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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 4
If
f
(
1
x
)
+
x
2
f
(
x
)
=
0
for
x
>
0
,
and
I
=
∫
x
1
/
x
f
(
z
)
d
z
,
1
2
≤
x
≤
2
then
I
is?
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0%
f
(
2
)
−
f
(
1
/
2
)
0%
f
(
1
/
2
)
−
f
(
2
)
0%
0
0%
None of these
Explanation
Substitute
z
=
1
t
I
=
∫
1
t
x
f
(
1
t
)
.
(
−
1
t
2
)
d
t
=
∫
1
x
x
(
−
t
2
.
f
(
t
)
)
(
−
1
t
2
)
=
∫
1
x
x
f
(
t
)
d
t
=
−
∫
x
1
x
f
(
t
)
d
t
=
−
I
∴
If
\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx
and
\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx
then
I_{1}+I_{2}
equals?
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0%
\displaystyle \frac{1}{3}
0%
\displaystyle -\frac{1}{3}
0%
0
0%
None of these
Explanation
I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx
\displaystyle x=\frac { -y-3 }{ 3 }
\displaystyle dx=-\frac { dy }{ 3 }
\displaystyle \Rightarrow I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx=-\int _{ -4 }^{ -5 } e^{ \left( y+5 \right) ^{ 2 } }dx=-I_{ 1 }
\displaystyle \Rightarrow I_{ 1 }+I_{ 2 }=0
If
f\left ( a+x \right )= f\left ( x \right )
, then
\forall
a> 0, n\epsilon N
the value of
\displaystyle \int_{0}^{n a}f\left ( x \right )dx
equals ?
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0%
\displaystyle \left ( n-1 \right )\int_{0}^{a}f\left ( x \right )dx
0%
\displaystyle \left ( 1-n \right )\int_{0}^{a}f\left ( x \right )dx
0%
\displaystyle n\int_{0}^{a}f\left ( x \right )dx
0%
None of the above
Explanation
I=\displaystyle \int _{ 0}^{ na }{ f\left( x \right) dx }
Substitute
x=a+t\Rightarrow dx=dt
I=\displaystyle \int _{ 0 }^{ \left( n-1 \right) a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+x \right) dx } =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( x \right) dx }
If
\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt
then
\displaystyle {f}'\left ( \frac{\pi }{3} \right )
is
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0%
nonexistent
0%
\displaystyle \pi /4
0%
\displaystyle \pi \sqrt{3/4}
0%
none of these
Explanation
\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ t }^{ 2 } } } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt
\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right]
\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 }
\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx
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0%
\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{6}\log _{2}e.
0%
\displaystyle -\frac{4}{3}\log _{3}e+\frac{1}{6}\log _{2}e.
0%
\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{3}\log _{2}e.
0%
\displaystyle -\frac{2}{3}\log _{3}e+\frac{1}{6}\log _{2}e.
Explanation
\displaystyle \int _{ 0 }^{ 1 } \frac { 2^{ x+1 }-3^{ x-1 } }{ 6^{ x } } dx=\int _{ 0 }^{ 1 } \left( 2.3^{ -x }-\frac { 1 }{ 3 } 2^{ -x } \right) dx
\displaystyle =\left[ -2\frac { 3^{ -x } }{ \log 3 } +\frac { 1 }{ 3 } \frac { 2^{ -x } }{ \log 2 } \right] ^{ 1 }
\displaystyle =-\frac { 2 }{ \log 3 } \left( \frac { 1 }{ 3 } -1 \right) +\frac { 1 }{ 3\log 2 } \left( \frac { 1 }{ 2 } -1 \right)
\displaystyle =\frac { 4 }{ 3 } \log _{ 3 } e-\frac { 1 }{ 6 } \log _{ 2 } e
Hence, option 'A' is correct.
