} 9y^2 &= 36 - 4x^2 \\ &= 4(9 - x^2) \\ y^2 &= \frac{4(9 - x^2)}{9} \\ y &= \sqrt{\frac{4}{9}}\sqrt{9 - x^2} \\ &= \frac23\sqrt{9 - x^2} \end{aligned}

\begin{aligned} 3y &= 6 - 2x = 2(3 - x) \\ y &= \frac23(3 - x) \end{aligned}

Between $x = 0$ and $x = 3$ the ellipse is above the line, so

\text{Area} = \int_0^3 (y_{\text{top}} - y_{\text{bottom}})\,dx

\begin{aligned} &= \frac23\bigg[\frac{x}{2}\sqrt{9 - x^2} + \frac92\sin^{-1}\frac{x}{3} \\ &\qquad + \frac{(3 - x)^2}{2}\bigg]_0^3 \\ &= \frac23\left[\left(0 + \frac92\cdot\frac{\pi}{2} + 0\right) - \frac92\right] \\ &= \frac23\left[\frac{9\pi}{4} - \frac92\right] \end{aligned}

Why?

Standard integral, with $a = 3$ ⁠:

\begin{aligned} &\int \sqrt{a^2 - x^2}\,dx \\ &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \end{aligned}

For the line part,

-\int (3 - x)\,dx = \frac{(3 - x)^2}{2}

since differentiating the right side gives $-(3 - x)$ ⁠.

Here $\frac{a^2}{2} = \frac{3^2}{2} = \frac92$ ⁠.

At $x = 3$ ⁠: $\sin^{-1}1 = \frac{\pi}{2}$ and the other terms are 0. At $x = 0$ ⁠: only the last term is left, and it is $\frac92$ ⁠.

\begin{aligned} x = 3&:\ \frac32\sqrt{9 - 9} + \frac92\sin^{-1}\frac33 \\ &\qquad + \frac{(3 - 3)^2}{2} \\ &= 0 + \frac92 \cdot \frac{\pi}{2} + 0 = \frac{9\pi}{4} \\[4pt] x = 0&:\ 0 + \frac92\sin^{-1}0 + \frac{(3 - 0)^2}{2} \\ &= 0 + 0 + \frac92 = \frac92 \end{aligned}

$\therefore$ Required area $= \left(\frac{3\pi}{2} - 3\right)$ sq. units.

Why?

Multiply out: $\frac23 \times \frac{9\pi}{4} = \frac{3\pi}{2}$ and $\frac23 \times \frac92 = 3$ ⁠.

\begin{aligned} \frac23 \times \frac{9\pi}{4} &= \frac{18\pi}{12} \\ &= \frac{18\pi \div 6}{12 \div 6} = \frac{3\pi}{2} \\[4pt] \frac23 \times \frac92 &= \frac{18}{6} = 3 \end{aligned}

Check: this is a quarter of the ellipse, $\frac14\pi ab = \frac{3\pi}{2}$ ⁠, minus the triangle with legs 3 and 2, of area 3.

(b)

The line $x - y + 1 = 0$ ⁠, i.e. $y = x + 1$ ⁠, meets the $x$ -axis at $x = -1$ ⁠. It is below the $x$ -axis for $-2 \le x < -1$ and above it for $-1 < x \le 3$ ⁠.

Why?

Put $y = 0$ in $y = x + 1$ ⁠: $x = -1$ ⁠. For $x < -1$ ⁠, $x + 1$ is negative, so the line is below the axis; for $x > -1$ it is above.

So the region is two triangles, one on each side of $x = -1$ ⁠.

The line

y = x + 1

⁠, the lines

x = -2

and

x = 3

and the

x

-axis; the region is shaded.

\begin{aligned} \text{Area} &= -\int_{-2}^{-1}(x + 1)\,dx \\ &\quad + \int_{-1}^{3}(x + 1)\,dx \end{aligned}

Why?

