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:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 1

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

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
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 2
The number of non-zero rows = 3
∴ P(A) = 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

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:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 3

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

Question 5.
\(\frac { 8x+13 }{\sqrt{4x+7}}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 4

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

Question 6.
\(\frac { 1 }{\sqrt{x+1}+\sqrt{x-1}}\)
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 5

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

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)
Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1 6
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

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

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

Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.1

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