\displaystyle \int_{1}^{2}\left ( x+\frac{1}{x} \right )^{3/2}\frac{x^{2}-1}{x^{2}}dx
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0%
\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}+\frac{8}{5}\sqrt{2}
0%
\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}
0%
\displaystyle \sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}
0%
\displaystyle \frac{3}{2}\sqrt{\left ( \frac{3}{2} \right )}-\frac{8}{5}\sqrt{2}
Explanation
Let
\displaystyle I=\int _{ 1 }^{ 2 } \left( x+\frac { 1 }{ x } \right) ^{ 3/2 }\frac { x^{ 2 }-1 }{ x^{ 2 } } dx
Put
\displaystyle x+\frac { 1 }{ x } =t\Rightarrow \left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx\Rightarrow dt
Therefore
\displaystyle I=\int _{ 2 }^{ 5/2 } t^{ 3/2 }dt=\frac { 2 }{ 5 } t^{ 5/2 }=\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 5/2 }-2^{ 5/2 } \right]
\displaystyle =\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 2 }\sqrt { \left( \frac { 5 }{ 2 } \right) } -2^{ 2 }\sqrt { 2 } \right] =\frac { 5 }{ 2 } \sqrt { \left( \frac { 5 }{ 2 } \right) } -\frac { 8 }{ 5 } \sqrt { 2 }
Hence, option 'B' is correct.
Evaluate the integrals
\displaystyle \int_{a}^{b}\frac{\log x}{x}dx
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\displaystyle \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}
0%
\displaystyle \frac{1}{4} \log\left ( ab\right )\cdot \log \frac{b}{a}
0%
\displaystyle \log\left ( ab\right )\cdot \log \frac{b}{a}
0%
\displaystyle -\frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}
Explanation
Substitute
\displaystyle \log x=t \therefore \left ( 1/x \right )dx=dt
and adjust the limits
\displaystyle \therefore I= \int t\:dt=\left [ \frac{1}{2}t^{2} \right ]_{\log a}^{\log b}
\displaystyle \therefore I= \frac{1}{2}\left [ \left ( \log b \right )^{2}-\left ( \log a \right )^{2} \right ]
\displaystyle = \frac{1}{2}\left [ \left ( \log b+\log a \right )\left ( \log b-\log a \right ) \right ]
\displaystyle \therefore I= \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}.
Given
\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,
the value of
\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx }
is?
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0%
{ e }^{ 4 }-e
0%
{ e }^{ 4 }-a
0%
2{ e }^{ 4 }-a
0%
2{ e }^{ 4 }-e-a
Explanation
Given
\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,
Let
I=\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx }
Put
\displaystyle \ln { \left( x \right) } ={ t }^{ 2 }\Rightarrow \frac { 1 }{ x } dx=2tdt
\therefore I=\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } .{ 2t }^{ 2 }dt={ \left( t{ e }^{ { t }^{ 2 } } \right) }_{ 1 }^{ 2 }-\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } dt={ 2e }^{ 4 }-e-a.
\displaystyle \int_{0}^{\pi /2}\frac{dx}{\sin x}
equals
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0%
0
0%
\displaystyle \frac{1}{2}
0%
1
0%
3/2
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { 1 }{ sin{ x } } dx } =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \csc { x } } dx
Multiply numerator and denominator by
cot{ x }+\csc { x }
, we get
\displaystyle \therefore I=-\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { -cot{ x }\csc { x } -\csc ^{ 2 }{ x } dx }{ cot{ x }+\csc { x } } dx }
Substitute
t=cot{ x }+\csc { x } \Rightarrow dt=\left( -\csc ^{ 2 }{ x } -cot{ x }\csc { x } \right) dx
\displaystyle \therefore I=-\int _{ 0 }^{ 1 }{ \dfrac { 1 }{ t } } dt=\left[ -\log { t } \right] _{ 0 }^{ 1 }=0
