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#### Question 33:

Evaluate the following integrals as limit of sums:

${\int }_{1}^{3}\left(3{x}^{2}+1\right)dx$                    [CBSE 2014]

#### Answer:

We have,

Here, a = 1, b = 3,  f(x) = 3x2 + 1 and $h=\frac{3-1}{n}=\frac{2}{n}⇒nh=2$

$=\underset{h\to 0}{\mathrm{lim}}\left[4nh+6×\frac{nh\left(nh-h\right)}{2}+3×\frac{\left(nh-h\right)nh\left(2nh-h\right)}{6}\right]\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left[4nh+3×nh\left(nh-h\right)+3×\frac{\left(nh-h\right)nh\left(2nh-h\right)}{6}\right]\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left[4×2+3×2×\left(2-h\right)+3×\frac{\left(2-h\right)×2×\left(2×2-h\right)}{6}\right]\phantom{\rule{0ex}{0ex}}=8+6×\left(2-0\right)+\frac{\left(2-0\right)×2×\left(4-0\right)}{2}\phantom{\rule{0ex}{0ex}}=8+12+8\phantom{\rule{0ex}{0ex}}=28$

4430.

.

#### Answer:

${\int }_{0}^{1}\frac{1}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{tan}}^{-1}x\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}-0\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$ode is 4430.

#### Answer:

${\int }_{0}^{\infty }{e}^{-x}dx\phantom{\rule{0ex}{0ex}}=-{\left[{e}^{-x}\right]}_{0}^{\infty }\phantom{\rule{0ex}{0ex}}=-\left(0-1\right)\phantom{\rule{0ex}{0ex}}=0+1\phantom{\rule{0ex}{0ex}}=1$ 4430.

#### Answer:

${\int }_{0}^{4}\frac{1}{\sqrt{16-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}\frac{1}{\sqrt{{4}^{2}-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{sin}}^{-1}\frac{x}{4}\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\left(\frac{\mathrm{\pi }}{2}-0\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$s 4430.

#### Answer:

${\int }_{0}^{1}\frac{1}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{tan}}^{-1}x\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}1-{\mathrm{tan}}^{-1}0\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}-0\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$

#### Question 22:

Evaluate each of the following integrals:

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{tan}xdx$

#### Answer:

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{tan}xdx\phantom{\rule{0ex}{0ex}}={\overline{)\mathrm{logsec}x}}_{0}^{\frac{\mathrm{\pi }}{4}}\phantom{\rule{0ex}{0ex}}=\mathrm{logsec}\frac{\mathrm{\pi }}{4}-\mathrm{logsec}0\phantom{\rule{0ex}{0ex}}=\mathrm{log}\sqrt{2}-\mathrm{log}1\phantom{\rule{0ex}{0ex}}=\mathrm{log}{2}^{\frac{1}{2}}-0\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{log}2$

#### Question 23:

$\underset{2}{\overset{3}{\int }}\frac{1}{x}dx$

#### Answer:

${\int }_{2}^{3}\frac{1}{x}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{log}}_{e}x\right]}_{2}^{3}\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{e}3-{\mathrm{log}}_{e}2\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{e}\left(\frac{3}{2}\right)$

#### Answer:

${\int }_{0}^{2}\sqrt{4-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{2}\sqrt{{2}^{2}-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\left[\frac{x}{2}\sqrt{4-{x}^{2}}+\frac{1}{2}×{2}^{2}{\mathrm{sin}}^{-1}\frac{x}{2}\right]}_{0}^{2}\phantom{\rule{0ex}{0ex}}={\left[\frac{x}{2}\sqrt{4-{x}^{2}}\right]}_{0}^{2}+2{\left[{\mathrm{sin}}^{-1}\frac{x}{2}\right]}_{0}^{2}\phantom{\rule{0ex}{0ex}}=0+2\left(\frac{\mathrm{\pi }}{2}-0\right)\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }$

4430.

