Application of Integrals — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 8, Application of Integrals (অনুকলৰ প্ৰয়োগ). It covers Exercise 8.1 and the Miscellaneous Exercise on Chapter 8, with each area computed by full integration and illustrated with a figure of the shaded region, so you can follow the reasoning and prepare confidently for your examination.
Summary
This chapter uses the definite integral to find the area of a region bounded by curves and lines. The area of the region bounded by the curve $y = f(x)$, the $x$-axis and the ordinates $x = a$ and $x = b$ (with $b > a$) is obtained by adding up thin vertical strips of height $y$ and width $dx$, giving Area $= \int_a^b y\,dx = \int_a^b f(x)\,dx$. When it is easier to use horizontal strips, the area of the region bounded by the curve $x = g(y)$, the $y$-axis and the lines $y = c$ and $y = d$ is Area $= \int_c^d x\,dy = \int_c^d g(y)\,dy$.
A definite integral gives a signed area, so care is needed when a curve dips below the $x$-axis. If $f(x) < 0$ on part of the interval, the integral over that part is negative, and we take its absolute value. When the curve is above the axis on one stretch and below it on another, the total area is the sum of the absolute values of the separate pieces: $A = |A_1| + A_2$. The chapter also records two standard results proved by integration — the area enclosed by the circle $x^2 + y^2 = a^2$ is $\pi a^2$, and the area enclosed by the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is $\pi a b$.
Summary: ASSEB Class 12 Mathematics Chapter 8 Application of Integrals explains how to find the area under a curve and between lines and arcs using definite integration, with vertical and horizontal strips, the treatment of regions below the $x$-axis, and the standard areas of a circle and an ellipse, with complete worked answers to Exercise 8.1 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 8.1
Throughout this exercise we use the standard result $\int \sqrt{a^2 – x^2}\,dx = \frac{x}{2}\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$, and $\sin^{-1}(1) = \frac{\pi}{2}$.
1. Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Answer: Comparing $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives $a^2 = 16$, $b^2 = 9$, so $a = 4$ and $b = 3$. The ellipse is symmetric about both axes, so its total area is four times the area of the part in the first quadrant. In that quadrant $y = \frac{3}{4}\sqrt{16 – x^2}$ with $x$ running from $0$ to $4$.
$$A = 4 \int_0^4 y\,dx = 4 \int_0^4 \frac{3}{4}\sqrt{16 – x^2}\,dx = 3 \int_0^4 \sqrt{16 – x^2}\,dx$$
$$A = 3\left[ \frac{x}{2}\sqrt{16 – x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} \right]_0^4 = 3\left[ (0 + 8 \sin^{-1} 1) – 0 \right] = 3 \times 8 \times \frac{\pi}{2} = 12\pi$$
Hence the required area is $12\pi$ square units. (This agrees with the standard result Area of an ellipse $= \pi a b = \pi \times 4 \times 3 = 12\pi$.)
2. Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Answer: Here $a^2 = 4$ and $b^2 = 9$, so $a = 2$ and $b = 3$. By symmetry the total area is four times the first-quadrant part, where $y = \frac{3}{2}\sqrt{4 – x^2}$ and $x$ runs from $0$ to $2$.
$$A = 4 \int_0^2 y\,dx = 4 \int_0^2 \frac{3}{2}\sqrt{4 – x^2}\,dx = 6 \int_0^2 \sqrt{4 – x^2}\,dx$$
$$A = 6\left[ \frac{x}{2}\sqrt{4 – x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_0^2 = 6\left[ (0 + 2 \sin^{-1} 1) – 0 \right] = 6 \times 2 \times \frac{\pi}{2} = 6\pi$$
Hence the required area is $6\pi$ square units (again $\pi a b = \pi \times 2 \times 3 = 6\pi$).
Choose the correct answer in the following Exercises 3 and 4.
3. Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is
(A) $\pi$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{4}$
Answer: The correct option is (A) $\pi$. In the first quadrant the circle gives $y = \sqrt{4 – x^2}$, and the required region lies between $x = 0$ and $x = 2$.
$$A = \int_0^2 \sqrt{4 – x^2}\,dx = \left[ \frac{x}{2}\sqrt{4 – x^2} + 2\sin^{-1}\frac{x}{2} \right]_0^2 = (0 + 2\sin^{-1} 1) – 0 = 2 \times \frac{\pi}{2} = \pi$$
So the area is $\pi$ square units, which is option (A). (It is exactly one quarter of the circle, whose full area is $\pi (2)^2 = 4\pi$.)
4. Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is
(A) $2$ (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$
Answer: The correct option is (B) $\frac{9}{4}$. The region is bounded on the left by the $y$-axis and on the right by the parabola, so we integrate horizontal strips with respect to $y$. From $y^2 = 4x$ we get $x = \frac{y^2}{4}$, and $y$ runs from $0$ to $3$.
$$A = \int_0^3 x\,dy = \int_0^3 \frac{y^2}{4}\,dy = \frac{1}{4}\left[ \frac{y^3}{3} \right]_0^3 = \frac{1}{4} \times \frac{27}{3} = \frac{27}{12} = \frac{9}{4}$$
Hence the area is $\frac{9}{4}$ square units, which is option (B).
Miscellaneous Exercise on Chapter 8
1. Find the area under the given curves and given lines:
(i) $y = x^2$, $x = 1$, $x = 2$ and $x$-axis
Answer: The area between the curve $y = x^2$, the $x$-axis and the ordinates $x = 1$ and $x = 2$ is
$$A = \int_1^2 y\,dx = \int_1^2 x^2\,dx = \left[ \frac{x^3}{3} \right]_1^2 = \frac{8}{3} – \frac{1}{3} = \frac{7}{3} \text{ square units.}$$
(ii) $y = x^4$, $x = 1$, $x = 5$ and $x$-axis
Answer: Since $y = x^4 \geq 0$ on $[1, 5]$, the area is
$$A = \int_1^5 x^4\,dx = \left[ \frac{x^5}{5} \right]_1^5 = \frac{5^5}{5} – \frac{1^5}{5} = \frac{3125 – 1}{5} = \frac{3124}{5} = 624.8 \text{ square units.}$$
2. Sketch the graph of $y = |x + 3|$ and evaluate $\int_{-6}^{0} |x + 3|\,dx$.
Answer: The graph of $y = |x + 3|$ is a V with its vertex at $(-3, 0)$: it is the straight line $y = -(x + 3)$ for $x \leq -3$ and the line $y = x + 3$ for $x \geq -3$. On the interval of integration the point $x = -3$ divides $[-6, 0]$ into two parts, so we split the integral there.
$$\int_{-6}^{0} |x + 3|\,dx = \int_{-6}^{-3} -(x + 3)\,dx + \int_{-3}^{0} (x + 3)\,dx$$
$$= -\left[ \frac{x^2}{2} + 3x \right]_{-6}^{-3} + \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0} = -\left[ -\frac{9}{2} – 0 \right] + \left[ 0 – \left( -\frac{9}{2} \right) \right] = \frac{9}{2} + \frac{9}{2} = 9$$
Hence $\int_{-6}^{0} |x + 3|\,dx = 9$. (Geometrically the shaded region is two right triangles, each of base $3$ and height $3$, giving $2 \times \frac{1}{2} \times 3 \times 3 = 9$ square units.)
3. Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.
Answer: On $[0, \pi]$ the curve $y = \sin x$ lies above the $x$-axis, while on $[\pi, 2\pi]$ it lies below the axis. The definite integral over $[\pi, 2\pi]$ is negative, so we take its absolute value and add.
$$A = \int_0^{\pi} \sin x\,dx + \left| \int_{\pi}^{2\pi} \sin x\,dx \right|$$
$$\int_0^{\pi} \sin x\,dx = [-\cos x]_0^{\pi} = (-\cos\pi) – (-\cos 0) = 1 + 1 = 2$$
$$\int_{\pi}^{2\pi} \sin x\,dx = [-\cos x]_{\pi}^{2\pi} = (-\cos 2\pi) – (-\cos\pi) = -1 – 1 = -2$$
Therefore $A = 2 + |-2| = 2 + 2 = 4$ square units.
Choose the correct answer in the following Exercises from 4 to 5.
4. Area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$ is
(A) $-9$ (B) $\frac{-15}{4}$ (C) $\frac{15}{4}$ (D) $\frac{17}{4}$
Answer: The correct option is (D) $\frac{17}{4}$. The curve $y = x^3$ is negative for $-2 \leq x < 0$ and positive for $0 < x \leq 1$, so we split the region at $x = 0$ and add the absolute values.
$$A = \left| \int_{-2}^{0} x^3\,dx \right| + \int_{0}^{1} x^3\,dx = \left| \left[ \frac{x^4}{4} \right]_{-2}^{0} \right| + \left[ \frac{x^4}{4} \right]_{0}^{1}$$
$$= \left| 0 – \frac{16}{4} \right| + \left( \frac{1}{4} – 0 \right) = 4 + \frac{1}{4} = \frac{17}{4} \text{ square units.}$$
So the area is $\frac{17}{4}$ square units, which is option (D). (If one wrongly evaluated $\int_{-2}^{1} x^3\,dx$ as a single signed integral, the result would be $-\frac{15}{4}$, the distractor (B); the area must be positive.)
