In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.
Solved example 24.24
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4.
Solution:
1. In the fig.24.32 below, the red curve represents
$\small{x=f(y)=\pm \frac{\sqrt y}{2}}$
 |
| Fig.24.32 |
2. We are asked to find the area of the blue region.
• Assuming a thin horizontal strip of width dy, this area is equal to:
$\small{\int_1^4{\left[f(y) \right]dy}~=~\int_1^4{\left[\frac{\sqrt y}{2} \right]dy}~=~\frac{7}{3}}$ sq.units
[Here we use $\small{\frac{\sqrt y}{2}}$
Instead of $\small{-\frac{\sqrt y}{2}}$
This is because, in the first quadrant, x values are +ve]
(The reader may write all steps related to the integration process)
Solved example 24.25
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.
Solution:
1. In the fig.24.33 below,
♦ the red curve represents
$\small{y=f(x)= \frac{3 x^2}{4}}$
♦ the green line represents
$\small{y=f(x)= \frac{3 x}{2}~+~6}$
 |
| Fig.24.33 |
• Solving the two equations, we get the points of intersection:
A(−2,3) and B(4,12)
2. We are asked to find the area of the blue region.
• This area is equal to:
$\small{\int_{-2}^4{\left[f(x)~-~g(x) \right]dx}~=~\int_{-2}^4{\left[\frac{3 x^2}{4}~-~\frac{3 x}{2}~-~6 \right]dx}~=~27}$ sq.units
(The reader may write all steps related to the integration process)
Solved example 24.26
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and the x-axis.
Solution:
1. In the fig.24.34 below,
♦ the red curve represents
$\small{y=f(x)= x^2}$
♦ the green line represents
$\small{y=f(x)= x~+~2}$
 |
| Fig.24.34 |
• Solving the two equations, we get the points of intersection:
A(−1,1) and B(2,4)
2. We are asked to find the area of the blue region.
• This area is equal to:
$\small{\int_{-1}^2{\left[f(x)~-~g(x) \right]dx}~=~\int_{-1}^2{\left[x^2~-~x~-~2 \right]dx}~=~\frac{9}{2}}$ sq.units
(The reader may write all steps related to the integration process)
Solved example 24.27
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
1. In the fig.24.35 below, the red curve represents
$\small{y=f(x)= \sin x}$
 |
| Fig.24.35 |
• Solving the equation, y = sin x = 0 we get the points of intersection with the x-axis as: 0, π, 2π, 3π, so on . . .
2. In our present case, the boundaries are x = 0 and x = 2π. So we need to find the sum: (magenta + blue)
3. Area of magenta region
$\small{\int_{0}^\pi{\left[f(x) \right]dx}~=~\int_{0}^\pi{\left[\sin x \right]dx}~=~2}$ sq.units
(The reader may write all steps related to the integration process)
4. Area of blue region
$\small{\int_{\pi}^{2\pi}{\left[f(x) \right]dx}~=~\int_{\pi}^{2\pi}{\left[\sin x \right]dx}~=~-2}$
(The reader may write all steps related to the integration process)
• Area cannot be −ve. So we take the absolute value. We can write:
Area of blue region = |−2| = 2 sq.units
5. Based on steps (3) and (4), we get the required area as:
(2 + 2) = 4 sq.units
Solved example 24.28
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Solution:
1. In the fig.24.36 below,
♦ Red curve represents $\small{y~=~f(x)~=~\pm 2 \sqrt{ax}}$
♦ Green line represents $\small{y~=~g(x)~=~mx}$
 |
| Fig.24.36 |
• We need only the x-coordinate of the point of intersection. By solving the two equations, we get the x-coordinate as: $\small{\frac{4a}{m^2}}$
2. We are asked to find the area of the blue region. This area is equal to:
$\small{\int_0^{\frac{4a}{m^2}}{\left[f(x) - g(x) \right]dx}~=~\int_0^{\frac{4a}{m^2}}{\left[2 \sqrt{ax}~-~mx \right]dx}}$
[Here we use $\small{2 \sqrt{ax}}$
Instead of $\small{-2 \sqrt{ax}}$
This is because, in the first quadrant, y values are +ve]
$\small{~=~2 \sqrt{a} \left[\frac{x^{3/2}}{3/2} \right]_0^{\frac{4a}{m^2}}~-~m \left[\frac{x^{2}}{2} \right]_0^{\frac{4a}{m^2}}}$
$\small{~=~4 \sqrt{a} \left[\frac{x^{3/2}}{3} \right]_0^{\frac{4a}{m^2}}~-~m \left[\frac{x^{2}}{2} \right]_0^{\frac{4a}{m^2}}}$
$\small{~=~4 \sqrt{a} \left[\frac{(4a)^{3/2}}{3(m^2)^{3/2}} \right]~-~m \left[\frac{(4a)^{2}}{2(m^2)^2} \right]}$
$\small{~=~ \left[\frac{32 a^2}{3m^3} \right]~-~ \left[\frac{8 a^2}{ m^3} \right]~=~\frac{8 a^2}{3 m^3}}$ sq.units
Solved example 24.29
Area bounded by the curve y = x3, the x-axis and the ordinates x = −2 and x = 1 is:
(A) −9 (B) −15/4 (C) 15/4 (D) 17/4
Solution:
1. In the fig.24.37 below, the red curve represents
$\small{x=f(y)=x^3}$
 |
| Fig.24.37 |
2. We are asked to find the area of (magenta + blue) region.
3. Area of magenta region
$\small{\int_{-2}^0{\left[f(x) \right]dx}~=~\int_{-2}^0{\left[x^3 \right]dx}~=~-4}$
(The reader may write all steps related to the integration process)
• Area cannot be −ve. So we take the absolute value. We can write:
Area of magenta region = |−4| = 4 sq.units
4. Area of blue region
$\small{\int_{0}^1{\left[f(x) \right]dx}~=~\int_{0}^1{\left[x^3 \right]dx}~=~\frac{1}{4}}$ sq.units
(The reader may write all steps related to the integration process)
5. Based on steps (3) and (4), we can write:
Required area = $\small{4+~\frac{1}{4}~=~\frac{17}{4}}$ sq.units
• So the correct option is (D)
The link below gives a few more solved examples:
Miscellaneous Exercise
In the next chapter, we will see differential equations.
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