#### Question 26:

Evaluate each of the following  integrals:

${\int }_{0}^{1}x{e}^{{x}^{2}}dx$                 [CBSE 2014]

#### Answer:

$I={\int }_{0}^{1}x{e}^{{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\int }_{0}^{1}{e}^{{x}^{2}}2xdx$

Put ${x}^{2}=z$

$⇒2xdx=dz$

When

When

$\therefore I=\frac{1}{2}{\int }_{0}^{1}{e}^{z}dz\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×{\overline{){e}^{z}}}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(e-{e}^{0}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(e-1\right)$

#### Question 27:

Evaluate each of the following integrals:

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{sin}2xdx$              [CBSE 2014]

#### Answer:

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{sin}2xdx\phantom{\rule{0ex}{0ex}}={\overline{)\frac{-\mathrm{cos}2x}{2}}}_{0}^{\frac{\mathrm{\pi }}{4}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}\left(\mathrm{cos}\frac{\mathrm{\pi }}{2}-\mathrm{cos}0\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}×\left(0-1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

#### Question 28:

Evaluate each of the following integrals:

${\int }_{e}^{{e}^{2}}\frac{1}{x\mathrm{log}x}dx$               [CBSE 2014]

#### Question 29:

Evaluate each of the following integrals:

${\int }_{0}^{\frac{\mathrm{\pi }}{2}}{e}^{x}\left(\mathrm{sin}x-\mathrm{cos}x\right)dx$                 [CBSE 2014]

#### Answer:

Disclaimer: The solution has been provided by taking the lower limit of integral as 0.

#### Question 30:

Solve each of the following integrals:

${\int }_{2}^{4}\frac{x}{{x}^{2}+1}dx$                 [CBSE 2014]

#### Question 31:

If find the value of k.

.

#### Question 32:

If write the value of a.

0.

#### Question 33:

If $f\left(x\right)={\int }_{0}^{x}t\mathrm{sin}tdt$, the write the value of $f\text{'}\left(x\right)$.                       [CBSE 2014]

#### Answer:

$f\left(x\right)={\int }_{0}^{x}t\mathrm{sin}tdt\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)={\overline{)t\left(-\mathrm{cos}t\right)}}_{0}^{x}-{\int }_{0}^{x}\frac{d}{dt}\left(t\right)×\left(-\mathrm{cos}t\right)dt\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=-\left(x\mathrm{cos}x-0\right)+{\int }_{0}^{x}\mathrm{cos}tdt\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=-x\mathrm{cos}x+{\overline{)\mathrm{sin}t}}_{0}^{x}$
$⇒f\left(x\right)=-x\mathrm{cos}x+\left(\mathrm{sin}x-0\right)\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=-x\mathrm{cos}x+\mathrm{sin}x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=-\left[x×\left(-\mathrm{sin}x\right)+\mathrm{cos}x×1\right]+\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-\left(-x\mathrm{sin}x\right)-\mathrm{cos}x+\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=x\mathrm{sin}x$

Thus, the value of $f\text{'}\left(x\right)$ is x sinx.

#### Question 34:

If ${\int }_{0}^{a}\frac{1}{4+{x}^{2}}dx=\frac{\mathrm{\pi }}{8}$, find the value of a.                                   [CBSE 2014]

#### Answer:

$⇒{\mathrm{tan}}^{-1}\frac{a}{2}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{2}=\mathrm{tan}\frac{\mathrm{\pi }}{4}=1\phantom{\rule{0ex}{0ex}}⇒a=2$

Thus, the value of a is 2.

#### Question 35:

Write the coefficient a, b, c of which the value of the integral is independent.

#### Answer:

Hence, the given integral is independent of b

#### Question 38:

where {x} denotes the fractional part of x.

s 4430.

