Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.1

Evaluate the following definite integrals:

Question 1.

\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)

Solution:

Question 2.

\(\int_{-2}^{3} \frac{1}{x+5} d x\)

Solution:

Question 3.

\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)

Solution:

Question 4.

\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)

Solution:

Question 5.

\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)

Solution:

Let I = \(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)

Let \(\frac{x}{(x+2)(x+3)}=\frac{A}{x+3}+\frac{B}{x+2}\)

∴ x = A(x + 2) + B(x + 3)

Put x + 3 = 0, i.e. x = -3, we get

-3 = A(-1) + B(0)

∴ A = 3

Put x + 2 = 0, i.e. x = -2, we get

-2 = A(0) + B(1)

∴ B = -2

Question 6.

\(\int_{1}^{2} \frac{d x}{x^{2}+6 x+5}\)

Solution:

Question 7.

If \(\int_{0}^{a}(2 x+1) d x\) = 2, find the real values of ‘a’.

Solution:

Let I = \(\int_{0}^{a}(2 x+1) d x\)

= \(\left[2 \cdot \frac{x^{2}}{2}+x\right]_{0}^{a}\)

= a^{2} + a – 0

= a^{2} + a

∴ I = 2 gives a^{2} + a = 2

∴ a^{2} + a – 2 = 0

∴ (a + 2)(a – 1) = 0 1

∴ a + 2 = 0 or a – 1 = 0

∴ a = -2 or a = 1.

Question 8.

If \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\) = 11, find ‘a’.

Solution:

Let I = \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\)

= \(\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{a}\)

= \(\left[x^{3}+x^{2}+x\right]_{1}^{a}\)

= (a^{3} + a^{2} + a) – (1 + 1 + 1)

= a^{3} + a^{2} + a – 3

∴ I = 11 gives a^{3} + a^{2} + a – 3 = 11

∴ a^{3} + a^{2} + a – 14 = 0

∴ (a^{3} – 8) + (a^{2} + a – 6) = 0

∴ (a – 2)(a^{2} + 2a + 4) + (a + 3)(a – 2) = 0

∴ (a – 2)(a^{2} + 2a + 4 + a + 3) = 0

∴ (a – 2)(a^{2} + 3a + 7) = 0

∴ a – 2 = 0 or a^{2} + 3a + 7 = 0

∴ a = 2 or a = \(\frac{-3 \pm \sqrt{9-28}}{2}\)

The latter two roots are not real.

∴ they are rejected.

∴ a = 2.

Question 9.

\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)

Solution:

Question 10.

\(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\)

Solution:

Let I = \(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\) = \(\int_{1}^{2} \frac{3 x}{(3 x)^{2}-1} d x\)

Put 3x = t

∴ 3 dx = dt

∴ dx = \(\frac{d t}{3}\)

When x = 1, t = 3 × 1 = 3

When x = 2, t = 3 × 2 = 6

Question 11.

\(\int_{1}^{3} \log x d x\)

Solution: