NCERT Exemplar Solutions Class 12 Maths Application of Integrals — Long Answer and Objective Q16–34

NCERT Exemplar Problems, Class 12 Mathematics — Application of Integrals, Long Answer questions 16 to 23 and Objective Type questions 24 to 34. All 19 questions solved.

Eight long-answer questions and eleven multiple-choice ones. The long answers are where the method meets awkward shapes: a parabola inside a circle (16, 19), a triangle given by its corners or by its sides (18, 20, 21), and two curves — $\sin x$ and $2\cos x$ — that dip below the axis and punish anyone who integrates straight across (17, 22). The objective questions are mostly the Short Answer band compressed into one line each, plus three that are pure geometry in disguise (27, 31, 32).

Two answers in the printed key are wrong, Questions 19 and 23, and one option is misprinted, Question 26. Each is worked through below so that you can see where the printed figure comes from and why it does not hold.

A note on the syllabus. The current NCERT textbook no longer covers the area between two curves. Most of this page does, and it remains in both the JEE Main and JEE Advanced syllabuses.

Key insight. An integral measures signed area; a question asking for area wants the unsigned kind. Whenever the curve crosses the axis inside the interval — Questions 17, 22 and 28 — split the integral at each crossing and add the absolute values. Integrating $\sin x$ from $0$ to $2\pi$ gives $0$, and the answer is $4$.

Question 16

Find the area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$.

Solution. Complete the square: $(x – 2)^2 + y^2 = 4$, the circle of radius $2$ centred at $(2, 0)$, passing through the origin. Substituting $y^2 = 2x$: $x^2 + 2x = 4x$, so $x = 0$ or $x = 2$. The curves meet at $(0, 0)$ and $(2, \pm 2)$.

For $0 < x < 2$ the circle lies outside the parabola: at $x = 1$, the circle’s upper half is at $\sqrt3 \approx 1.73$ and the parabola at $\sqrt2 \approx 1.41$. So between $x = 0$ and $x = 2$ there are two thin crescents, one above the axis and one below, each enclosed by both curves. Their total area is

$$A = 2\int_0^2\left(\sqrt{4x – x^2} – \sqrt{2x}\right)dx$$

The first integral is the area of a quarter of the circle — the part above the axis and left of its centre — which is $\frac14\pi(2)^2 = \pi$. The second is $\sqrt2\cdot\frac23 x^{3/2}\Big|_0^2 = \sqrt2\cdot\frac23\cdot2\sqrt2 = \frac83$. So

The other reading. “The region bounded by” two curves is ambiguous when one sits partly inside the other. The region inside both — inside the circle and inside the parabola — has area $4\pi – 2\left(\pi – \frac83\right) = 2\pi + \frac{16}{3}$. The book’s answer is the crescents, so that is the one given above; in an examination, say which region you mean.

Question 17

Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Solution. $\sin x \ge 0$ on $[0, \pi]$ and $\le 0$ on $[\pi, 2\pi]$. Integrating straight across gives $0$ — the two arches cancel. The area is the sum of their sizes:

$$A = \int_0^\pi\sin x\,dx + \left|\int_\pi^{2\pi}\sin x\,dx\right| = \Big[-\cos x\Big]_0^\pi + \left|\Big[-\cos x\Big]_\pi^{2\pi}\right| = 2 + |-2|$$

Question 18

Find the area of region bounded by the triangle whose vertices are $(-1, 1)$, $(0, 5)$ and $(3, 2)$, using integration.

Solution. Call the vertices $A(-1, 1)$, $B(0, 5)$, $C(3, 2)$. The three sides are

• $AB$: slope $\frac{5 – 1}{0 + 1} = 4$, so $y = 4x + 5$;
• $BC$: slope $\frac{2 – 5}{3 – 0} = -1$, so $y = 5 – x$;
• $AC$: slope $\frac{2 – 1}{3 + 1} = \frac14$, so $y = \frac{x + 5}{4}$.

The vertex $B$ is the highest point, so the upper boundary changes at $x = 0$: it is $AB$ on $[-1, 0]$ and $BC$ on $[0, 3]$. The lower boundary is $AC$ throughout.

$$A = \int_{-1}^{0}\left(4x + 5 – \frac{x + 5}{4}\right)dx + \int_0^3\left(5 – x – \frac{x + 5}{4}\right)dx = \int_{-1}^{0}\frac{15(x + 1)}{4}\,dx + \int_0^3\frac{15 – 5x}{4}\,dx$$

$$= \frac{15}{4}\cdot\frac12 + \frac14\left(45 – \frac{45}{2}\right) = \frac{15}{8} + \frac{45}{8}$$

Check. The determinant formula gives $\frac12\left|-1(5 – 2) + 0(2 – 1) + 3(1 – 5)\right| = \frac12\cdot 15$. The question asks for integration, so the check is not the answer — but it is the fastest way to know the integration is right.