#### Question 44:

If  denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

equals

(a) π/2
(b) π/4
(c) π/6
(d) π/8

#### Answer:

(d) $\mathrm{\pi }$/8

equals

(a) 0
(b) 1/2
(c) 2
(d) 3/2

(c) 2

#### Question 3:

The value of is
(a) $\frac{{\mathrm{\pi }}^{2}}{4}$

(b) $\frac{{\mathrm{\pi }}^{2}}{2}$

(c) $\frac{3{\mathrm{\pi }}^{2}}{2}$

(d) $\frac{{\mathrm{\pi }}^{2}}{3}$

π24

#### Question 4:

The value of $\underset{0}{\overset{2\mathrm{\pi }}{\int }}\sqrt{1+\mathrm{sin}\frac{x}{2}}dx$ is
(a) 0
(b) 2
(c) 8
(d) 4

(c) 8

#### Question 5:

The value of the integral is
(a) 0
(b) π/2
(c) π/4
(d) none of these

(c) π/4

equals

(a) log 2 − 1
(b) log 2
(c) log 4 − 1
(d) − log 2

#### Answer:

(b) log 2

$=\frac{1}{2×\frac{1}{2}}{\left[\mathrm{log}\left|\frac{t+\frac{1}{2}-\frac{1}{2}}{t+\frac{1}{2}+\frac{1}{2}}\right|\right]}_{1}^{\infty }\phantom{\rule{0ex}{0ex}}={\left[\mathrm{log}\left|\frac{t}{t+1}\right|\right]}_{1}^{\infty }\phantom{\rule{0ex}{0ex}}={\left[\mathrm{log}\left|\frac{\frac{t}{t}}{\frac{t}{t}+\frac{1}{t}}\right|\right]}_{1}^{\infty }\phantom{\rule{0ex}{0ex}}={\left[\mathrm{log}\left|\frac{1}{1+\frac{1}{t}}\right|\right]}_{1}^{\infty }\phantom{\rule{0ex}{0ex}}=\mathrm{log}\frac{1}{1+0}-\mathrm{log}\frac{1}{1+1}\phantom{\rule{0ex}{0ex}}=\mathrm{log}\left(1\right)-\mathrm{log}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=0-\left(-\mathrm{log}2\right)\phantom{\rule{0ex}{0ex}}=\mathrm{log}2$

equals
(a) 2
(b) 1
(c) π/4
(d) π2/8

(a) 2

#### Question 8:

equals

(a) $\mathrm{log}\left(\frac{2}{3}\right)$

(b) $\mathrm{log}\left(\frac{3}{2}\right)$

(c) $\mathrm{log}\left(\frac{3}{4}\right)$

(d) $\mathrm{log}\left(\frac{4}{3}\right)$

#### Answer:

(d) $\mathrm{log}\left(\frac{4}{3}\right)$

#### Question 9:

equals

(a) $\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(b) $\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(c)

(d)

3tan1(13)

#### Question 10:

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\sqrt{\frac{1-x}{1+x}}dx=$

(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{\mathrm{\pi }}{2}-1$

(c) $\frac{\mathrm{\pi }}{2}+1$

(d) π + 1

#### Answer:

Disclaimer: None of the given option is correct.

#### Question 11:

(a) $\frac{\mathrm{\pi }}{\sqrt{{a}^{2}-{b}^{2}}}$

(b) $\frac{\mathrm{\pi }}{ab}$

(c) $\frac{\mathrm{\pi }}{{a}^{2}+{b}^{2}}$

(d) (a + b) π

#### Answer:

$={\int }_{0}^{\mathrm{\pi }}\frac{1+{\mathrm{tan}}^{2}\frac{x}{2}}{a\left(1+{\mathrm{tan}}^{2}\frac{x}{2}\right)+b\left(1-{\mathrm{tan}}^{2}\frac{x}{2}\right)}\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\mathrm{\pi }}\frac{1+{\mathrm{tan}}^{2}\frac{x}{2}}{\left(a+b\right)+\left(a-b\right){\mathrm{tan}}^{2}\frac{x}{2}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\mathrm{\pi }}\frac{{\mathrm{sec}}^{2}\frac{x}{2}}{\left(a+b\right)+\left(a-b\right){\mathrm{tan}}^{2}\frac{x}{2}}dx$