The value of
\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}+2x}}
is
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0%
\pi /6
0%
\pi /3
0%
\pi /2
0%
\pi
Explanation
Let
\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }+2x } } dx } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { \left( x+1 \right) }^{ 2 }-1 } } dx }
Substitute
u=x+1\Rightarrow du=dx
\displaystyle\therefore I=\int _{ 1 }^{ 2 }{ \frac { 1 }{ u\sqrt { { u }^{ 2 }-1 } } du }
Substitute
\displaystyle s=\sqrt { { u }^{ 2 }-1 } \Rightarrow ds=\frac { u }{ \sqrt { { u }^{ 2 }-1 } } du
\displaystyle\therefore I=\int _{ 0 }^{ \sqrt { 3 } }{ \frac { 1 }{ { s }^{ 2 }+1 } ds } ={ \left[ \tan ^{ -1 }{ s } \right] }_{ 0 }^{ \sqrt { 3 } }=\frac { \pi }{ 3 }
Value of
\displaystyle \int_{0}^{\pi /4}\left ( \sqrt{\tan x}-\sqrt{\cot x} \right )\: dx
is
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0%
\sqrt{2}\log \left ( \sqrt{2}-1 \right )
0%
\sqrt{2}\log \left ( \sqrt{2}+1 \right )
0%
\log \left ( \sqrt{2}+1 \right )
0%
\log \left ( \sqrt{2}-1 \right )
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \sqrt { \tan { x } } -\sqrt { cot{ x } } \right) } dx=-\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { cos{ x }-sin{ x } }{ \sqrt { cos{ x }sin{ x } } } dx }
Substitute
\left( sin{ x+cos{ x } } \right) =t\Rightarrow 2sin{ x }cos{ x }={ t }^{ 2 }-1
\displaystyle \therefore I=\sqrt { 2 } \int _{ 1 }^{ \sqrt { 2 } }{ \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } \\ =\left[ \sqrt { 2 } \log { \left( t+\sqrt { { t }^{ 2 } } -1 \right) } \right] _{ 0 }^{ \sqrt { 2 } }\\ =\sqrt { 2 } \log { \left( \sqrt { 2 } -1 \right) }
The value of
\displaystyle \int_{a}^{b}\displaystyle \frac{\log x}{x}\: dx
is
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\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )
0%
\displaystyle \frac{1}{2}\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )
0%
\log \left ( a^{2}-b^{2} \right )
0%
\left ( a+b \right )\log \left ( a+b \right )
Explanation
\displaystyle I=\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx={ \left[ \log { x } .\log { x } \right] }_{ a }^{ b }-\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx\\ \Rightarrow 2I={ \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b }={ \left( \log { b } \right) }^{ 2 }-{ \left( \log { a } \right) }^{ 2 }
\displaystyle \Rightarrow I=\frac { 1 }{ 2 } \left( \log { b } +\log { a } \right) \left( \log { b } -\log { a } \right) =\frac { 1 }{ 2 } \log { ab } \log { \frac { b }{ a } }
If
f\left ( x \right )= A\sin \left ( \dfrac {\pi x}{2} \right )\: +\: B,f{}'\left ( \dfrac 12 \right )= \sqrt{2}
and
\displaystyle \int_{0}^{1}f\left ( x \right )dx= \displaystyle \frac{2A}{\pi }
,
then the constants
A
and
B
are
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0%
A=\dfrac {\pi}{2}, B=\dfrac {\pi}{2}
0%
A=\dfrac {2}{\pi}, B=3\pi
0%
A=0, B=-4\pi
0%
A=\dfrac {4}{\pi}, B=0
Explanation
f(x)=A\sin{(\pi x/2)}+B
\implies f^{'}(x)=\dfrac{A\pi}{2}\cos{(\pi x/2)}
\implies f^{'}(1/2)=\dfrac{A\pi}{2}\cos{(\pi/4)=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}}
\implies\sqrt{2}=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}
Hence,
A=\dfrac{4}{\pi}
\displaystyle\int_{0}^{1} f(x)dx=\left[-\dfrac{2A}{\pi}\cos{(\pi x/2)}+Bx\right]_{0}^{1}
\implies\dfrac{2A}{\pi}=\dfrac{2A}{\pi}+B
Hence,
B=0
Hence, answer is option-(D).
