Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.1

Question 1.
Integrate the following with respect to x.
$$\sqrt { 3x+5 }$$
Solution:

Question 2.
(9x² – $$\frac { 4 }{x^2}$$)²
Solution:
A = $$\left[\begin{array}{cccc} -2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7 \end{array}\right]$$
The order of A is 3 × 4
∴ P(A) < 3
Let us transform the matrix A to an echelon form

The number of non-zero rows = 3
∴ P(A) = 3

Question 3.
(3 + x) (2 – 5x)
Solution:
∫(3 + x) (2 – 5x) dx
= ∫(6 – 15x + 2x – 5x²) dx
= ∫(6 – 13x – 5x²) dx
= 6x – $$\frac { 13x^2 }{2}$$ – $$\frac { 5x^3 }{3}$$ + c

Question 4.
√x (x³ – 2x + 3)
Solution:

Question 5.
$$\frac { 8x+13 }{\sqrt{4x+7}}$$
Solution:

Question 6.
$$\frac { 1 }{\sqrt{x+1}+\sqrt{x-1}}$$
Solution:

Question 7.
Given f'(x) = x + b, f(1) = 5 and f(2) = 13, then find f(x)
Solution:
f(x) = ∫f'(x) dx
f(x) = ∫(x + b) dx = $$\frac { x^2 }{2}$$ + bx + c
f (1) = 5 ⇒ $$\frac { (1)^2 }{2}$$ + b(1) + c = 5
$$\frac { 1 }{2}$$ + b + c = 5 ⇒ b + c = 5 – 1/2
b + c = $$\frac { 9 }{2}$$ ⇒ 2b + 2c = 9 …….. (1)
f(2) = 13 ⇒ $$\frac { (2)^2 }{2}$$ + b(2) + c = 13
2 + 2b + c = 13
2b + c = 11 ……… (2)
Solving eqn (1) & (2)

Substitute c = -2 in eqn (2)
2b – 2 = 11 ⇒ 2b = 11 + 2
b = 13/2
f(x) = $$\frac { x^2 }{2}$$ + $$\frac { 13x }{2}$$ – 2

Question 8.
Given f'(x) = 8x³ – 2x and f (2) = 8, then find f(x)
Solution:
f(x) = ∫f'(x)dx
= ∫(8x³ – 2x)dx = 8($$\frac { x^4 }{4}$$) -2($$\frac { x^2 }{2}$$) + c
f(x) = 2x4 – x² + c
f (2) = 8 ⇒ 2(2)4 – (2)² + c = 8
32 – 4 + c = 8 ⇒ c = -20
∴ f (x) = 2x4 – x² – 20