$=\frac{2}{\left(a-b\right)}{\int }_{0}^{\infty }\frac{1}{{\left(\sqrt{\frac{a+b}{a-b}}\right)}^{2}+{t}^{2}}dt\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(a-b\right)}×\sqrt{\frac{a-b}{a+b}}{\left[{\mathrm{tan}}^{-1}\frac{t}{\sqrt{\frac{a+b}{a-b}}}\right]}_{0}^{\infty }\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{{a}^{2}-{b}^{2}}}\left[\frac{\mathrm{\pi }}{2}-0\right]\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{{a}^{2}-{b}^{2}}}\left[\frac{\mathrm{\pi }}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{\sqrt{{a}^{2}-{b}^{2}}}$

is

(a) π/3
(b) π/6
(c) π/12
(d) π/2

#### Question 13:

Given that the value of $\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left({x}^{2}+4\right)\left({x}^{2}+9\right)},$ is

(a) $\frac{\mathrm{\pi }}{60}$

(b) $\frac{\mathrm{\pi }}{20}$

(c) $\frac{\mathrm{\pi }}{40}$

(d) $\frac{\mathrm{\pi }}{80}$

(a) 1
(b) e − 1
(c) e + 1
(d) 0

(a) 1

#### Question 15:

is equal to

(a) $\frac{\mathrm{\pi }}{12}$

(b) $\frac{\mathrm{\pi }}{6}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) $\frac{\mathrm{\pi }}{3}$

#### Answer:

(a) $\frac{\mathrm{\pi }}{12}$

${\int }_{1}^{\sqrt{3}}\frac{1}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{tan}}^{-1}x\right]}_{1}^{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{12}$
30.

#### Question 16:

(a) $\frac{\mathrm{\pi }}{12}+\mathrm{log}\left(2\sqrt{2}\right)$

(b) $\frac{\mathrm{\pi }}{2}+\mathrm{log}\left(2\sqrt{2}\right)$

(c) $\frac{\mathrm{\pi }}{6}+\mathrm{log}\left(2\sqrt{2}\right)$

(d) $\frac{\mathrm{\pi }}{3}+\mathrm{log}\left(2\sqrt{2}\right)$

#### Question 17:

The value of the integral is

(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{6}$

(d) $\frac{\mathrm{\pi }}{3}$

is equal to

(a) 1
(b) 2
(c) − 1
(d) − 2

(b) 2

is equal to

(a)

(b)

(c)

(d) π

(a)

The value of is
(a) 1
(b) e − 1
(c) 0
(d) − 1

(b) e − 1

#### Question 21:

If then a equals

(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{1}{2}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) 1

#### Answer:

(b) $\frac{1}{2}$

4430.

#### Question 22:

If equals

(a) 4a2
(b) 0
(c) 2a2
(d) none of these

(b) 0

#### Question 23:

The value of is

(a) $\frac{{\mathrm{\pi }}^{4}}{2}$

(b) $\frac{{\mathrm{\pi }}^{4}}{4}$

(c) 0

(d) none of these

(c) 0

#### Question 24:

is equal to

(a) loge 3

(b) ${\mathrm{log}}_{e}\sqrt{3}$

(c) $\frac{1}{2}\mathrm{log}\left(-1\right)$

(d) log (−1)

#### Answer:

(b) ${\mathrm{log}}_{e}\sqrt{3}$

is equal to

(a) −2
(b) 2
(c) 0
(d) 4

(b) 2

#### Question 26:

The derivative of is

(a)

(b)

(c) (ln x)−1 x (x − 1)

(d)