The value of the integral
\displaystyle \int_{0}^{\pi /4}\displaystyle \frac{\sin x+\cos x}{3+\sin 2x}dx
is
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0%
\log 2
0%
\log 3
0%
\left ( 1/4 \right )\log 3
0%
\left ( 1/8 \right )\log 3
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \pi /4 } \frac { \sin x+\cos x }{ 3+\sin 2x } dx=\int _{ 0 }^{ \pi /4 } \frac { \sin { x } +\cos { x } }{ 4-{ \left( \sin { x } -\cos { x } \right) }^{ 2 } } dx
Substitute
\sin { x } -\cos { x } =t
\displaystyle I=\int _{ -1 }^{ 0 }{ \frac { dt }{ 4-{ t }^{ 2 } } } =\frac { 1 }{ 4 } \log { 3 }
The value of
\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}
is
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\displaystyle \frac{\pi }{1+\alpha ^{2}}
if
\alpha > 1
0%
\displaystyle \frac{\pi }{\alpha ^{2}-1}
if
\alpha > 1
0%
\displaystyle \frac{\pi }{1+\alpha ^{2}}
if
\alpha < 1
0%
\displaystyle \frac{\pi }{\alpha ^{2}-1}
if
\alpha < 1
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } =\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } }
Substitute
\displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt
\displaystyle I=\int _{ 0 }^{ \infty } \dfrac { 1 }{ 1-2\alpha \left( \dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } \right) +\alpha ^{ 2 } } .\dfrac { 2t }{ 1+{ t }^{ 2 } } dt
\displaystyle =\int _{ 0 }^{ \infty } \dfrac { 2t }{ 1+{ t }^{ 2 }-2\alpha \left( 1-{ t }^{ 2 } \right) +\alpha ^{ 2 }\left( 1+{ t }^{ 2 } \right) } dt
\displaystyle =\frac { \pi }{ { \alpha }^{ 2 }-1 }
if
\alpha >1
If
I= \displaystyle \int_{1}^{2}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}-1}}
then
I
is equal to
Report Question
0%
1
0%
1/2
0%
1/\sqrt{2}
0%
1/\sqrt{3}
Explanation
As form of integrand is
\displaystyle \dfrac { 1 }{ L\sqrt { Q } }
Substitute
\displaystyle L=\dfrac { 1 }{ t } \Rightarrow x+1=\dfrac { 1 }{ t }
\displaystyle I=\int _{ \dfrac { 1 }{ 2 } }^{ \dfrac { 1 }{ 3 } }{ \dfrac { \left( \dfrac { -1 }{ { t }^{ 2 } } \right) dt }{ \dfrac { 1 }{ t } \sqrt { { \left( \dfrac { 1 }{ t } -1 \right) }^{ 2 }-1 } } } =\int _{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ \dfrac { dt }{ \sqrt { 1-2t } } }
\displaystyle ={ \left[ \dfrac { { \left( 1-2t \right) }^{ \dfrac { 1 }{ 2 } } }{ \left( -2 \right) \left( \dfrac { 1 }{ 2 } \right) } \right] }_{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ = }\dfrac { 1 }{ \sqrt { 3 } }
The value of
\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{e^{x}+e^{-x}}
is
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0%
\tan ^{-1}e
0%
\tan ^{-1}\left ( e \right )-\pi /4
0%
\tan ^{-1}\left ( e \right )-\tan ^{-1}\left ( 1/e \right )
0%
\tan ^{-1}\left ( 1/e \right )+\pi /4
Explanation
Let
\displaystyle I=\int _{ 0 }^{ 1 } \frac { dx }{ e^{ x }+e^{ -x } } =\int _{ 0 }^{ 1 } \frac { { e }^{ x }dx }{ e^{ 2x }+1 }
Substitute
{ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt
\displaystyle I=\int _{ 1 }^{ e }{ \frac { t }{ { t }^{ 2 }+1 } dt } ={ \left[ \tan ^{ -1 }{ t } \right] }_{ 1 }^{ e }=\tan ^{ -1 }{ e } -\frac { \pi }{ 4 }
The value of
\displaystyle \int_{8}^{15}\displaystyle \frac{dx}{\left ( x-3 \right )\sqrt{x+1}}
is
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0%
\log \displaystyle \frac{5}{3}
0%
\displaystyle \frac{1}{2}\log \displaystyle \frac{5}{3}
0%
2\log \displaystyle \frac{5}{3}
0%
\sqrt{2}\log \left ( \sqrt{2}-1 \right )
Explanation
