Referring to the AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals Exercise 8b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Application of Integrals Solutions Exercise 8b
Question 1.
Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x-axis
Solution:
(i) y = x2, x = 1, x = 2 and x-axis
Required Area A = \(\int_1^2 y d x=\int_1^2 x^2 d x=\left[\frac{x^3}{3}\right]_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3} \text { sq. units }\)
(ii) y = x4, x = 1, x = 5 and x-axis
Required Area A =\(\int_1^5 y d x=\int_1^5 x^4 d x=\left[\frac{x^5}{5}\right]_1^5=\frac{(5)^5}{5}-\frac{1}{5}=(5)^4-\frac{1}{5}=625-\frac{1}{5}=624.8 \text { sq. units }\)
Question 2.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^0|x+3| d x\).
Solution:
| X | -6 | -5 | -4 | -3 | -2 | -1 | 0 |
| Y | 3 | 2 | 1 | 0 | 1 | 2 | 3 |
(x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0
∴ \(\int_{-6}^0|(x+3)| d x=-\int_{-6}^{-3}(x+3) d x+\int_{-3}^0(x+3) d x\)
= \(-\left[\frac{x^2}{2}+3 x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3 x\right]_{-3}^0\)
= \(-\left[\left(\frac{(-3)^2}{2}+3(-3)\right)-\left(\frac{(-6)^2}{2}+3(-6)\right)\right]+\left[0-\left(\frac{(-3)^2}{2}+3(-3)\right)\right]=-\left[-\frac{9}{2}\right]-\left[-\frac{9}{2}\right]\) = 9 sq. units
Question 3.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
Area bounded by the sine curve = Area OAB + Area BCD
Area (OAB) + Area(BCD) = \(\int_0^\pi \sin x d x+\left|\int_\pi^{2 \pi} \sin x d x\right|\)
= \([-\cos x]_0^\pi+\left|[-\cos x]_\pi^{2 \pi}\right|=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi|\)
= 1 + 1 + |(-1 – 1)| = 2 + |-2| = 2 + 2 = 4 sq. units
Question 4.
Find the area cut off between the line y = 0 and the parabola y = x2 – 4x + 3
Solution:
Solving the given equations, we have x2 – 4x + 3 = 0
⇒ (x – 1)(x – 3) = 0 ⇒ x = 1, 3
Also the given curve lies below the x-axis
Required area A = \(\int_1^3-y d x=-\int_1^3\left(x^2-4 x+3\right) d x=-\left[\frac{x^3}{3}-\frac{4 x^2}{2}+3 x\right]_1^3\)
= \(\left[9-18+9-\frac{1}{3}+2-3\right]=-\left[-\frac{1}{3}-1\right]=\frac{4}{3} \text { sq. units }\)
Question 5.
Find the area enclosed between the curve y = x2, X-axis and the lines x = -1, x = 2
Solution:
The area bounded by the curve y = x2, the x-axis and the lines x = -1, x = 2 is
A = \(\int_{-1}^2 \mathrm{ydx}=\int_{-1}^2 \mathrm{x}^2 \mathrm{dx}=\left[\frac{\mathrm{x}^3}{3}\right]_{-1}^2=\left(\frac{8}{3}\right)-\left(\frac{-1}{3}\right)=\frac{8}{3}+\frac{1}{3}=\frac{9}{3}\) = 3 sq. units
Question 6.
Find the area bounded between the curve y2 = 2x + 1 and x = 0.
Solution:
Solving y2 – 1 = 2x and x = 0, we get y2 – 1 = 0 ⇒ y= ±1
The parabola y2 – 1 = 2x meets the X-axis at y = 1 and y = -1.
As the curve is symmetric about the X-axis and lies left side to the Y-axis, the area bounded by the curve and the Y-axis is
A = \(2 \int_0^1(-x) d y=2 \int_0^1-\left(\frac{y^2-1}{2}\right) d y=\int_0^1-\left(y^2-1\right) d y=\int_0^1\left(1-y^2\right) d y=\left[y-\frac{y^3}{3}\right]_0^1=1-\frac{1}{3}=\frac{2}{3} \text { sq. units }\)
Question 7.
Find the area enclosed by the line y = 3x and curve y = 6x – x2.
Solution:
The given curves are y = 3x …………(1); y = 6x – x2 …..(2)
Solving (1), (2) we get 3x = 6x – x2 ⇒ x2 – 3x = 0 ⇒ x = 0,3
The upper boundary curve is y = 6x – x2
The lower boundary curve is y = 3x
∴ Required Area A = \(\int_0^3\left(\left(6 x-x^2\right)-3 x\right) d x=\int_0^3\left(3 x-x^2\right) d x=\left[\frac{3 x^2}{2}-\frac{x^3}{3}\right]_0^3=\frac{27}{2}-9=\frac{9}{2} \text { sq.n units }\)
Question 8.
Find the area between curve y = x3 + 3 and lines y = 0, x = -1, x = 2
Solution:
The area bounded by the curve y = x3 + 3, the x-axis and the lines x = -1, x = 2 is
A = \(\int_{-1}^2 y d x=\int_{-1}^2\left(x^3+3\right) d x=\left[\frac{x^4}{4}+3 x\right]_{-1}^2=\left[\left(\frac{16}{4}+6\right)-\left(\frac{1}{4}-3\right)\right]=\frac{40}{4}+\frac{11}{4}=\frac{51}{4} \text { sq. units }\)
Question 9.
Find the area between curve y2 = 3x and line x = 3.
Solution:
Solving x = 3 and y2 = 3x we have y2 = 3(3) = 9 ⇒ y = ±3
∴ Required area A
\(=\int_{-3}^3\left[3-\frac{y^2}{3}\right] d y=2 \int_0^3\left[3-\frac{y^2}{3}\right] d y=2\left[3 y-\frac{y^3}{9}\right]_0^3\) = 2(9 – 3) = 12 sq. units.
Question 10.
Find the area between curve y = x2 and the line y = 2x.
Solution:
The given curves are y = x2 …. (1); y = 2x … (2)
Solving (1), (2) we have x2 = 2x ⇒ x2 – 2x = 0
⇒ x(x – 2) = 0 ⇒ x = 0, 2
The upper boundary curve is y = 2x,
the lower boundary curve is y = x2.
∴ The area enclosed between the curves is
A = \(\int_0^2\left(2 x-x^2\right) d x=\left[2 \cdot \frac{x^2}{2}-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}\)