Below the axis, $x + 1 < 0$ ⁠, so that integral is negative. An area is positive, so we take it with a minus sign (its modulus). That is why the integral is split at $x = -1$ ⁠.

Watch out: $\int_{-2}^{3}(x + 1)\,dx$ in one go gives $\frac{15}{2}$ ⁠, which is wrong, because the part below the axis cancels part of the area above.

\begin{aligned} &= -\left[\frac{(x + 1)^2}{2}\right]_{-2}^{-1} \\ &\quad + \left[\frac{(x + 1)^2}{2}\right]_{-1}^{3} \end{aligned}

Why?

\int (x + 1)\,dx = \frac{(x + 1)^2}{2} + C

Check: differentiating the right side gives $x + 1$ ⁠. For a definite integral the constant cancels, so it is left out in the brackets.

\begin{aligned} &= -\left(0 - \frac12\right) + (8 - 0) \\ &= \frac12 + 8 \end{aligned}

Why?

Values of the bracket at the limits:

\begin{aligned} x = -1&:\ \frac{(-1 + 1)^2}{2} = \frac{0^2}{2} \\ &\quad = 0 \\ x = -2&:\ \frac{(-2 + 1)^2}{2} = \frac{(-1)^2}{2} \\ &\quad = \frac12 \\ x = 3&:\ \frac{(3 + 1)^2}{2} = \frac{4^2}{2} \\ &\quad = \frac{16}{2} = 8 \end{aligned}

Upper minus lower in each bracket:

\begin{aligned} -\left(0 - \frac12\right) &= -\left(-\frac12\right) = \frac12 \\ 8 - 0 &= 8 \\ \frac12 + 8 &= \frac12 + \frac{16}{2} = \frac{17}{2} \end{aligned}

Check with triangles: $\frac12 \times 1 \times 1 = \frac12$ below the axis, and $\frac12 \times 4 \times 4 = 8$ above it.

$\therefore$ Required area $= \dfrac{17}{2}$ sq. units.

Where this comes from

Chapter 8, Application of Integrals: Area: ellipse and a line; under a line.

Question type: (a) A curve and a line + Ellipses (combined); (b) A curve that dips below the axis

How to spot it: click to see

(a) 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: The set is inside the ellipse and above the line $2x + 3y = 6$ ⁠; they meet at $(3, 0)$ and $(0, 2)$ ⁠. From 0 to 3 the ellipse is on top, so subtract the line $\frac23(3 - x)$ ⁠.

(a) Type 4. Ellipses: read off $a$ and $b$ ⁠, then times four

Spot it: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ⁠, with numbers under $x^2$ and $y^2$ ⁠.

Here: For the top edge, read $a = 3$ ⁠, $b = 2$ from $\frac{x^2}{9} + \frac{y^2}{4} = 1$ ⁠: $y = \frac23\sqrt{9 - x^2}$ ⁠, integrated from 0 to 3 with the square-root card.

(b) Type 5. A curve that dips below the axis: split where it crosses

Spot it: part of the curve is below the $x$ -axis between the limits, like $\sin x$ from 0 to $2\pi$ ⁠, or a line that crosses the axis.

Here: The line $y = x + 1$ crosses the axis at $x = -1$ ⁠, inside $[-2, 3]$ ⁠. Split there: the piece from $-2$ to $-1$ is below, so take it with a minus sign ( $\frac12$ ⁠), then add 8.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Application of Integrals in past papers: question 7 of 17 · see them all

Q11: Area of {4x2+9y2≤36, 2x+3y≥6}\{4x^2 + 9y^2 \le 36,\ 2x + 3y \ge 6\}{4x2+9y2≤36, 2x+3y≥6}⁠, or under x−y+1=0x - y + 1 = 0x−y+1=0 from x=−2x = -2x=−2 to x=3x = 3x=3

Q11: Area of $\{4x^2 + 9y^2 \le 36,\ 2x + 3y \ge 6\}$ ⁠, or under $x - y + 1 = 0$ from $x = -2$ to $x = 3$