Inter 2nd Year Maths Exercise 8b Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 8 Application of Integrals Solutions Exercise 8b will help students to clear their doubts quickly.

Inter 2nd Year Maths Application of Integrals Solutions Exercise 8b

Application of Integrals Exercise 8b Solutions

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) Given curve y = x² and lines x = 1, x = 2
⇒ x² = y
Area of ABCD w.r.to the x-axis is given by

Area = \(\int_1^2 \mathrm{y} \mathrm{dx}=\int_1^2 \mathrm{x}^2 \mathrm{dx}=\left[\frac{\mathrm{x}^3}{3}\right]_1^2\)
= \(\frac{2^3}{3}-\frac{1^3}{3}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\) sq. units

(ii) Given curve y = x⁴
x = 1, x = 5 and X-axis

Region of ABCD w.r.to the x-axis is
Area = \(\int_1^5 y d x\)
= \(\int_1^5 x^4 d x\)
= \(\left[\frac{x^5}{5}\right]_1^5\)
= \(\frac{5^5}{5}-\frac{1^5}{5}\)
= \(\frac{3125-1}{5}\)
= 624.8 sq. units

Question 2.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^0|x+3| d x\).
Solution:
Given \(\int_{-6}^0|x+3| d x\)
f(x) = |x + 3|
x + 3 if x + 3 ≥ 0 ⇒ x ≥ -3
-(x + 3) if x + 3 < o ⇒ x ≤ -3

Area of ABC + Area of ADE with respect to X-axis = \(\int_{-6}^{-3} y d x+\int_{-3}^0 y d x\)
= \(\int_{-6}^{-3}|x+3| d x+\int_{-3}^0|x+3| d x\)
= \(\int_{-6}^{-3}-(x+3) d x+\int_{-3}^0(x+3) d x\)
= \(\left[\frac{x^2}{2}+3 x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3 x\right]_{-3}^0\)
= \(-\left[\left(\frac{9}{2}-9\right)-(18-18)\right]+\left[0-\left(\frac{9}{2}-9\right)\right]\)
= \(\frac{9}{2}+\frac{9}{2}\)
= 9 sq. units.

Question 3.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
The given curve is y = sin x

Area = Area of OAB + Area of BCD = 2 Area of OAB
= \(2 \int_0^\pi \sin x d x\)
= \(2[-\cos x]_0^\pi\)
= 2[-cos π + cos 0]
= 2(1 + 1)
= 4
∴ Total Area from 0 to 2π = 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:

Given curve is y = x² – 4x + 3, y = 0
⇒ x² – 4x + 3 = 0
⇒ x² – 3x – x + 3 = 0
⇒ (x – 3) (x – 1) = 0
⇒ x = 1, x = 3

∴ The Area = \(\int_1^3-(y) d x\)
= \(\int_1^3-\left(x^2-4 x+3\right) d x\)
= \(-\left[\frac{x^3}{3}-4 \frac{x^2}{2}+3 x\right]_1^3\)
= \(\left[\left(\frac{3^3}{3}-4 \frac{3^2}{2}+3(3)\right)-\left(\frac{1}{3}-\frac{4}{2}+3\right)\right]\)
= \(-\left[(9-18+9)-\left(\frac{1}{3}-2+3\right)\right]\)
= \(-\left[0-\left(\frac{1}{3}+1\right)\right]\)
= \(-\left[-\left(\frac{1+3}{3}\right)\right]\)
= \(\frac {4}{3}\) sq.units

Question 5.
Find the area bounded by the parabola y = x², the x-axis, and the lines x = -1, x = 2.
Solution:
Given y = x², x-axis x = -1, x = 2

∴ The required area A = \(\int_{-1}^2 x^2 d x\)
= \(\frac{1}{3}\left[x^3\right]_{-1}^2\)
= \(\frac {1}{2}\)[8 + 1]
= 3 sq. units

Question 6.
Find the area bounded between the curves y² = 2x + 1 and x = 0
Solution:
Given y² = 2x + 1 and x = 0.
⇒ x = \(\frac{y^2-1}{2}\) ……(1)

The required area = \(-\int_{-1}^1 x d y\)
= \(-\int_{-1}^1 \frac{y^2-1}{2} d y\)
= \(-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-1}^1+\frac{1}{2}[y]_{-1}^1\)
= \(-\frac {1}{6}\)(1 + 1) + \(\frac {1}{2}\)(2)
= \(\frac {2}{3}\) sq.units

Question 7.
Find the area enclosed by the line y = 3x and the curve y = 6x – x²
Solution:

Given curves are y = 3x …….(1)
y = 6x – x² ………(2)
Solve equations (1) and (2)
3x = 6x – x²
⇒ x(x – 3) = 0
⇒ x = 0 and x = 3
y = 3x

y = 6x – x²

∴ The required area = \(\int_0^3\left[\left(6 x-x^2\right)-3 x\right] d x\)
= \(\int_0^3\left(3 x-x^2\right) d x\)
= \(\left[\frac{3 x^2}{2}-\frac{x^3}{3}\right]_0^3\)
= \(\frac{27}{2}-\frac{27}{3}\)
= \(\frac {9}{2}\) sq.units.

Question 8.
Find the area between curve y = x³ + 3 and lines y = 0, x = -1, x = 2.
Solution:
Given curves are y = x³ + 3, y = 0, x = -1, x = 2
The bounded area = \(\int_{-1}^2 y d x\)
= \(\int_{-1}^2\left(x^3+3\right) d x\)
= \(\left[\frac{x^4}{4}+3 x\right]_{-1}^2\)
= \(\left[\left(\frac{2^4}{4}+3(2)\right)-\left(\frac{(-1)^4}{4}+3(-1)\right)\right]\)
= \(\left(\frac{16}{4}+6\right)-\left(\frac{1}{4}-3\right)\)
= 10 – \(\left(\frac{1-12}{4}\right)\)
= 10 + \(\frac {11}{4}\)
= \(\frac {51}{4}\) sq.units

Question 9.
Find the area between the curve y² = 3x and the line x = 3.
Solution:
Given y² = 3x and x = 3
⇒ y = ±3
∴ y = 3 and y = -3
Required area = \(\int_{-3}^3 y d x\)
= \(2 \int_0^3 \sqrt{3} \sqrt{x} d x\)
= \(2 \sqrt{3}\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^3\)
= \(\frac{4 \sqrt{3}}{3} 3 \sqrt{3}\)
= 12 sq. units

Question 10.
Find the area between the curve y = x² and the line y = 2x.
Solution:
Given y = x² ……..(1)
y = 2x …….(2)
From (1) and (2)
x² = 2x
∴ x = 0, x = 2

The required area = \(\int_0^2\left(2 x-x^2\right) d x\)
= \(\left(x^2-\frac{x^3}{3}\right)_0^2\)
= \(2^2-\frac{2^3}{3}\)
= 4 – \(\frac {8}{3}\)
= \(\frac {4}{3}\) sq.units