#### Answer:

(c) (ln x)−1 x (x − 1)

Using Newton Leibnitz formula

$\begin{array}{c}{f}^{\text{'}}\left(x\right)=\frac{1}{{\mathrm{log}}_{e}{x}^{3}}\left(3{x}^{2}\right)-\frac{1}{{\mathrm{log}}_{e}{x}^{2}}\left(2x\right)\\ =\frac{3{x}^{2}}{3\mathrm{ln}x}-\frac{2x}{2\mathrm{ln}x}\\ =\frac{{x}^{2}}{\mathrm{ln}x}-\frac{x}{\mathrm{ln}x}\\ =\frac{1}{\mathrm{ln}x}x\left(x-1\right)\\ ={\left(\mathrm{ln}x\right)}^{-1}x\left(x-1\right)\end{array}$

#### Question 27:

If then the value of I10 + 90I8 is

(a) $9{\left(\frac{\mathrm{\pi }}{2}\right)}^{9}$

(b) $10{\left(\frac{\mathrm{\pi }}{2}\right)}^{9}$

(c) ${\left(\frac{\mathrm{\pi }}{2}\right)}^{9}$

(d) $9{\left(\frac{\mathrm{\pi }}{2}\right)}^{8}$

#### Question 28:

(a) $\frac{15}{16}$

(b) $\frac{3}{16}$

(c) $-\frac{3}{16}$

(d) $-\frac{16}{3}$

#### Answer:

Disclaimer: The question given is not correct because the function provided does not converge in the given domain.

is equal to

(a)

(b)

(c)

(d)

(c) ln(3/2)

#### Question 30:

The value of the integral is
(a) 4
(b) 2
(c) −2
(d) 0

(a) 4

is equal to

(a) 0
(b) 1
(c) π/2
(d) π/4

(d) π/4

equals to

(a) π
(b) π/2
(c) π/3
(d) π/4

(d) π/4

is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

(c) π/2

is equal to

(a) π/4
(b) π/2
(c) π
(d) 1

(d) 1

is equal to

(a) π
(b) π/2
(c) 0
(d) 2π

(c) 0

The value of is

(a) π/4
(b) π/8
(c) π/2
(d) 0

(a) π/4

(a) π ln 2
(b) −π ln 2
(c) 0
(d)

#### Answer:

(a) π ln 2

Substitute x = tan θ
dx = sec2 θ dθ.
when,
x = 0  ⇒ θ = 0

Let us consider,

is equal to

(a)

(b) 0

(c)

(d)

#### Question 39:

If f (a + bx) = f (x), then $\underset{a}{\overset{b}{\int }}$ x f (x) dx is equal to

(a)

(b)

(c)

(d)

(d)

The value of is

(a) 1
(b) 0
(c) −1
(d) π/4

(b) 0

#### Question 41:

The value of is

(a) 2

(b) $\frac{3}{4}$

(c) 0

(d) −2

(c) 0

The value of is

(a) 0
(b) 2
(c) π
(d) 1

#### Answer:

(c) $\mathrm{\pi }$

#### Answer:

I =

using partial fraction,

putting the values of A,B and C we get

#### Answer:

${\int }_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\sqrt{\frac{1-x}{1+x}×\frac{1-x}{1-x}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\frac{1-x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\frac{1}{\sqrt{1-{x}^{2}}}dx-{\int }_{0}^{1}\frac{x}{\sqrt{1-{x}^{2}}}dx\phantom{\rule{0ex}{0ex}}={\left[{\mathrm{sin}}^{-1}x\right]}_{0}^{1}+{\left[\sqrt{1-{x}^{2}}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}-1$