Question 19

Draw a rough sketch of the region $\left\{(x, y) : y^2 \le 6ax \text{ and } x^2 + y^2 \le 16a^2\right\}$. Also find the area of the region sketched using method of integration.

Solution. Take $a > 0$. The region is the part of the disc of radius $4a$ that lies inside the parabola $y^2 = 6ax$. They meet where $x^2 + 6ax = 16a^2$, i.e. $(x + 8a)(x – 2a) = 0$; only $x = 2a$ is admissible (the parabola needs $x \ge 0$), giving the points $(2a, \pm 2\sqrt3\,a)$.

The region is symmetric about the $x$-axis. Above the axis, from $x = 0$ to $2a$ the boundary is the parabola, and from $2a$ to $4a$ it is the circle:

$$A = 2\left[\int_0^{2a}\sqrt{6ax}\,dx + \int_{2a}^{4a}\sqrt{16a^2 – x^2}\,dx\right]$$

The first integral is $\sqrt{6a}\cdot\frac23(2a)^{3/2} = \frac23\sqrt{12}\,a^2\cdot 2 = \frac{8\sqrt3}{3}a^2$.

The second, by the standard formula,

$$\left[\frac x2\sqrt{16a^2 – x^2} + 8a^2\sin^{-1}\frac{x}{4a}\right]_{2a}^{4a} = 8a^2\cdot\frac\pi2 – \left(a\cdot 2\sqrt3\,a + 8a^2\cdot\frac\pi6\right) = \frac{8\pi}{3}a^2 – 2\sqrt3\,a^2$$

Adding and doubling:

$$A = 2\left[\frac{8\sqrt3}{3} – 2\sqrt3 + \frac{8\pi}{3}\right]a^2 = 2\left[\frac{2\sqrt3}{3} + \frac{8\pi}{3}\right]a^2$$

The printed key gives $\frac43\left(\sqrt3 + 2\pi\right)a^2$, and it is wrong. That value is about $10.69a^2$ — but the parabola part alone, $2\cdot\frac{8\sqrt3}{3}a^2 \approx 9.24a^2$, plus the circular cap, $2\left(\frac{8\pi}{3} – 2\sqrt3\right)a^2 \approx 9.83a^2$, already comes to $19.06a^2$, and a direct count of grid points inside both curves agrees. The key’s figure is what you get by halving the $\pi$ term — that is, by forgetting to double the circular part. If your answer has $4\pi$ in it, trust it.

Question 20

Compute the area bounded by the lines $x + 2y = 2$, $y – x = 1$ and $2x + y = 7$.

Solution. Pair the lines to find the vertices:

• $x + 2y = 2$ and $y = x + 1$: $3x + 2 = 2$, so $(0, 1)$;
• $y = x + 1$ and $2x + y = 7$: $3x + 1 = 7$, so $(2, 3)$;
• $x + 2y = 2$ and $2x + y = 7$: $x = 4$, $y = -1$, so $(4, -1)$.

The lowest side, $y = \frac{2 – x}{2}$, runs the full width $[0, 4]$. The upper boundary is $y = x + 1$ on $[0, 2]$ and $y = 7 – 2x$ on $[2, 4]$:

$$A = \int_0^2\left(x + 1 – \frac{2 – x}{2}\right)dx + \int_2^4\left(7 – 2x – \frac{2 – x}{2}\right)dx = \int_0^2\frac{3x}{2}\,dx + \int_2^4\left(6 – \frac{3x}{2}\right)dx = 3 + 3$$

Question 21

Find the area bounded by the lines $y = 4x + 5$, $y = 5 – x$ and $4y = x + 5$.

Solution. These are the three sides of Question 18’s triangle: $y = 4x + 5$ and $y = 5 – x$ meet at $(0, 5)$; $y = 4x + 5$ and $4y = x + 5$ at $(-1, 1)$; $y = 5 – x$ and $4y = x + 5$ at $(3, 2)$. The integral is therefore the same one:

$$A = \int_{-1}^{0}\left(4x + 5 – \frac{x + 5}{4}\right)dx + \int_0^3\left(5 – x – \frac{x + 5}{4}\right)dx = \frac{15}{8} + \frac{45}{8}$$

The book sets the same triangle twice, once by its corners and once by its sides. The second version is the harder one, because finding the vertices is now part of the work.