\displaystyle I=\int _{ 8 }^{ 15 } \frac { dx }{ \left( x-3 \right) \sqrt { x+1 } }
Substitute
\sqrt { x+1 } =t\Rightarrow x+1={ t }^{ 2 }
\displaystyle I=\int _{ 3 }^{ 4 }{ \frac { 2t }{ \left( { t }^{ 2 }-4 \right) t } dt } ={ \left[ \frac { 2 }{ 2.2 } \log { \left( \frac { t-2 }{ t+2 } \right) } \right] }_{ 3 }^{ 4 }
\displaystyle =\frac { 1 }{ 2 } \left[ \log { \frac { 1 }{ 3 } } -\log { \frac { 1 }{ 5 } } \right] =\frac { 1 }{ 2 } \log { \frac { 5 }{ 3 } }
If
b > a,
and
\displaystyle I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,
then
I
equals
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0%
\displaystyle \frac{\pi}{2} (b -a)
0%
\pi (b -a)
0%
\pi/2
0%
2\pi (b-a)
Explanation
Substitute
b -x = t^{2}
so that
\displaystyle I = \int _{\sqrt{b-a}}^{0}\sqrt{\frac{b-t^{2}-a}{t^{2}}} (-2t) dt
\displaystyle= 2 \int_{0}^{c} \sqrt{c^{2}-t^{2}} dt
, where
c= \sqrt{b-a}
=\displaystyle 2 \left [ \frac{1}{2} t \sqrt{c^{2}-t^{2}}+ \frac{c^{2}}{2} \sin ^{-1} \left( \frac{t}{c}\right) \right] _{0}^{c}
=0 + c^{2} \sin^{-1} (1) -0
\displaystyle =\frac{\pi}{2}(b-a)
Value of
\displaystyle \int_{0}^{2a}\dfrac{x^{3/2}}{\sqrt{2a-x}} dx
is
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0%
\displaystyle \frac{3\pi a^{2}}{2}
0%
\pi a^{3}
0%
\sqrt{2}\pi a^{3}
0%
2\pi a^{3}
Explanation
Let
\displaystyle I=\int _{ 0 }^{ 2a }{ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \sqrt { 2a-x } } dx }
Substitute
\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx
\displaystyle I=2\int _{ 0 }^{ \sqrt { 2a } }{ \frac { { u }^{ 4 } }{ \sqrt { 2a-{ u }^{ 2 } } } du }
Substitute
u=\sqrt { 2a } \sin { t } \Rightarrow du=\sqrt { 2a } \cos { t } dt
\displaystyle I=2\sqrt { 2a } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sqrt { 2 } { a }^{ \frac { 3 }{ 2 } }\sin ^{ 4 }{ t } \right) dt } =8{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sin ^{ 4 }{ t } dt }
Using reduction formulae
\displaystyle \int { \sin ^{ m }{ x } dx } =\frac { -\cos { x } \sin ^{ m-1 }{ x } }{ m } +\frac { m-1 }{ m } \int { \sin ^{ m-2 }{ x } dx }
\displaystyle I={ \left[ -2{ a }^{ 2 }\sin ^{ 3 }{ t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ t } dt }
\displaystyle =0+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \frac { 1 }{ 2 } -\frac { 1 }{ 2 } \cos { 2t } \right) dt }
\displaystyle ={ \left[ 3{ a }^{ 2 }t-3{ a }^{ 2 }\sin { t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { 3\pi { a }^{ 2 } }{ 2 }
Value of
\displaystyle \int_{0}^{25}\displaystyle \frac{1}{\sqrt{4+\sqrt{x}}}\: dx
is
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0%
2\left ( \sqrt{29}-1 \right )
0%
2\left ( \sqrt{29}-5 \right )
0%
3 \sqrt{29}-1
0%
none of these
Explanation
Let
\displaystyle I=\int _{ 0 }^{ 25 }{ \frac { 1 }{ \sqrt { \sqrt { x } +4 } } dx }
Substitute
\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx
\displaystyle I=2\int _{ 0 }^{ 5 }{ \frac { u }{ \sqrt { u+4 } } du }
Substitute
s=u+4\Rightarrow ds=du
\displaystyle I=2\int _{ 4 }^{ 9 }{ \frac { s-4 }{ \sqrt { s } } ds } =2\int _{ 4 }^{ 9 }{ \left( \sqrt { s } -\frac { 4 }{ \sqrt { s } } \right) ds }
\displaystyle={ \left[ \frac { 4{ s }^{ 3/2 } }{ 3 } \right] }_{ 4 }^{ 9 }-{ \left[ 16\sqrt { s } \right] }_{ 4 }^{ 9 }=36-\frac { 32 }{ 3 } -48-32=-\frac { 164 }{ 3 }
\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx
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0%
is equal to zero
0%
is equal to one
0%
is equal to
\displaystyle \frac{1}{2}
0%
