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Definite Integration

JEE Mathematics · 230 questions · Page 22 of 23 · Click an option or "Show Solution" to reveal answer

Q211
∫0π[cot⁡x]dx,\int\limits_0^\pi {\left[ {\cot x} \right]dx,} where [.]\left[ . \right] denotes the greatest integer function, is equal to:
A 11
B −1-1
C −π2 - {\pi \over 2}
D π2 {\pi \over 2}
Correct Answer
Option C
Solution

Let

I=∫0π[cot⁡x]dx      ...(1)I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)}
=∫0π[cot⁡(π−x)]dx= \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx
=∫0π[−cot⁡x]dx      ...(2)= \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)}

Adding two values of

II

in

eqns(1)  &  (2),e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),

We get

2I=∫0π([cot⁡x]+[−cot⁡x])dx2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx}
=∫0π(−1)dx= \int_0^\pi {\left( { - 1} \right)dx}
[\left[ {} \right.

As

[x]+[−x]=−1,  if  \left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,
x∉zx \notin z

and

[x]+[−x]=0,  if  x∈z\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z
]\left. {} \right]
=[−x]0π=−π⇒I=−π2= \left[ { - x} \right]_0^\pi = - \pi \Rightarrow I = - {\pi \over 2}
Q212
If ∫12dx(x2−2x+4)32=kk+5,\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = {k \over {k + 5}}, then k is equal to :
A 1
B 2
C 3
D 4
Correct Answer
Option A
Solution

Given, I =

∫12dx(x2−2x+4)32\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}}

=

∫12dx[(x−1)2+3]32\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}}

Let x −- 1 =

3\sqrt 3

tanθ\theta ⇒\Rightarrow

   \,\,\,

dx =

3\sqrt 3

sec2θ\theta dθ\theta When x = 1, then θ\theta = 0 and when x = 2, θ\theta =

π6{\pi \over 6}

I =

∫0π63sec⁡2θ dθ(3tan⁡2θ+3)32\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}}

=

∫0π63sec⁡2θ dθ(3sec⁡2θ)32\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}}

=

∫0π63sec⁡2θ dθ33sec⁡3θ\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}}

=

13{1 \over 3}
∫0π6dθsec⁡θ\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}}

=

13∫0π6cos⁡θ dθ{1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta }

=

13[sin⁡θ]θπ6{1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}

=

13×12{1 \over 3} \times {1 \over 2}

=

16{1 \over 6}
∴   \therefore\,\,\,

According to the question,

kk+5{k \over {k + 5}}

=

16{1 \over 6}

⇒\Rightarrow

   \,\,\,

6k = k + 5 ⇒\Rightarrow

   \,\,\,

k = 1

Q213
Let f : R → \to R be a continuously differentiable function such that f(2) = 6 and f'(2) = 148{1 \over {48}}. If ∫6f(x)4t3dt\int\limits_6^{f\left( x \right)} {4{t^3}} dt = (x - 2)g(x), then lim⁡x→2g(x)\mathop {\lim }\limits_{x \to 2} g\left( x \right) is equal to :
A 18
B 36
C 12
D 24
Correct Answer
Option A
Solution

Given

∫6f(x)4x3dx=g(x).(x−2)\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)
⇒g(x)=\Rightarrow g(x) =
∫0f(x)4x3dxx−2{{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}

∴\therefore

lim⁡x→2g(x)=\mathop {\lim }\limits_{x \to 2} g(x) =
lim⁡x→2∫0f(x)4x3dxx−2\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}

At x = 2 this limit is in

00{0 \over 0}

form. So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.

⇒lim⁡x→24f3(x).f′(x)1=4×63×148=18\Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18
Q214
∫−3π2−π2[(x+π)3+cos⁡2(x+3π)]dx\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx is equal to
A π432{{{\pi ^4}} \over {32}}
B π432+π2{{{\pi ^4}} \over {32}} + {\pi \over 2}
C π2{\pi \over 2}
D π4−1{\pi \over 4} - 1
Correct Answer
Option C
Solution
I=∫−3π2−π2[(x+π)3+cos⁡2(x+3π)] dxI = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx

Put

x+π=tx + \pi = t
I=∫−π2π2(t3+cos⁡2t)dtI = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt}
=2∫−π2π2cos⁡2tdt= 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt
[\left[ {} \right.

using the property of even and odd function

]\left. {} \right]
=∫0π2(1+cos⁡2t)dt=π2+0= \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0
Q215
Let I=∫01sin⁡xxdxI = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx} and J=∫01cos⁡xxdx.J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} . Then which one of the following is true?
A 1>231 > {2 \over 3} and J>2J > 2
B 1<231 < {2 \over 3} and J<2J < 2
C 1<231 < {2 \over 3} and J>2J > 2
D 1>231 > {2 \over 3} and J<2J < 2
Correct Answer
Option B
Solution