5. The area bounded by the curve $y = x\,|x|$, $x$-axis and the ordinates $x = -1$ and $x = 1$ is given by
(A) $0$ (B) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $\frac{4}{3}$
[Hint: $y = x^2$ if $x > 0$ and $y = -x^2$ if $x < 0$].
Answer: The correct option is (C) $\frac{2}{3}$. Using the hint, $y = -x^2$ on $[-1, 0]$ (below the axis) and $y = x^2$ on $[0, 1]$ (above the axis). Taking the absolute value on the negative part,
$$A = \int_{-1}^{0} |{-x^2}|\,dx + \int_{0}^{1} x^2\,dx = \int_{-1}^{0} x^2\,dx + \int_{0}^{1} x^2\,dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1}$$
$$= \left( 0 – \left( -\frac{1}{3} \right) \right) + \left( \frac{1}{3} – 0 \right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \text{ square units.}$$
Hence the area is $\frac{2}{3}$ square units, which is option (C).
Additional Questions and Answers
Multiple Choice Questions
1. The area of the region bounded by the curve $y = f(x)$, the $x$-axis and the lines $x = a$ and $x = b$ (with $b > a$) is
(A) $\int_a^b y\,dy$ (B) $\int_a^b y\,dx$ (C) $\int_a^b x\,dy$ (D) $\int_a^b f(y)\,dx$
Answer: (B) $\int_a^b y\,dx$. The area is the sum of vertical strips of height $y = f(x)$ and width $dx$.
2. The area enclosed by the circle $x^2 + y^2 = a^2$ is
(A) $\pi a$ (B) $2\pi a$ (C) $\pi a^2$ (D) $4\pi a^2$
Answer: (C) $\pi a^2$. Integration gives $4 \int_0^a \sqrt{a^2 – x^2}\,dx = \pi a^2$, the familiar area of a circle of radius $a$.
3. The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is
(A) $\pi a b$ (B) $\pi(a + b)$ (C) $\pi a^2 b^2$ (D) $2\pi a b$
Answer: (A) $\pi a b$. This reduces to $\pi a^2$ (the circle) when $a = b$.
4. The area bounded by the curve $y = x^2$, the $x$-axis and the lines $x = 0$ and $x = 3$ is
(A) $3$ (B) $6$ (C) $9$ (D) $27$
Answer: (C) $9$. $\int_0^3 x^2\,dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} = 9$.
5. The area bounded by the line $y = x$, the $x$-axis and the ordinates $x = 0$ and $x = 4$ is
(A) $8$ (B) $16$ (C) $4$ (D) $32$
Answer: (A) $8$. $\int_0^4 x\,dx = \left[ \frac{x^2}{2} \right]_0^4 = 8$ (also the area of a right triangle with base $4$ and height $4$).
6. The area of the region bounded by $y = \cos x$, the $x$-axis between $x = 0$ and $x = \frac{\pi}{2}$ is
(A) $0$ (B) $1$ (C) $2$ (D) $\frac{1}{2}$
Answer: (B) $1$. $\int_0^{\pi/2} \cos x\,dx = [\sin x]_0^{\pi/2} = 1 – 0 = 1$.
7. If a curve $y = f(x)$ lies entirely below the $x$-axis on $[a, b]$, then the area between the curve and the axis equals
(A) $\int_a^b f(x)\,dx$ (B) $\left| \int_a^b f(x)\,dx \right|$ (C) a negative number (D) $0$
Answer: (B) $\left| \int_a^b f(x)\,dx \right|$. When $f(x) < 0$ the integral is negative, so the area is its absolute value.
8. The area of the region bounded by the curve $y = \sqrt{x}$, the $x$-axis and the lines $x = 0$ and $x = 4$ is
(A) $\frac{16}{3}$ (B) $\frac{8}{3}$ (C) $\frac{4}{3}$ (D) $8$
Answer: (A) $\frac{16}{3}$. $\int_0^4 \sqrt{x}\,dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \times 8 = \frac{16}{3}$.