can not be evaluated
Explanation
I = \displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx
Substitute
\displaystyle x = \frac{1}{t}\Rightarrow dx =-\frac{1}{t^2}dt
I = \displaystyle \int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln (1/t)}{1/t} .\frac{-dt}{t^2}
\quad \displaystyle =\int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln t}{t} dt
\quad \displaystyle =\int_{\infty }^0 f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x} dx=-I
\Rightarrow 2I = 0\Rightarrow I = 0
If
\displaystyle \int_{0}^{\pi /3}\frac{\cos }{3+4\sin x}dx=K\log \frac{\left ( 3+2\sqrt{3} \right )}{3}
then K is
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0%
\displaystyle \frac{1}{2}
0%
\displaystyle \frac{1}{3}
0%
\displaystyle \frac{1}{4}
0%
\displaystyle \frac{1}{8}
Explanation
\displaystyle I=\int _{ 0 }^{ \pi /3 }{ \cfrac { \cos { x } }{ 3+4\sin { x } } dx }
Substituting
t=3+4\sin { x } \Rightarrow dt=4\cos { x }
\displaystyle I=\frac { 1 }{ 4 } \int _{ 3 }^{ 3+2\sqrt { 3 } }{ \frac { 1 }{ t } dt } ={ \left[ \frac { \log { t } }{ 4 } \right] }_{ 3 }^{ 3+2\sqrt { 3 } }
\displaystyle=\frac { \log { \left( 3+2\sqrt { 3 } \right) } -\log { 3 } }{ 4 } =\frac { 1 }{ 4 } \log { \left( \frac { 3+2\sqrt { 3 } }{ 3 } \right) } \Rightarrow K=\frac { 1 }{ 4 }
Suppose that F(x) is an antiderivative of f(x)
\displaystyle =\frac{\sin x}{x},x> 0
then
\displaystyle \int_{1}^{3}\frac{\sin 2x}{x}
can be expressed as
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0%
F(6) - F(2)
0%
\displaystyle \frac{1}{2}\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )
0%
\displaystyle \frac{1}{2}\left ( F\left ( 3 \right )-F\left ( 1 \right ) \right )
0%
\displaystyle 2\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )
Explanation
\displaystyle F\left ( x \right )=\int \frac{\sin x}{x}dx
Now
\displaystyle I = \int_{1}^{3}\frac{\sin 2x}{x}dx\left [ put2x=t \right ]=\int_{2}^{6}\frac{2}{2}\frac{\sin }{t}dt=\left [ F\left ( x \right ) \right ]_{2}^{6}=F(6)-F(2)
If
0 < \alpha < 1
and
\displaystyle I=\int _{-1}^{1} \frac{dx}{\sqrt{1-2\alpha x+\alpha^{2}}}
then
I
equals
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0%
1/\alpha
0%
2/\alpha
0%
3/\alpha
0%
none of these
Explanation
\displaystyle I =\int_{-1}^{1} (1 -2\alpha x + \alpha^{2} )^{-1/2} dx
\displaystyle =\left. \frac{2(1-2\alpha x+\alpha^{2})^{1/2}}{-2\alpha}\right ]_{-1}^{1}
\displaystyle =-\frac{1}{\alpha}[([(1 -2\alpha + \alpha^{2})^{1/2} -(1 + 2 \alpha + \alpha ^{2})^{1/2} ]
=-\dfrac{1}{\alpha}[\sqrt{(\alpha-1)^2}-\sqrt{(\alpha+1)^2}]
=-\dfrac1\alpha\left[(1-\alpha)-(1+\alpha)\right]\ldots(\because0<\alpha<1)
=2
Ans: D
\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx
has the value equal to
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0%
0
0%
\displaystyle \frac{3}{4}
0%
\displaystyle \frac{5}{4}
0%
2
Explanation
\displaystyle I=\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ x } } \sin { \left( x-\frac { 1 }{ x } \right) dx }
Put
\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=\frac { 1 }{ { t }^{ 2 } } dt
\displaystyle I=\int _{ 2 }^{ \frac { 1 }{ 2 } }{ t\sin { \left( \frac { 1 }{ t } -t \right) \left( \frac { -1 }{ { t }^{ 2 } } \right) dt } } =\int _{ 2 }^{ \frac { 1 }{ 2 } }{ \frac { 1 }{ t } \sin { \left( t-\frac { 1 }{ t } \right) dt } }
\displaystyle =-\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ t } } \sin { \left( t-\frac { 1 }{ t } \right) dt } =-I.