Question 22

Find the area bounded by the curve $y = 2\cos x$ and the $x$-axis from $x = 0$ to $x = 2\pi$.

Solution. $\cos x$ changes sign at $\frac\pi2$ and $\frac{3\pi}{2}$, so the interval splits into three pieces: positive on $\left[0, \frac\pi2\right]$, negative on $\left[\frac\pi2, \frac{3\pi}2\right]$, positive on $\left[\frac{3\pi}{2}, 2\pi\right]$.

$$A = \Big[2\sin x\Big]_0^{\pi/2} + \left|\Big[2\sin x\Big]_{\pi/2}^{3\pi/2}\right| + \Big[2\sin x\Big]_{3\pi/2}^{2\pi} = 2 + |-4| + 2$$

Question 23

Draw a rough sketch of the given curve $y = 1 + |x + 1|$, $x = -3$, $x = 3$, $y = 0$ and find the area of the region bounded by them, using integration.

Solution. Remove the modulus at $x = -1$:

$$y = \begin{cases} -x, & x \le -1 \\ x + 2, & x \ge -1 \end{cases}$$

The graph is a V with its corner at $(-1, 1)$, always above the $x$-axis. Between $x = -3$ and $x = 3$:

$$A = \int_{-3}^{-1}(-x)\,dx + \int_{-1}^{3}(x + 2)\,dx = \left[-\frac{x^2}{2}\right]_{-3}^{-1} + \left[\frac{x^2}{2} + 2x\right]_{-1}^{3} = 4 + \left(\frac{21}{2} + \frac32\right) = 4 + 12$$

The printed key gives 15, and it is wrong. The region is two trapezia, which can be checked without integrating: on the left, parallel sides $y(-3) = 3$ and $y(-1) = 1$, width $2$, area $4$; on the right, parallel sides $1$ and $y(3) = 5$, width $4$, area $12$. The total is $16$.

Question 24

The area of the region bounded by the $y$-axis, $y = \cos x$ and $y = \sin x$, $0 \le x \le \frac\pi2$, is

Solution. The curves cross at $x = \frac\pi4$. The region touching the $y$-axis lies between $x = 0$ and $x = \frac\pi4$, where $\cos x$ is on top:

$$\int_0^{\pi/4}(\cos x – \sin x)\,dx = \Big[\sin x + \cos x\Big]_0^{\pi/4} = \sqrt2 – 1$$

Question 25

The area of the region bounded by the curve $x^2 = 4y$ and the straight line $x = 4y – 2$ is

Solution. The line is $y = \frac{x + 2}{4}$ and the parabola $y = \frac{x^2}{4}$. They meet where $x^2 = x + 2$, at $x = -1$ and $x = 2$, and the line is on top between them:

$$\int_{-1}^{2}\frac{x + 2 – x^2}{4}\,dx = \frac14\cdot\frac92 = \frac98$$

This is a quarter of Question 6’s integral from the Short Answer band, and the $\frac{(\beta – \alpha)^3}{6}$ shortcut gives it at once: $\frac14\cdot\frac{27}{6}$.

Question 26

The area of the region bounded by the curve $y = \sqrt{16 – x^2}$ and $x$-axis is

Solution. The curve is the upper half of the circle of radius $4$, so the area is $\frac12\pi(4)^2 = 8\pi$. By integration, $\int_{-4}^{4}\sqrt{16 – x^2}\,dx = \left[\frac x2\sqrt{16 – x^2} + 8\sin^{-1}\frac x4\right]_{-4}^{4} = 8\pi$.

The book prints option (A) as “8 sq units”, without the $\pi$. The key names (A), and none of the four options as printed is $8\pi$, so the $\pi$ has simply been dropped from (A). It is restored above.

Question 27

Area of the region in the first quadrant enclosed by the $x$-axis, the line $y = x$ and the circle $x^2 + y^2 = 32$ is

Solution. The region is a sector: it is bounded by two radii — the $x$-axis and $y = x$, which make an angle of $\frac\pi4$ at the origin — and the arc of the circle of radius $\sqrt{32}$. Its area is $\frac12 r^2\theta = \frac12\cdot 32\cdot\frac\pi4 = 4\pi$.

