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Definite Integration

JEE Mathematics · 230 questions · Page 3 of 23 · Click an option or "Show Solution" to reveal answer

Q21
The value of integral, ∫36x9−x+xdx\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx is
A 12{1 \over 2}
B 32{3 \over 2}
C 22
D 11
Correct Answer
Option B
Solution
I=∫36x9−x+xdx         ...(1)I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)
I=∫369−x9−x+xdx         ...(2)I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)
[ \left[ \, \right.

using

∫abf(x)dx=∫abf(a+b−x)dx\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} }
 ]\left. \, \right]

Adding equation

(1)(1)

and

(2)(2)
2I=∫36dx=3⇒I=322I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}
Q22
The value of ∫1a[x]f′(x)dx,a>1\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1 where [x]{\left[ x \right]} denotes the greatest integer not exceeding xx is
A af(a)−{f(1)+f(2)+.............f([a])}af\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( {\left[ a \right]} \right)} \right\}
B [a]f(a)−{f(1)+f(2)+...........f([a])}\left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( {\left[ a \right]} \right)} \right\}
C [a]f([a])−{f(1)+f(2)+...........f(a)}\left[ a \right]f\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( a \right)} \right\}
D af([a])−{f(1)+f(2)+.............f(a)}af\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( a \right)} \right\}
Correct Answer
Option B
Solution

Let

a=k+ha = k + h

where

kk

is an integer such that

[a]=k\left[ a \right] = k

and

0≤h<10 \le h < 1

∴\therefore

∫1a[x]f′(x)dx=∫121f′(x) dx\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx
            +∫232f′(x)dx+....\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....
            ∫k−1k(k−1)dx+∫kk+hkf′(x)dx\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx
{f(2)−f(1)}+2{f(3)−f(2)}\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}
             \,\,\,\,\,\,\,\,\,\,\,\,\,
+3{f(4)−f(3)}+ ........ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,
             \,\,\,\,\,\,\,\,\,\,\,\,\,
+(k−1){f(k)−f(k−1)}+ \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}
             \,\,\,\,\,\,\,\,\,\,\,\,\,
+k{f(k+h)−f(k)}+ k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}
=−f(1)−f(2)−f(3).......= - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......
             \,\,\,\,\,\,\,\,\,\,\,\,\,
−f(k)+kf(k+h)- f\left( k \right) + kf\left( {k + h} \right)
=[a]f(a)−{f(1)+f(2)= \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.
             \,\,\,\,\,\,\,\,\,\,\,\,\,
+f(3)+........f([a])}\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}
Q23
∫0πxf(sin⁡x)dx\int\limits_0^\pi {xf\left( {\sin x} \right)dx} is equal to
A π∫0πf(cos⁡x)dx\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx}
B  π∫0πf(sinx)dx\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx}
C π2∫0π/2f(sinx)dx{\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx}
D π∫0π/2f(cos⁡x)dx\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}
Correct Answer
Option D
Solution
I=∫0πxf(sin⁡ x)dxI = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx}
=∫0π(π−x)f(sin⁡x)dx= \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx}
=π∫0πf(sin⁡x)dx−1= \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1}
⇒2I=ππ0f(sin⁡x)dx\Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx
I=π2∫0πf(sin⁡x)dxI = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx}
=π∫0π/2f(sin⁡x)dx= \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}
=π∫0π/2f(cos⁡x)dx= \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}
Q24
Let F(x)=f(x)+f(1x),F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right), where f(x)=∫lxlog⁡t1+tdt,f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,} Then F(e)F(e) equals
A 11
B 22
C 1/21/2
D 00
Correct Answer
Option C
Solution

Given

f(x)=f(x)+f(1x),f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),

where

f(x)=∫1xlog⁡ t1+t dtf\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt

∴\therefore

F(e)=f(e)+f(1e)F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)
⇒F(e)\Rightarrow F\left( e \right)
=∫1elog⁡ t1+tdt+∫11/elog⁡t1+tdt   ....(A)= \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)

Now for solving,

I=∫11/elog⁡t1+tdtI = \int_1^{1/e} {{{\log t} \over {1 + t}}dt}

∴\therefore Put

1t=z⇒−1t2dt=dz{1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz
⇒dt=−dzz2\Rightarrow dt = - {{dz} \over {{z^2}}}

and limit for

t=1⇒z=1t = 1 \Rightarrow z = 1

and for

t=1/e⇒z=et = 1/e \Rightarrow z = e

∴\therefore

I=∫1elog⁡(1z)1+1z(−dzz2)I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)
=∫1e(log⁡1−log⁡z).zz+1(−dzz2)= \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)}
=∫1e−log⁡z(z+1)(−dzz)= \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)}

[ as

log⁡1=0\log 1 = 0

]

=∫1elog⁡zz(z+1)dz= \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz

∴\therefore

I=∫1elog⁡ tt(t+1)dtI = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt}
[\left[ {} \right.

