Question 1

Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
2. Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Accepted Answer

1. The given ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

This ellipse has semi-major axis $a = 4$ (since $16 = 4^2$) and semi-minor axis $b = 3$ (since $9 = 3^2$).

Area of ellipse = $\pi a b = \pi \times 4 \times 3 = 12\pi$.

2. The given ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Here, semi-major axis $a = 3$ (since $9 = 3^2$) and semi-minor axis $b = 2$ (since $4 = 2^2$).

Area of ellipse = $\pi a b = \pi \times 3 \times 2 = 6\pi$. — The standard form of ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The area enclosed by the ellipse is given by $\pi a b$.

For question 1:
- $a^2 = 16 \Rightarrow a = 4$
- $b^2 = 9 \Rightarrow b = 3$
- Area = $\pi \times 4 \times 3 = 12\pi$

For question 2:
- $a^2 = 9 \Rightarrow a = 3$
- $b^2 = 4 \Rightarrow b = 2$
- Area = $\pi \times 3 \times 2 = 6\pi$

Question 2

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

Accepted Answer

The circle is $x^2 + y^2 = 4$, radius $r=2$. The area in the first quadrant bounded by $x=0$ and $x=2$ is the quarter circle sector from $x=0$ to $x=2$. Since $x=2$ is the rightmost point of the circle, the area is one quarter of the circle area.

Area of full circle = $\pi r^2 = \pi 	imes 2^2 = 4\pi$

Area in first quadrant bounded by $x=0$ and $x=2$ = $\frac{1}{4} 	imes 4\pi = \pi$

Hence, the answer is (A) $\pi$. — The circle $x^2 + y^2 = 4$ has radius 2. The first quadrant portion bounded by $x=0$ and $x=2$ corresponds to the quarter circle sector. The area of the full circle is $4\pi$, so the quarter circle area is $\pi$. Since $x=2$ is the extreme right boundary of the circle, the area bounded is exactly the quarter circle area.

Question 3

Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is

Accepted Answer

Given curve: $y^2 = 4x$ or $x = \frac{y^2}{4}$.

Bounded by $y$-axis ($x=0$), curve, and line $y=3$.

Area = $\int_{y=0}^{3} x \, dy = \int_0^3 \frac{y^2}{4} dy = \frac{1}{4} \int_0^3 y^2 dy = \frac{1}{4} \left[ \frac{y^3}{3} ight]_0^3 = \frac{1}{4} 	imes \frac{27}{3} = \frac{27}{12} = \frac{9}{4}$.

Hence, answer is (B) $\frac{9}{4}$. — The curve $y^2=4x$ can be rewritten as $x=\frac{y^2}{4}$. The area bounded by the curve, $y$-axis ($x=0$), and line $y=3$ is the integral of $x$ with respect to $y$ from 0 to 3. Evaluating the integral gives $\frac{9}{4}$.

Question 4

Find the area under the given curves and given lines:
(i) y = x^2, x = 1, x = 2 and x-axis
(ii) y = x^4, x = 1, x = 5 and x-axis

Accepted Answer

Solution:
(i) Area under y = x^2 from x = 1 to x = 2 and x-axis:
Area = ∫ from 1 to 2 of x^2 dx = [x^3/3] from 1 to 2 = (8/3) - (1/3) = 7/3.

(ii) Area under y = x^4 from x = 1 to x = 5 and x-axis:
Area = ∫ from 1 to 5 of x^4 dx = [x^5/5] from 1 to 5 = (3125/5) - (1/5) = (3124/5) = 624.8. — To find the area under the curve y = f(x) between x = a and x = b, integrate f(x) dx from a to b.

(i) For y = x^2, integrate x^2 dx from 1 to 2:
∫ x^2 dx = x^3/3
Evaluate at 2 and 1:
(2^3)/3 - (1^3)/3 = (8/3) - (1/3) = 7/3.

(ii) For y = x^4, integrate x^4 dx from 1 to 5:
∫ x^4 dx = x^5/5
Evaluate at 5 and 1:
(5^5)/5 - (1^5)/5 = 3125/5 - 1/5 = 3124/5 = 624.8.

Question 5

Sketch the graph of y = |x + 3| and evaluate ∫ from -6 to 0 of |x + 3| dx.

Accepted Answer

Solution:
The function y = |x + 3| is V-shaped with vertex at x = -3.

For x < -3, x + 3 < 0, so |x + 3| = -(x + 3).
For x ≥ -3, x + 3 ≥ 0, so |x + 3| = x + 3.

