Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.1

Question 1.

Evaluate \(\int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x\)

Solution:

Question 2.

Evaluate \(\int\left(1+x+\frac{x^{2}}{2 !}\right) d x\)

Solution:

Question 3.

Evaluate \(\int \frac{3 x^{3}-2 \sqrt{x}}{x} d x\)

Solution:

Question 4.

Evaluate ∫(3x^{2} – 5)^{2} dx

Solution:

∫(3x^{2} – 5)^{2} dx

= ∫(9x^{4} – 30x^{2} + 25) dx

= 9∫x^{4} dx – 30∫x^{2} dx + 25∫1 dx

= 9(\(\frac{x^{5}}{5}\)) – 30(\(\frac{x^{3}}{3}\)) + 25x + c

= \(\frac{9x^{5}}{5}\) – 10x^{3} + 25x + c.

Question 5.

Evaluate \(\int \frac{1}{x(x-1)} d x\)

Solution:

Question 6.

If f'(x) = x^{2} + 5 and f(0) = -1, then find the value of f(x).

Solution:

By the definition of integral

f(x) = ∫f'(x) dx

= ∫(x^{2} + 5) dx

= ∫x^{2} dx + 5∫1 dx

= \(\frac{x^{3}}{3}\) + 5x + c

Now, f(0) = -1 gives

f(0) = 0 + 0 + c = -1

∴ c = -1

∴ from (1), f(x) = \(\frac{x^{3}}{3}\) + 5x – 1.

Question 7.

If f(x) = 4x^{3} – 3x^{2} + 2x + k, f(0) = -1 and f(1) = 4, find f(x).

Solution:

By the definition of integral

f(x) = ∫f'(x) dx

= ∫(4x^{3} – 3x^{2} + 2x + k) dx

= 4∫x^{3} dx – 3∫x^{2} dx + 2∫x dx + k∫1 dx

= 4(\(\frac{x^{4}}{4}\)) – 3(\(\frac{x^{3}}{3}\)) + 2(\(\frac{x^{2}}{2}\)) + kx + c

∴ f(x) = x^{4} – x^{3} + x^{2} + kx + c

Now, f(0) = 1 gives

f(0) = 0 – 0 + 0 + 0 + c = 1

∴ c = 1

∴ from (1), f(x) = x^{4} – x^{3} + x^{2} + kx + 1

Further f(1) = 4 gives

f(1) = 1 – 1 + 1 + k + 1 = 4

∴ k = 2

∴ from (2), f(x) = x^{4} – x^{3} + x^{2} + 2x + 1.

Question 8.

If f(x) = \(\frac{x^{2}}{2}\) – kx + 1, f(0) = 2 and f(3) = 5, find f(x).

Solution:

By the definition of integral