By property

∫abf(t)dt\int_a^b {f\left( t \right)dt}
                   \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
=∫abf(x)dx= \int_a^b {f\left( x \right)dx}
]\left. {} \right]

Equation

(A)(A)

becomes

F(e)=∫1elog⁡t1+tdt+∫1elog⁡tt(1+t)dtF\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt
=∫1et.log⁡t+log⁡tt(1+t)dt= \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt
=∫1e(log⁡t)(t+1)t(1+t)= \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}}
⇒F(e)=∫1elog⁡ttdt\Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt}

Let

log⁡t=x\log t = x

∴\therefore

1tdt=dx{1 \over t}dt = dx
[\left[ {} \right.

for limit

t=1,x=0t = 1,x = 0

and

t=e,x=log⁡ e=1t = e,x = \log \,e = 1
]\left. {} \right]

∴\therefore

F(e)=∫01x dxF\left( e \right) = \int_0^1 {x\,dx}
⇒F(e)=[x22]01\Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1
⇒F(e)=12\Rightarrow F\left( e \right) = {1 \over 2}
Q25
The solution for xx of the equation ∫2xdttt2−1=π2\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} is
A 32{{\sqrt 3 } \over 2}
B 222\sqrt 2
C 22
D None
Correct Answer
Option D
Solution
∫2xdttt2−1=π2\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}

∴\therefore

[sec⁡−1t]2x=π2\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}
[\left[ {} \right.

As

∫dxxx2−1=sec⁡−1x\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x
]\left. {} \right]
⇒sec⁡−1x−sec⁡−12=π2\Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}
⇒sec⁡−1x−π4=π2\Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}
⇒sec⁡−1x=π2+π4\Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}
⇒sec⁡−1x=3π4\Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}
⇒x=sec⁡3π4\Rightarrow x = \sec {{3\pi } \over 4}
⇒x=−2\Rightarrow x = - \sqrt 2
Q26
Consider the integral I=∫010[x]e[x]ex−1dxI = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} , where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
A 45 (e −- 1)
B 45 (e + 1)
C 9 (e + 1)
D 9 (e −- 1)
Correct Answer
Option A
Solution
I=∫010[x] . e[x]+1−xdxI = \int_0^{10} {[x]\,.\,{e^{[x] + 1 - x}}} dx
=∫010 dx+∫12e2−xdx+∫232 . e3−xdx+∫343.e4−xdx+.........+∫9109 e10−xdx= \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx}
=−{(1−e)+2(1−e)+3(1−e)+.......+9(1−e)}= - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\}
=45(e−1)= 45(e - 1)
Q27
The value of ∫018log⁡(1+x)1+x2dx\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx is
A π8log⁡2{\pi \over 8}\log 2
B π2log⁡2{\pi \over 2}\log 2
C log⁡2\log 2
D πlog⁡2\pi \log 2
Correct Answer
Option D
Solution
I=∫018log⁡(1+x)1+x2dxI = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx

put

x=tan⁡ θ,x = \tan \,\theta ,

∴\therefore

dxdθ=sec⁡2θ⇒dx=sec⁡2θdθ{{dx} \over {d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta d\theta

∴\therefore

I=8∫0π/4log⁡(1+tan⁡θ)1+tan⁡2θ.sec⁡2θdθI = 8\int\limits_0^{\pi /4} {{{\log \left( {1 + \tan \theta } \right)} \over {1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta d\theta
I=8∫0π/4log⁡(1+tan⁡θ)dθ ...(i)I = 8\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta \,...\left( i \right)
=8∫0π/4log⁡[1+tan⁡(π4−θ)]dθ= 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]} d\theta
=8∫0π/4log⁡[1+1−tan⁡θ1+tan⁡θ]dθ= 8\int\limits_0^{\pi /4} {\log \left[ {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right]} d\theta
=8∫0π/4log⁡[21+tan⁡θ]dθ= 8\int\limits_0^{\pi /4} {\log \left[ {{2 \over {1 + \tan \theta }}} \right]} d\theta
=8∫0π/4[log⁡2−log⁡(1+tan⁡θ)]dθ= 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]d\theta }
I=8.(log⁡2)[x]0π/4−8I = 8.\left( {\log 2} \right)\left[ x \right]_0^{\pi /4} - 8
         ∫0π/4log⁡(1+tan⁡θ)dθ\,\,\,\,\,\,\,\,\,\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta
I=8.π4.log⁡2−II = 8.{\pi \over 4}.\log 2 - I
[\left[ {} \right.