Evaluate the integral:
∫ from -6 to 0 |x + 3| dx = ∫ from -6 to -3 -(x + 3) dx + ∫ from -3 to 0 (x + 3) dx

First integral:
∫ from -6 to -3 -(x + 3) dx = ∫ from -6 to -3 (-x - 3) dx = [-x^2/2 - 3x] from -6 to -3
= [-(9/2) - 3(-3)] - [-(36/2) - 3(-6)] = (-4.5 + 9) - (-18 + 18) = 4.5 - 0 = 4.5

Second integral:
∫ from -3 to 0 (x + 3) dx = [x^2/2 + 3x] from -3 to 0 = (0 + 0) - (9/2 - 9) = 0 - (4.5 - 9) = 4.5

Total area = 4.5 + 4.5 = 9. — The absolute value function splits the integral at the point where the inside expression is zero, i.e., at x = -3.

From -6 to -3, |x + 3| = -(x + 3).
From -3 to 0, |x + 3| = x + 3.

Calculate each integral separately and add:

1) ∫ from -6 to -3 -(x + 3) dx = ∫ (-x - 3) dx
2) ∫ from -3 to 0 (x + 3) dx

Evaluating both integrals and summing gives the total area under the curve from -6 to 0 as 9.

Question 6

Find the area bounded by the curve y = sin x between x = 0 and x = 2π. Choose the correct answer in the following

Accepted Answer

Solution:
The curve y = sin x oscillates between positive and negative values between 0 and 2π.
Area bounded means the total area between the curve and x-axis.

Calculate:
Area = ∫ from 0 to π sin x dx - ∫ from π to 2π sin x dx
= [-cos x] from 0 to π - [-cos x] from π to 2π
= (-cos π + cos 0) - (-cos 2π + cos π)
= (-(-1) + 1) - (-(1) + (-1))
= (1 + 1) - (-1 -1) = 2 - (-2) = 4.

Hence, the correct answer is 4. — The sine function is positive from 0 to π and negative from π to 2π.
To find the total area bounded by the curve and x-axis, integrate the positive part and subtract the negative part (or take absolute value).

Area = ∫_0^π sin x dx - ∫_π^{2π} sin x dx
= [-cos x]_0^π - [-cos x]_π^{2π}
= (1 + 1) - (-1 - 1) = 4.

Therefore, the total area bounded by y = sin x from 0 to 2π is 4.

Question 7

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) $\frac{17}{4}$

Accepted Answer

To find the area bounded by the curve y = x^3, the x-axis, and the lines x = -2 and x = 1, we calculate the definite integral of y = x^3 from -2 to 1.

Step 1: Set up the integral:
Area = \int_{-2}^{1} x^3 \, dx

Step 2: Integrate:
\int x^3 \, dx = \frac{x^4}{4} + C

Step 3: Evaluate definite integral:
= \left[ \frac{x^4}{4} \right]_{-2}^{1} = \frac{1^4}{4} - \frac{(-2)^4}{4} = \frac{1}{4} - \frac{16}{4} = \frac{1}{4} - 4 = -\frac{15}{4}

Step 4: Since area cannot be negative, take absolute value:
Area = \frac{15}{4}

Hence, the correct answer is option (C) $\frac{15}{4}$. — The integral of x^3 from -2 to 1 is negative because the curve lies below the x-axis for x < 0 and above for x > 0. The definite integral gives the net area (signed area). To find the actual area bounded by the curve and the x-axis, we take the absolute value of the integral. Thus, the area is $\frac{15}{4}$.

Question 8

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 $x < 0$].

Accepted Answer

Given y = x|x|, which can be rewritten as: - For x > 0, y = x^2 - For x < 0, y = -x^2 We need to find the area bounded by the curve, x-axis, and lines x = -1 and x = 1. Step 1: Express the area as sum of two integrals: Area = \int_{-1}^{0} |y| \, dx + \int_{0}^{1} |y| \, dx Since for x < 0, y = -x^2 (below x-axis), area contribution is positive: \int_{-1}^{0} -y \, dx = \int_{-1}^{0} -(-x^2) \, dx = \int_{-1}^{0} x^2 \, dx For x > 0, y = x^2 (above x-axis), area is: \int_{0}^{1} x^2 \, dx Step 2: Calculate each integral: \int_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3} \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3} Step 3: Total area: Area = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} Therefore, the correct answer is option (C) $\frac{2}{3}$. — The function y = x|x| behaves differently on negative and positive sides of x. On negative side, it is negative of x squared, and on positive side, it is x squared. Since area is always positive, we take absolute values accordingly. Integrating the positive functions over the respective intervals and summing gives the total area.

Question 9

What is the formula to find the area bounded by the curve $y = f(x)$, the $x$-axis, and the ordinates $x = a$ and $x = b$?

Accepted Answer

$\int_a^b f(x) \, dx$ — The area bounded by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$ is given by integrating the function $f(x)$ with respect to $x$ from $a$ to $b$, i.e., $\int_a^b f(x) \, dx$.

Question 10

The area bounded by the curve $x = g(y)$, the $y$-axis, and the lines $y = c$ and $y = d$ is given by which integral?

Accepted Answer

$\int_c^d g(y) \, dy$ — When the curve is expressed as $x = g(y)$, the area bounded by the curve, the $y$-axis, and the horizontal lines $y = c$ and $y = d$ is found by integrating $g(y)$ with respect to $y$ from $c$ to $d$, i.e., $\int_c^d g(y) \, dy$.

Question 11

If the curve $y = f(x)$ lies below the $x$-axis between $x = a$ and $x = b$, what is the correct way to express the area bounded by the curve, $x$-axis, and ordinates $x = a$ and $x = b$?

Accepted Answer

$\left| \int_a^b f(x) \, dx \right|$ — When the curve lies below the $x$-axis, the definite integral $\int_a^b f(x) \, dx$ is negative. However, the area is always positive, so we take the absolute value $\left| \int_a^b f(x) \, dx \right|$.

Question 12

Consider a curve $y = f(x)$ that lies partly above and partly below the $x$-axis between $x = a$ and $x = b$. If $A_1$ is the integral over the part below the axis and $A_2$ over the part above, what is the total area bounded by the curve and the $x$-axis?

Accepted Answer

$|A_1| + A_2$ — Since $A_1$ is negative (area below $x$-axis) and $A_2$ is positive (area above $x$-axis), the total area is the sum of the absolute value of $A_1$ and $A_2$, i.e., $|A_1| + A_2$.

Question 13

Find the area enclosed by the circle $x^{2} + y^{2} = a^{2}$ using integration with respect to $x$.

Accepted Answer

\pi a^{2} — Given: Circle equation $x^{2} + y^{2} = a^{2}$
Find: Area enclosed by the circle
Formula: $A = 4 \int_0^a y \, dx$, where $y = \sqrt{a^{2} - x^{2}}$
Solution:
Step 1: Substitute $y$ in integral: $A = 4 \int_0^a \sqrt{a^{2} - x^{2}} \, dx$
Step 2: Use formula for integral $\int \sqrt{a^{2} - x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C$
Step 3: Evaluate definite integral:
$A = 4 \left[ \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} ight]_0^a = 4 	imes \frac{a^{2}}{2} 	imes \frac{\pi}{2} = \pi a^{2}$
Answer: $\pi a^{2}$
Note: Students often forget to multiply by 4 due to symmetry.

Question 14

Using horizontal strips, find the area enclosed by the circle $x^{2} + y^{2} = a^{2}$.

Accepted Answer

\pi a^{2} — Given: Circle equation $x^{2} + y^{2} = a^{2}$
Find: Area enclosed by the circle using horizontal strips
Formula: $A = 4 \int_0^a x \, dy$, where $x = \sqrt{a^{2} - y^{2}}$
Solution:
Step 1: Substitute $x$ in integral: $A = 4 \int_0^a \sqrt{a^{2} - y^{2}} \, dy$
Step 2: Use integral formula $\int \sqrt{a^{2} - y^{2}} \, dy = \frac{y}{2} \sqrt{a^{2} - y^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{y}{a} + C$
Step 3: Evaluate definite integral:
$A = 4 \left[ \frac{y}{2} \sqrt{a^{2} - y^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{y}{a} ight]_0^a = 4 	imes \frac{a^{2}}{2} 	imes \frac{\pi}{2} = \pi a^{2}$
Answer: $\pi a^{2}$
Note: Symmetry is used to multiply the first quadrant area by 4.

Question 15

Find the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ using integration with respect to $x$.

Accepted Answer

\pi a b — Given: Ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Find: Area enclosed by the ellipse
Formula: $A = 4 \int_0^a y \, dx$, where $y = \frac{b}{a} \sqrt{a^{2} - x^{2}}$
Solution:
Step 1: Substitute $y$ in integral: $A = 4 \int_0^a \frac{b}{a} \sqrt{a^{2} - x^{2}} \, dx = \frac{4b}{a} \int_0^a \sqrt{a^{2} - x^{2}} \, dx$
Step 2: Use integral formula $\int \sqrt{a^{2} - x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} + C$
Step 3: Evaluate definite integral:
$A = \frac{4b}{a} \left[ \frac{a^{2}}{2} 	imes \frac{\pi}{2} ight] = \pi a b$
Answer: $\pi a b$
Note: Symmetry is used to multiply the first quadrant area by 4.

Mathematics

Mathematics — Study Notes

8.1 Introduction

8.2 Area under Simple Curves

Example 1: Area enclosed by the circle x² + y² = a²

Practice Questions — Mathematics

All 7 Chapters in Mathematics Part-II