Fill in the Blanks
1. The area bounded by the curve $y = f(x)$, the $x$-axis and the lines $x = a$ and $x = b$ is ________.
Answer: $\int_a^b f(x)\,dx$
2. When a region is bounded by a curve $x = g(y)$ and the $y$-axis between $y = c$ and $y = d$, we integrate using ________ strips.
Answer: horizontal
3. The area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is ________.
Answer: $6\pi$ (since $a = 3$, $b = 2$ and area $= \pi a b$)
4. The area of a circle of radius $5$, found by integration, is ________.
Answer: $25\pi$
5. If the definite integral of $f$ over $[a, b]$ turns out to be negative, then the area equals its ________.
Answer: absolute value
True or False
1. The definite integral $\int_a^b f(x)\,dx$ always gives the geometric area between the curve and the $x$-axis.
Answer: False. It gives a signed area; it equals the geometric area only when $f(x) \geq 0$ throughout $[a, b]$.
2. The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi a b$.
Answer: True.
3. Area is always a non-negative quantity.
Answer: True. Area is measured as a positive number, which is why we take absolute values of negative integrals.
4. To find an area using horizontal strips, we integrate $x$ with respect to $y$.
Answer: True. The area is $\int_c^d x\,dy$, where $x$ is expressed as a function of $y$.
5. The area bounded by $y = \sin x$ between $x = 0$ and $x = 2\pi$ is $0$.
Answer: False. The signed integral is $0$, but the area is $4$ square units, because the trough on $[\pi, 2\pi]$ contributes a positive area.
Short Answer Questions
1. Find the area of the region bounded by the line $y = 3x$, the $x$-axis and the ordinate $x = 2$.
Answer: $A = \int_0^2 3x\,dx = \left[ \frac{3x^2}{2} \right]_0^2 = \frac{3 \times 4}{2} = 6$ square units.
2. Find the area of the region in the first quadrant bounded by the curve $y = x^2$, the $y$-axis and the lines $y = 1$ and $y = 4$.
Answer: Here we use horizontal strips, with $x = \sqrt{y}$. $A = \int_1^4 x\,dy = \int_1^4 \sqrt{y}\,dy = \left[ \frac{2}{3} y^{3/2} \right]_1^4 = \frac{2}{3}(8 – 1) = \frac{14}{3}$ square units.
3. Find the area bounded by the curve $y = \cos x$, the $x$-axis, between $x = 0$ and $x = \pi$.
Answer: On $[0, \frac{\pi}{2}]$ the cosine is positive and on $[\frac{\pi}{2}, \pi]$ it is negative, so $A = \int_0^{\pi/2} \cos x\,dx + \left| \int_{\pi/2}^{\pi} \cos x\,dx \right| = [\sin x]_0^{\pi/2} + \left| [\sin x]_{\pi/2}^{\pi} \right| = 1 + |{-1}| = 2$ square units.
4. Find the area of the region bounded by the curve $y = \sqrt{x}$, the $x$-axis and the line $x = 9$.
Answer: $A = \int_0^9 \sqrt{x}\,dx = \left[ \frac{2}{3} x^{3/2} \right]_0^9 = \frac{2}{3} \times 27 = 18$ square units.
Key Terms
| Term | Meaning |
|---|---|
| Application of Integrals | Using the definite integral to compute areas of regions bounded by curves and lines. |
| Area under a curve | The region enclosed between a curve $y = f(x)$, the $x$-axis and two ordinates $x = a$, $x = b$, equal to $\int_a^b f(x)\,dx$ when $f \geq 0$. |
| Ordinate | A vertical line $x = a$ (or the $y$-coordinate of a point) that bounds a region on the left or right. |
| Vertical strip | A thin element of area of height $y$ and width $dx$; summing them gives $\int_a^b y\,dx$. |
| Horizontal strip | A thin element of area of width $x$ and height $dy$; summing them gives $\int_c^d x\,dy$. |
| Elementary area | The small area $dA = y\,dx$ of a single strip, located at an arbitrary position in the region. |
| Definite integral | $\int_a^b f(x)\,dx$, the limit of a sum; it gives the signed area between the curve and the $x$-axis. |
| Signed area | The value of a definite integral, which is negative where the curve lies below the $x$-axis. |
| Absolute value of area | The positive measure taken when an integral is negative, so that $\text{Area} = |A_1| + A_2$ for a curve crossing the axis. |
| Area of a circle | The region enclosed by $x^2 + y^2 = a^2$, equal to $\pi a^2$ by integration. |
| Area of an ellipse | The region enclosed by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, equal to $\pi a b$. |