We know that

sin⁡xx<1,{{\sin x} \over x} < 1,

for

x∈(0,1)x \in \left( {0,1} \right)
⇒sin⁡xx<x\Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x

on

x∈(0,1)x \in \left( {0,1} \right)
⇒∫01sin⁡xxdx<∫01xdx=[2x3/23]01\Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1
⇒∫01sin⁡xxdx<23⇒I<23\Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}

Also

cos⁡xx<1x{{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}

for

x∈(0,1)x \in \left( {0,1} \right)
⇒∫01cos⁡xxdx<∫01x−1/2dx\Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} }
              =[2x]01=2\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2
⇒∫01cos⁡xxdx<2\Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2}
⇒J<2\Rightarrow J < 2
Q216
If ∫0πxf(sin⁡x)dx=A∫0π/2f(sin⁡x)dx,\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } then AA is
A 2π2\pi
B π\pi
C π4{\pi \over 4}
D 00
Correct Answer
Option B
Solution

Let

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

∴\therefore

2I=π∫2πf(sin⁡x)dx2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx
=π.2∫0π2f(sin⁡x)dx= \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx

∴\therefore

I=π∫0π2f(sin⁡x)dxI = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx
⇒A=π\Rightarrow A = \pi
Q217
The integral ∫24log⁡ x2log⁡x2+log⁡(36−12x+x2)dx\int\limits_2^4 {{{\log \,{x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}dx} is equal to :
A 11
B 66
C 22
D 44
Correct Answer
Option A
Solution
I=∫24log⁡x2log⁡x2+log⁡(36−12x+x2)I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}}
I=∫24log⁡x2log⁡x2+log⁡(6−x)2          ...(i)I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)
I=∫24log⁡(6−x)2log⁡(6−x)2+log⁡x2        ...(ii)I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}} \,\,\,\,\,\,\,\,...\left( {ii} \right)

Adding

(1)(1)

and

(2)(2)
2I=∫24dx=[x]24=2⇒1−12I = \int\limits_2^4 {dx = \left[ x \right]_2^4} = 2 \Rightarrow 1 - 1
Q218
The value of ∫−π/2π/2dx[x]+[sin⁡x]+4,\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} , where [t] denotes the greatest integer less than or equal to t, is
A 112(7π−5){1 \over {12}}\left( {7\pi - 5} \right)
B 112(7π+5){1 \over {12}}\left( {7\pi + 5} \right)
C 310(4π−3){3 \over {10}}\left( {4\pi - 3} \right)
D 320(4π−3){3 \over {20}}\left( {4\pi - 3} \right)
Correct Answer
Option D
Solution
I=∫−π2π2dx[x]+[sin⁡x]+4{\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}}
=∫−π2−1dx−2−1+4+∫−10dx−1−1+4= \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}}
                 +∫01dx0+0+4+∫1π2dx1+0+4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} }
∫−π2−1dx1+∫−10dx2+∫01dx4+∫1π2dx5\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}}
(−1+π2)+12(0+1)+14+15(π2−1)\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)
−1+12+14−15+π2+π10- 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}
−20+10+5−420+6π10{{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}
−920+3π5{{ - 9} \over {20}} + {{3\pi } \over 5}
Q219
Let I=∫ab(x4−2x2)dx.{\rm I} = \int\limits_a^b {\left( {{x^4} - 2{x^2}} \right)} dx. If I is minimum then the ordered pair (a, b) is -
A (2,−2)\left( {\sqrt 2 , - \sqrt 2 } \right)
B (0,2)\left( {0,\sqrt 2 } \right)
C (−2,2)\left( { - \sqrt 2 ,\sqrt 2 } \right)
D (−2,0)\left( { - \sqrt 2 ,0} \right)
Correct Answer
Option C
Solution

Let f(x) = x2(x2 −- 2) As long as f(x) lie below the x-axis, define integral will remain negative, so correct value of (a, b) is (−-

2\sqrt 2

,

2\sqrt 2

) for minimum of I

Q220
If ∫0π3tan⁡θ2k sec⁡θ dθ=1−12,(k>0),\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right), then value of k is :
A 4
B 12{1 \over 2}
C 1
D 2
Correct Answer
Option D
Solution
∫0π3tan⁡θ2k sec⁡θ dθ=1−12,(k>0),\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),
⇒  ∫0π/3tan⁡θ2ksec⁡θ dθ=1−12(k>0)\Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)
⇒  12k(−2cos⁡θ)0π/3=1−12\Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}
⇒  12k(−2+2)=1−12\Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} \right) = 1 - {1 \over {\sqrt 2 }}
⇒  2(2−1)2k=2−12\Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }}
⇒  k=2\Rightarrow \,\,k = 2
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