Definite Integration — Page 23 | JEE Mathematics

Q221

The value of integral

\int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}

is :

A

\pi \sqrt 2

B

\pi \left( {\sqrt 2 - 1} \right)

C

{\pi \over 2}\left( {\sqrt 2 + 1} \right)

D

2\pi \left( {\sqrt 2 - 1} \right)

Correct Answer

Option B

Solution

Let

I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}

also let K =

{x \over {1 + \sin x}}

Multiplying numerator and denominator by (1 $-$ sin x) we get;

K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}

= x(1 $-$ sin x) sec2x = x sec2x $-$ x sin x sec2x = x sec2x $-$ x tan x sec x Now,

I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} }

$\Rightarrow$ I =

\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}

-

\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}}

$\Rightarrow$ I =

\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}

-

\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}

- \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}

$\Rightarrow$ I =

- {{3\pi } \over 4} - {\pi \over 4}

+ {{3\pi } \over 4}\left( {\sqrt 2 } \right)

+ {\pi \over 4}\left( {\sqrt 2 } \right)

- \log \left( {\sqrt 2 + 1} \right)

+ \log \left( {\sqrt 2 + 1} \right)

$\Rightarrow$ I = - $\pi$ +

{\sqrt 2 \pi }

$\Rightarrow$ I =

\pi \left( {\sqrt 2 - 1} \right)

Q222

The value of the integral

\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx

(where [x] denotes the greatest integer less than or equal to x) is

A 0

B 4

C 4

-

sin 4

D sin 4

Correct Answer

Option A

Solution

I $=$

\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx

{\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx}

\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.

as

\left. \begin{array}{ll}\, \\ \, \end{array} x \ne n\pi \right)

{\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0}

Q223

The value of

\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx

is :

A

4 \over 3

B

-

4 \over 3

C 0

D

2 \over 3

Correct Answer

Option A

Solution

\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx

The period of

\left| {\cos x} \right|

=

{\pi \over 2}

$\therefore$ I = 2

\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx

as in the range 0 to

{\pi \over 2}

\left| {\cos x} \right|

is positive. So,

\left| {\cos x} \right|

=

cosx

$\therefore$ I = 2

\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx}

= 2

\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx

I =

{1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}

I =

{1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]

I =

{1 \over 2}\left[ { - {1 \over 3} + 3} \right]

=

{1 \over 2}\left( {{8 \over 3}} \right)

=

{4 \over 3}

Q224

The value of the integral

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

is :

A 0

B

{3 \over 4}

C

{3 \over 8}

\pi

D

{3 \over 16}

\pi

Correct Answer

Option C

Solution

Let I =

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

......(1) $\Rightarrow$ I =

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx

[ As

\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx

] =

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx

=

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx

.......(2) Adding equation (1) and (2) we get, 2I = 2

\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx

2I = 4

\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx

I = 2

\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx

=

{{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}

[ Using Gamma function ] =

{{3\pi } \over 8}

Q225

The value of the integral

\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,

where [x] denotes the greatest integer less than or equal to x, is :

A 6

B 3

C 7

D

{1 \over 3}

Correct Answer

Option B

Solution

Let I =

\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}}

=

\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)

Using,

\int\limits_a^b {f\left( {a + b - x} \right)dx\,}

=

\,\,\int\limits_a^b {f(x)\,\,dx}

I =

\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)

Adding (1) and (2) 2I =

\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx

$\Rightarrow$

\,\,\,

2I =

\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}

= 6 = I = 3

Q226

If

2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,

then

\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx

is equalto :

A log4

B

{\pi \over 2}

+ log2

C log2

D

{\pi \over 2}

-

log4

Correct Answer

Option C

Solution

Given,

2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx

=

\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx

$\Rightarrow$

\,\,\,

2\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx = \int\limits_0^1 {{\pi \over 2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx

$\Rightarrow$

\,\,\,

\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx}

$\Rightarrow$

\,\,\,

\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx}

Let I =

\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx

=

\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx

=

{\pi \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1

=

{\pi \over 4} - {1 \over 2}\log 2

\therefore\,\,\,

\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx

=

{\pi \over 2} - 2\left( {{\pi \over 4} - {1 \over 2}\log 2} \right)

=

{\pi \over 2} - {\pi \over 2} + \log 2

= log2

Q227

If

\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}

for some positive real number a, then a is equal to :

A 7

B 8

C

{{15} \over 2}

D

{{17} \over 2}

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}

$\Rightarrow$

\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}

= {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}

= {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}

$\Rightarrow$

{{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120

$\Rightarrow$ 2a2 + 3a $-$ 119 $=$ 0 $\Rightarrow$ 2a2 + 17a $-$ 14a $-$ 119 $=$ 0 $\Rightarrow$ (a $-$ 7) (2a + 17) $=$ 0 $\Rightarrow$ a $=$ 7, $-$

{{17} \over 2}

Q228

The integral

\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}

equals :

A

{\pi \over {40}}

B

{1 \over {20}}{\tan ^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)

C

{1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)} \right)

D

{1 \over 5}\left( {{\pi \over 4}{{-\tan }^{ - 1}}\left( {{1 \over {3\sqrt 3 }}} \right)} \right)

Correct Answer

Option C

Solution

I $=$

\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}

{\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}}

Put tan5x $=$ t

{\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)

Q229

Let f be a differentiable function from R to R such that

\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},

for all

x,y \in

R. If

f\left( 0 \right) = 1

then

\int\limits_0^1 {{f^2}} \left( x \right)dx

is equal to :

A 1

B 2

C

{1 \over 2}

D 0

Correct Answer

Option A

Solution

\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}

\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}

\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}

\Rightarrow \left| {f'\left( x \right)} \right| \le 0

\Rightarrow f'\left( x \right) = 0

\Rightarrow f\left( x \right) =

constant as

f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1

\int\limits_0^1 {{f^2}} \left( x \right)dx = 1

Q230

If

{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;

{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx

and

{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx;

then

A I2 > I3 > I1

B I2 > I1 > I3

C I3 > I2 > I1

D I3 > I1 > I2

Correct Answer

Option C

Solution

Given,

{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;

{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx

and

{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx

For x

\in

(0, 1) $\Rightarrow$ x > x2 or $-$ x < $-$ x2 and x2 > x3 or $-$ x2 < $-$ x3 $\therefore$

{e^{ - {x^2}}}

<

{e^{ - {x^3}}}

and

{e^{ - x}}

<

{e^{ - {x^2}}}

$\Rightarrow$

{e^{ - x}} < {e^{ - {x^2}}} < {e^{ - {x^3}}}

$\Rightarrow$

{e^{ - {x^3}}} > {e^{ - {x^2}}} > {e^{ - x}}

$\Rightarrow$

{I_3} > {I_2} > {I_1}