Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b)

I. Evaluate the following definite integrals.

Question 1.
$$\int_0^a(a^2x-x^3) d x$$
Solution: Question 2.
$$\int_2^3 \frac{2 x}{1+x^2} d x$$
Solution:
I = [ln|1 + x²|]³2
= ln 10 – ln 5
= ln(10/5)
= ln 2 Question 3.
$$\int_0^\pi \sqrt{2+2 \cos \theta} d \theta$$
Solution: Question 4.
$$\int_0^\pi\sin^3x\cos^3xd x$$
Solution: Question 5.
$$\int_0^2|1-x|d x$$
Solution: Question 6.
$$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x$$
Solution:
$$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x dx}{1+e^x}$$ ………….. (i)
cos x is even function
ex is neither even nor odd. Adding (i) and (ii) we get Question 7.
$$\int_0^1\frac{dx}{\sqrt{3-2x}}$$
Solution:
3 – 2x = t²
-2dx = 2t dt
dx = -t dt
3 – (2.1) = t²
1 = t²
3 – 2.0 = t²

Question 8.
$$\int_0^a(\sqrt{a}-\sqrt{x})^2$$
Solution: Question 9.
$$\int_0^{\pi / 4} \sec^4\theta d\theta$$
Solution:
Let ∫sec4 θ dθ = $$\int_0^{\pi / 4} \sec^2\theta(1+\tan^2\theta)d\theta$$
Put tan θ = y
sec² θ dθ = dy
θ = $$\frac{\pi}{4}$$ ⇒ y= 1
θ = 0 ⇒ y = 0
I = $$\int_0^a$$(1 + y²)dy = [y + $$\frac{y^3}{3}$$]¹0
= 1 + $$\frac{1}{3}=\frac{4}{3}$$ Question 10.
$$\int_0^3\frac{x}{\sqrt{x^2+16}}$$
Solution: Question 11.
I = $$\int_0^1$$x.e-x²dx
Solution:
I = $$\int_0^a$$x.e-x²dx
⇒ -x² = t
⇒ -2x dx = dt
2x dx = -dt
x = 1 ⇒ t = 0
x = 0 ⇒ t = 1
I = $$\frac{1}{2}$$$$\int_0^a$$-et dt
= $$\frac{1}{2}$$[-et]-10
= $$\frac{1}{2}$$[e0 – e-1]
= $$\frac{1}{2}$$(1 – $$\frac{1}{e}$$)

Question 12.
I = $$\int_1^5\frac{dx}{\sqrt{2x-1}}$$
Solution: II. Evaluate the following integrals:

Question 1.
I = $$\int_0^4\frac{x^2}{1+x}$$
Solution: Question 2.
$$\int_{-1}^2 \frac{x^2}{x^2+2}$$
Solution: Question 3.
I = $$\int_0^1 \frac{x^2}{x^2+2}$$
Solution: Question 4.
$$\int_0^{\pi / 2}$$x² sin x dx
Solution: Question 5.
$$\int_0^4$$|2-x|dx
Solution: Question 6.
$$\int_0^{\pi / 2}\frac{\sin^5 x}{\sin^5 x+\cos^5 x}$$
Solution:   Question 7.
$$\int_0^{\pi / 2}\frac{\sin^2 x-\cos^2 x}{\sin^3 x+\cos^3 x}$$dx
Solution: Question 8. Solution: Question 9. Solution: Question 10. Solution:  Question 11. Solution: Question 12. Solution:  Question 13. Solution: Question 14. Solution: Question 15. Solution: III. Evaluate the following integrals :

Question 1.
$$\int_0^{\pi / 2}\frac{dx}{4+5\cos x}$$
Solution: Question 2.
$$\int_a^b\sqrt{(x-a)(b-x)}$$dx
Solution: Question 3. Solution:  Question 4. Solution: Question 5. Solution:  Question 6.
$$\int_0^a$$x(a – x)ndx
Solution:
I = $$\int_0^a$$x(a – x)ndx ………… (i)
$$\int_0^a$$f(x) dx = $$\int_0^a$$f(a – x)dx
I = $$\int_0^a$$(a – x).(x)ndx ………… (ii)
I = $$\int_0^a$$axn dx – xn+1 dx Question 7.
$$\int_0^2x\sqrt{2-x}$$dx
Solution:
I = $$\int_0^2x.\sqrt{2-x}$$dx
$$\int_0^2$$f(x)dx = $$\int_0^2$$f(a – x)dx
= $$\int_0^2$$(2 – x).√x dx
= $$\int_0^2$$((2√x – x√x)) dx Question 8.
$$\int_0^{\pi}$$x sin³ x dx
Solution:   Question 9.
$$\int_0^{\pi}\frac{x}{1+\sin x}$$dx
Solution:
$$\int_0^{\pi}\frac{x}{1+\sin x}$$dx …………… (i) Question 10. Solution:  Question 11.
$$\int_0^1\frac{log(1+x)}{1+x^2}$$dx
Solution:
Put x = tan θ
dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = x ⇒ θ = $$\frac{\pi}{4}$$  Question 12.
$$\int_0^{\pi}\frac{x \sin x}{1+\cos^2 x}$$dx
Solution:  Question 13.
$$\int_0^{\pi/2}\frac{\sin^2 x}{\cos x+\sin x}$$dx
Solution:  Question 14.
$$\int_0^{\pi}\frac{1}{3+2\cos x}$$dx
Solution: Question 15.
$$\int_0^{\pi/4}$$log(1 + tan x)dx
Solution: Question 16.
$$\int_{-1}^{3/2}$$|x sin πx| dx
Solution:
We know that |x. sin πx| = x . sin πx
where -1 ≤ x ≤ 1
and |x . sin πx| = – x.sin πx where 1 < x ≤ 3/2  Question 17.
$$\int_0^1sin^{-1}\frac{2x}{1+x^2}$$dx
Solution:
$$\int_0^1sin^{-1}\frac{2x}{1+x^2}$$dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = 1 ⇒ θ π/4 Question 18.
$$\int_0^1$$x tan-1x dx
Solution:
$$\int_0^1$$x. tan-1x dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0; Question 19.
$$\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}$$dx
Solution:  Question 20.
Suppose that f : R → R is continuous periodic function and T is the period of it. Let a ∈ R. Then prove that for any positive integer n, Solution: 