Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.8 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

Using second fundamental theorem, evaluate the following:

Question 1.

\(\int_{0}^{1}\) e^{2x} dx

Solution:

Question 2.

\(\int_{0}^{1/4}\) \(\sqrt { 1 -4x}\) dx

Solution:

Question 3.

\(\int_{0}^{1}\) \(\frac { xdx }{x^2+1}\)

Solution:

Question 4.

\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)

Solution:

\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)

= {log |1 + ex|}\(_{0}^{3}\)

= log |1 + e³| – log |1 + e°|

= log |1 + e³| – log |1 + 1|

= log |1 + e³| – log |2|

= log |\(\frac { 1+e^3 }{2}\)|

Question 5.

\(\int_{0}^{1}\) xe^{x²} dx

Solution:

Question 6.

\(\int_{1}^{e}\) \(\frac { dx }{x(1+logx)^3}\)

Solution:

Question 7.

\(\int_{-1}^{1}\) \(\frac { 2x+3 }{x^2+3x+7}\) dx

Solution:

Question 8.

\(\int_{0}^{π/2}\) \(\sqrt { 1 +cosx} \) dx

Solution:

Question 9.

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

Solution:

Evaluate the following

Question 10.

\(\int_{1}^{4}\) f(x) dx where f(x) = \(\left\{\begin{array}{l}

4 x+3,1 \leq x \leq 2 \\

3 x+5,2 \end{array}\right.\)

Solution:

\(\int_{1}^{4}\) f(x) dx

= \(\int_{1}^{2}\) f(x) dx + \(\int_{2}^{4}\) f(x) dx

= \(\int_{1}^{2}\) (4x + 3) dx + \(\int_{2}^{4}\) (3x + 5) dx

(8 + 6) – [5] + [24 + 20] – [6 + 10]

= 14 – 5 + 44 – 16

= 58 – 21

= 37

Question 11.

\(\int_{0}^{2}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}

3-2 x-x^{2}, & x \leq 1 \\

x^{2}+2 x-3, & 1<x \leq 2

\end{array}\right.\)

Solution:

\(\int_{0}^{2}\) f(x) dx

= \(\int_{0}^{1}\) f(x) dx + \(\int_{1}^{2}\) f(x) dx

= \(\int_{0}^{1}\) (3 – 2x – x²) dx + \(\int_{1}^{2}\) (x² + 2x – 3) dx

Question 12.

\(\int_{-1}^{1}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}

x, & x \geq 0 \\

-x, & x<0

\end{array}\right.\)

Solution:

\(\int_{-1}^{1}\) f(x) dx

\(\int_{-1}^{0}\) f(x) dx + \(\int_{0}^{1}\) f(x) dx

= \(\int_{-1}^{0}\) (-x) dx + \(\int_{0}^{1}\) x dx

Question 13.

f(x) = \(\left\{\begin{array}{l}

c x, \quad 0<x<1 \\

0, \text { otherwise }

\end{array}\right.\) find ‘c’ if \(\int_{0}^{1}\) f(x) dx = 2

Solution:

Given

f(x) = \(\left\{\begin{array}{l}

c x, \quad 0<x<1 \\

0, \text { otherwise }

\end{array}\right.\)

⇒ \(\int_{0}^{1}\) f(x) dx = 2

⇒ \(\int_{0}^{1}\) cx dx = 2

c[ \(\frac { x^2 }{ 2 }\) ]\(_{0}^{1}\) = 2

c[ \(\frac { 1 }{ 2 }\) – 0 ] = 2

\(\frac { 1 }{ 2 }\) = 2

⇒ c = 4