By integration, the line meets the circle at $(4, 4)$, and $\int_0^4 x\,dx + \int_4^{\sqrt{32}}\sqrt{32 – x^2}\,dx = 8 + (4\pi – 8) = 4\pi$.

Question 28

Area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is

Solution. $\cos x$ changes sign at $\frac\pi2$: $\int_0^{\pi/2}\cos x\,dx = 1$ and $\left|\int_{\pi/2}^{\pi}\cos x\,dx\right| = 1$. The straight integral from $0$ to $\pi$ is $0$, which is not among the options for a reason.

Question 29

The area of the region bounded by parabola $y^2 = x$ and the straight line $2y = x$ is

Solution. Both curves are naturally $x$ in terms of $y$, so integrate along $y$. They meet where $y^2 = 2y$, at $y = 0$ and $y = 2$; between them the line $x = 2y$ is to the right of the parabola $x = y^2$:

$$\int_0^2\left(2y – y^2\right)dy = 4 – \frac83 = \frac43$$

Question 30

The area of the region bounded by the curve $y = \sin x$ between the ordinates $x = 0$, $x = \frac\pi2$ and the $x$-axis is

Solution. $\int_0^{\pi/2}\sin x\,dx = \Big[-\cos x\Big]_0^{\pi/2} = 0 – (-1) = 1$. No sign change on this interval.

Question 31

The area of the region bounded by the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ is

Solution. By symmetry, four times the first-quadrant part:

$$4\int_0^5\frac45\sqrt{25 – x^2}\,dx = \frac{16}{5}\cdot\frac{25\pi}{4} = 20\pi$$

This is $\pi ab$ with $a = 5$, $b = 4$. Options (B) and (C) carry $\pi^2$, which cannot appear in an area bounded by a conic — a quick way to rule them out.

Question 32

The area of the region bounded by the circle $x^2 + y^2 = 1$ is

Solution. $4\int_0^1\sqrt{1 – x^2}\,dx = 4\cdot\frac\pi4 = \pi$, the area of a unit circle.

Question 33

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2$ and $x = 3$ is

Solution. The $x$-axis is the implied fourth side. On $[2, 3]$ the line is positive:

$$\int_2^3(x + 1)\,dx = \left[\frac{x^2}{2} + x\right]_2^3 = \frac{15}{2} – 4 = \frac72$$

Question 34

The area of the region bounded by the curve $x = 2y + 3$ and the $y$-axis, $y = 1$ and $y = -1$ is

Solution. Bounded by the $y$-axis and horizontal lines, so integrate along $y$. On $[-1, 1]$, $x = 2y + 3$ is between $1$ and $5$, always to the right of the axis:

$$\int_{-1}^{1}(2y + 3)\,dy = \Big[y^2 + 3y\Big]_{-1}^{1} = 4 – (-2) = 6$$

Common mistakes

• Integrating straight across a sign change in Questions 17, 22 and 28. The definite integral of $\sin x$ over $[0, 2\pi]$ is $0$; the area is $4$. Find every zero of the curve inside the interval first.
• Forgetting to double the circular part in Question 19. The region is symmetric about the $x$-axis, and both pieces — parabola and circle — need the factor of 2. Doubling only one is how the key’s $2\pi$ arises.
• Using one upper boundary across a whole triangle in Questions 18, 20 and 21. A triangle’s upper (or lower) side changes at the middle vertex, so the integral splits there.
• Not saying which region you mean in Question 16. When one curve sits partly inside another, “the region bounded by” them has more than one reading. State it, and the examiner can follow your working whichever was intended.
• Removing a modulus at the wrong point in Question 23. $|x + 1|$ changes form at $x = -1$, not at $x = 0$.
• Choosing an option with $\pi^2$ for a conic’s area, as in Question 31. Areas bounded by circles and ellipses carry $\pi$ to the first power.

Practise next

• Exemplar Short Answer Q1–15, the first band of the unit, if setting up “upper minus lower” was the difficulty here.
• NCERT textbook Miscellaneous Exercise on Application of Integrals — its $y = |x + 3|$ and $y = \sin x$ questions are the textbook versions of Questions 23 and 17.
• NCERT textbook Exercise 8.1 for the ellipse and circle areas behind Questions 31 and 32.
• Exemplar Integrals Long Answer Q35–47, where the $\int\sqrt{a^2 – x^2}\,dx$ form used in Questions 19 and 26 is one of several standard integrals in play.