#### Answer:

${\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1+\mathrm{cos}x}}{{\left(1-\mathrm{cos}x\right)}^{\frac{5}{2}}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1+\mathrm{cos}x}}{{\left(1-\mathrm{cos}x\right)}^{\frac{5}{2}}}×\frac{\sqrt{1-\mathrm{cos}x}}{\sqrt{1-\mathrm{cos}x}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sin}x}{{\left(1-\mathrm{cos}x\right)}^{3}}dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\left[{\left(1-\mathrm{cos}x\right)}^{-2}\right]}_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}\left[1-4\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$

#### Question 23:

Evaluate the following integrals:

${\int }_{2}^{4}\frac{{x}^{2}+x}{\sqrt{2x+1}}dx$

#### Answer:

Let I = ${\int }_{2}^{4}\frac{{x}^{2}+x}{\sqrt{2x+1}}dx$

Put 2x + 1 = z2

$⇒2dx=2zdz\phantom{\rule{0ex}{0ex}}⇒dx=zdz$

When

When

$\therefore I={\int }_{\sqrt{5}}^{3}\frac{{\left(\frac{{z}^{2}-1}{2}\right)}^{2}+\frac{{z}^{2}-1}{2}}{z}×zdz\phantom{\rule{0ex}{0ex}}⇒I={\int }_{\sqrt{5}}^{3}\frac{{z}^{4}-2{z}^{2}+1+2{z}^{2}-2}{4}dz\phantom{\rule{0ex}{0ex}}⇒I=\frac{1}{4}{\int }_{\sqrt{5}}^{3}\left({z}^{4}-1\right)dz\phantom{\rule{0ex}{0ex}}⇒I=\frac{1}{4}×{\overline{)\left(\frac{{z}^{5}}{5}-z\right)}}_{\sqrt{5}}^{3}$
$⇒I=\frac{1}{4}\left[\left(\frac{243}{5}-3\right)-\left(\frac{25\sqrt{5}}{5}-\sqrt{5}\right)\right]\phantom{\rule{0ex}{0ex}}⇒I=\frac{1}{4}×\frac{228}{5}-\frac{1}{4}×4\sqrt{5}\phantom{\rule{0ex}{0ex}}⇒I=\frac{57}{5}-\sqrt{5}$

ode is 4430.

#### Answer:

${\int }_{0}^{4}xdx\phantom{\rule{0ex}{0ex}}={\left[\frac{{x}^{2}}{2}\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=8-0\phantom{\rule{0ex}{0ex}}=8$

#### Answer:

${\int }_{0}^{2}\left(2{x}^{2}+3\right)dx\phantom{\rule{0ex}{0ex}}={\left[\frac{2{x}^{3}}{3}+3x\right]}_{0}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{3}+6=\frac{34}{3}$

#### Question 26:

Evaluate the following definite integrals:

${\int }_{0}^{\frac{\mathrm{\pi }}{2}}{x}^{2}\mathrm{sin}xdx$                         [CBSE 2014]

#### Answer:

Applying integration by parts, we have

Again applying integration by parts, we have

$I=0+2\left[x\mathrm{sin}x{|}_{0}^{\frac{\mathrm{\pi }}{2}}-{\int }_{0}^{\frac{\mathrm{\pi }}{2}}1×\mathrm{sin}xdx\right]\phantom{\rule{0ex}{0ex}}⇒I=2\left(\frac{\mathrm{\pi }}{2}\mathrm{sin}\frac{\mathrm{\pi }}{2}-0\right)-2{\int }_{0}^{\frac{\mathrm{\pi }}{2}}\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}⇒I=2\left(\frac{\mathrm{\pi }}{2}-0\right)-2\left(-\mathrm{cos}x\right){|}_{0}^{\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒I=\mathrm{\pi }+2\left(\mathrm{cos}\frac{\mathrm{\pi }}{2}-\mathrm{cos}0\right)\phantom{\rule{0ex}{0ex}}⇒I=\mathrm{\pi }+2\left(0-1\right)\phantom{\rule{0ex}{0ex}}⇒I=\mathrm{\pi }-2$

#### Answer:

#### Answer:

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.