From equation

(i)(i)
]\left. {} \right]
⇒2I=2πlog⁡2,\Rightarrow 2I = 2\pi \log 2,

∴\therefore

I=πlog⁡2I = \pi \log 2
Q28
Statement-1 : The value of the integral ∫π/6π/3dx1+tan⁡ x\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} is equal to π/6\pi /6 Statement-2 : ∫abf(x)dx=∫abf(a+b−x)dx.\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.
A Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C Statement- 1 is true; Statement-2 is False.
D Statement-1 is false; Statement-2 is true.
Correct Answer
Option D
Solution

Let

I=∫π/6π/3dx1+tan⁡ xI = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}}
=∫π/6π/3dxtan⁡(π2−x)= \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {\sqrt {\tan \left( {{\pi \over 2} - x} \right)} }}}
=∫π/6π/3tan⁡ x dx1+tan⁡ x            ...(i)= \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Also, Given,

II
=∫π/6π/3tan⁡ x dx1+tan⁡ x            ...(ii)= \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

By adding

(1)(1)

and

(2),(2),

we get

2I=∫π/6π/3dx2I = \int\limits_{\pi /6}^{\pi /3} {dx}
⇒I=12[π3−π6]\Rightarrow I = {1 \over 2}\left[ {{\pi \over 3} - {\pi \over 6}} \right]
=π12,= {\pi \over {12}},

statements -

11

is false

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

It is fundamental property.

Q29
The integral ∫0π1+4sin⁡2x2−4sin⁡x2 dx\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx equals:
A 43−44\sqrt 3 - 4
B 43−4−π34\sqrt 3 - 4 - {\pi \over 3}
C π−4\pi - 4
D 2π3−4−43{{2\pi } \over 3} - 4 - 4\sqrt 3
Correct Answer
Option B
Solution

Let

I=∫0π1+4sin⁡2x2−4sin⁡x2dxI = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx
=∫0π∣2sin⁡x2−1∣dx= \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx
=∫0π/3(1−2sin⁡x2)dx+∫π/3π(2sin⁡x2−1)dx= \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \over 2} - 1} \right)} dx
[\left[ {} \right.

As

sin⁡x2=12⇒x2=π6⇒x=π3,\sin {x \over 2} = {1 \over 2} \Rightarrow {x \over 2} = {\pi \over 6} \Rightarrow x = {\pi \over 3},
x2{x \over 2}
           =5π6⇒x=5π3\,\,\,\,\,\,\,\,\,\,\, = {{5\pi } \over 6} \Rightarrow x = {{5\pi } \over 3}
]\left. {} \right]
=[x+4cos⁡x2]0π/3+[−4cos⁡x2−x]π/3π= \left[ {x + 4\cos {x \over 2}} \right]_0^{\pi /3} + \left[ { - 4\cos {x \over 2} - x} \right]_{\pi /3}^\pi
=π3+422−4+(0−π+432+π3)= {\pi \over 3} + 4{{\sqrt 2 } \over 2} - 4 + \left( {0 - \pi + 4{{\sqrt 3 } \over 2} + {\pi \over 3}} \right)
=43−4−π3= 4\sqrt 3 - 4 - {\pi \over 3}
Q30
lim⁡n→∞((n+1)(n+2)...3nn2n)1n\mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n}}} is equal to:
A 9e2{9 \over {{e^2}}}
B 3 log⁡ 3−23\,\log \,3 - 2
C 18e4{{18} \over {{e^4}}}
D 27e2{{27} \over {{e^2}}}
Correct Answer
Option D
Solution
y=lim⁡n→∞((n+1)(n+2)...3nn2n)1ny = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n {}}}}
ln⁡ y=lim⁡n→∞1nln⁡(1+1n)(1+2n)\ln \,y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\ln \left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)
                            ....(1+2n)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {1 + {2 \over n}} \right)
ln⁡ y=lim⁡n→∞1n[ln⁡(1+1n)+ln⁡(1+2n)\ln {\mkern 1mu} y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\ln \left( {1 + {1 \over n}} \right) + \ln \left( {1 + {2 \over n}} \right)} \right.
                     +....+ln⁡(1+2nn)]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + .... + \ln \left( {1 + {{2n} \over n}} \right)} \right]
=lim⁡n→∞1n∑r=12nln⁡(1+rn)= \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^{2n} {\ln \left( {1 + {r \over n}} \right)}
=∫02ln⁡(1+x)dx= \int_0^2 {\ln \left( {1 + x} \right)dx}

Let

1+x=t⇒dx=dt1 + x = t \Rightarrow dx = dt

when

x=0,t=1x = 0,t = 1
             x=2,  t=3\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,t = 3
ln⁡ y=∫13ln⁡t d t=[t ln⁡ t−t13]\ln \,y = \int_1^3 {\ln t\,d\,t = } \left[ {t\,\ln \,t - \left. {t_1^3} \right]} \right.
           =ln⁡(33e2)=ln⁡(27e2)\,\,\,\,\,\,\,\,\,\,\, = \ln \left( {{{{3^3}} \over {{e^2}}}} \right) = \ln \left( {{{27} \over {{e^2}}}} \right)
⇒y=27e2\Rightarrow y = {{27} \over {{e^2}}}
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