\Rightarrow 2I=0\quad \Rightarrow I=0
The value of the definite integral
\displaystyle \int_{1}^{\infty}\left ( e^{x+1}+e^{3-x} \right )^{-1}dx
is
Report Question
0%
\displaystyle \frac{\pi }{4e^{2}}
0%
\displaystyle \frac{\pi }{4e}
0%
\displaystyle \frac{1}{e^{2}}\left ( \frac{\pi }{2}-\tan ^{-1}\frac{1}{e} \right )
0%
\displaystyle \frac{\pi }{2e^{2}}
Explanation
I=\int _{ 1 }^{ \infty }{ \cfrac { 1 }{ { e }^{ x+1 }+{ e }^{ 3-x } } dx } =\int _{ 1 }^{ \infty }{ \cfrac { { e }^{ x-1 } }{ \left( { e }^{ x }-ie \right) \left( { e }^{ x }+ie \right) } dx }
Substituting
{ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt
I=\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( t-ie \right) \left( t+ie \right) } dt } =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( e-it \right) \left( e+it \right) } dt } \\ =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \left( \cfrac { i }{ 2\left( et+i{ e }^{ 2 } \right) } -\cfrac { i }{ 2\left( et-i{ e }^{ 2 } \right) } \right) } dt\\ ={ \left[ \cfrac { i\log { \left( et+i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } -\cfrac { i\log { \left( et-i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } \right] }_{ e }^{ \infty }\\ ={ \left[ \cfrac { \tan ^{ -1 }{ \left( { e }^{ x-1 } \right) } }{ { e }^{ 2 } } \right] }_{ e }^{ \infty }=\cfrac { \pi }{ 4{ e }^{ 2 } }
\displaystyle \int ^{\displaystyle \frac{3 \pi}{10}}_{\displaystyle \frac{\pi}{5}} \frac{sin x}{sin x + cos x}dx
is equal to
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0%
\pi
0%
\displaystyle \frac{\pi}{2}
0%
\displaystyle \frac{\pi}{4}
0%
\displaystyle \frac{\pi}{20}
Explanation
Let
\displaystyle I=\int { \frac { \sin { x } }{ \sin { x } +\cos { x } } dx }
Multiply numerator and denominator by
I=\int { \dfrac { \sin { x } }{ \sin { x } +\cos { x } } dx } \\ \csc ^{ 3 }{ x }
\displaystyle I=\int { \frac { \csc ^{ 2 }{ x } }{ \csc ^{ 2 }{ x } +\cot { x } \csc ^{ 2 }{ x } } dx } =\int { \frac { \csc ^{ 2 }{ x } }{ \cot ^{ 3 }{ x } +\cot ^{ 2 }{ x } +\cot { x } +1 } dx }
Put
t=\cot { x } \Rightarrow dt=-\csc ^{ 2 }{ x } dx
\displaystyle I=-\int { \frac { 1 }{ { t }^{ 3 }+{ t }^{ 2 }+t+1 } dt } =-\int { \frac { 1-t }{ 2\left( { t }^{ 2 }+1 \right) } dt } -\int { \frac { 1 }{ 2\left( t+1 \right) } dt }
\displaystyle =\frac { 1 }{ 2 } \int { \frac { t }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ t+1 } dt }
\displaystyle =\frac { 1 }{ 4 } \log { \left( { t }^{ 2 }+1 \right) } -\frac { 1 }{ 2 } \tan ^{ -1 }{ t } -\frac { 1 }{ 2 } \log { \left( t+1 \right) }
\displaystyle =\frac { 1 }{ 2 } \left( x-\log { \left( \sin { x } +\cos { x } \right) } \right)
Hence
\displaystyle \int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } } \frac { sinx }{ sinx+cosx } dx=\frac { \pi }{ 20 }
If
\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt
where
x > 0
, then the value(s) of
x
satisfying the equation,
f(x) +f(1/x)=2
is
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0%
2
0%
e
0%
\displaystyle e^{-2}
0%
\displaystyle e^{2}
Explanation
\displaystyle f\left( x \right)=\int _{ 1 }^{ x }{ \frac { \ln { t } }{ 1+t } dt }
\displaystyle \Rightarrow f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \ln { t } }{ 1+t } dt }
Substituting
t=\dfrac { 1 }{ u } \Rightarrow dt=\left( -\dfrac { 1 }{ { u }^{ 2 } } \right) du
Therefore
\displaystyle f\left( \dfrac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \dfrac { \ln { \left( \dfrac { 1 }{ u } \right) \left( -1 \right) } }{ \left( 1+\dfrac { 1 }{ u } \right) { u }^{ 2 } } du }
\displaystyle=\int _{ 1 }^{ x }{ \frac { \ln { u } }{ u\left( u+1 \right) } du } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }
Now,
\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \cfrac { \ln { t } }{ \left( 1+t \right) } dt } +\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }
\displaystyle=\int _{ 1 }^{ x }{ \frac { \left( 1+t \right) \ln { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t } dt } =\frac { 1 }{ 2 } { \left( \ln { x } \right) }^{ 2 }
Hence
\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =2\Rightarrow x={ e }^{ \pm 2 }
Choose a function
f(x)
such that it is integrable over every interval on the real line
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0%
f(x) = [x]
0%
f(x)=x|x|
0%
f(x)=[sinx]
0%
f(x)=\dfrac{|x-1|}{x-1}
\displaystyle \int_{0}^{\infty }\frac{x}{\left ( 1+x \right )\left ( 1+x^{2} \right )}dx
Report Question
0%
\displaystyle \frac{\pi }{4}
0%
\displaystyle \frac{\pi }{2}
0%
is sme as
\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( 1+x \right )\left ( 1+x^{2} \right )}
0%
cannot be evaluated
Explanation
Let
I=\int _{ 0 }^{ \infty }{ \cfrac { xdx }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } }
Using partial fraction
=\int _{ 0 }^{ \infty }{ \left( \cfrac { x+1 }{ 2\left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2\left( 1+x \right) } \right) dx } \\ ={ \left( \lim _{ b\rightarrow \infty }{ \cfrac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+x \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }x } \right) }_{ 0 }^{ b }\\ =\lim _{ b\rightarrow \infty }{ \left( \cfrac { 1 }{ 2 } \log { \left( 1+b \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+b \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }b \right) } \\ =\cfrac { \pi }{ 4 }
State true or false:
The average value of the function
f(x) = sin^2xcos^3x
on the interval
[ -\pi ,\pi ]
is 0.
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0%
True
0%
False
Explanation
\int _{ -\pi }^{ \pi }{ f\left( x \right) dx } =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos ^{ 3 }{ x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos { x } dx } -\int _{ -\pi }^{ \pi }{ \sin ^{ 4 }{ x } \cos { x } dx }
\displaystyle ={ \left[ \frac { \sin ^{ 3 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }-{ \left[ \frac { \sin ^{ 5 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }=0
I_{1}
is equal to
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0%
\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta
0%
\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta
0%
\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta
0%
\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta
Explanation
\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx
x=sin\theta, dx=cos\theta d\theta
\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta
\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta
Ans:
A
Evaluate
\displaystyle \int_{0}^{\pi /2} \frac{dx}{2+\sin 2x}
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0%
\displaystyle \frac{2\pi }{{3}}
0%
\displaystyle \frac{\pi }{{3}}
0%
\displaystyle \frac{2\pi }{{5}}
0%
None of these
Explanation
\displaystyle \int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\sin { 2x } } } =\int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\frac { 2\tan { x } }{ 1+\tan ^{ 2 }{ x } } } } dx
\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 2x }{ \frac { \sec ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } +\tan { x+1 } } dx }
Put
\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=dt
\displaystyle I=\frac { 1 }{ 2 } \int { \frac { 1 }{ { { t }^{ 2 }+t+1 } } } dt.=\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( t+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 3 }{ 2 } } } =\frac { 2\pi }{ 3 }
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect
Explanation
Reason is true
Assertion
Put
\displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^{2}}dt
\displaystyle l=-\int_{3}^{1/3}tcosec ^{99}\left ( \frac{1}{t}-t \right )\frac{1}{t^{2}}dt
\displaystyle l=-\int_{1/3}^{3}\frac{1}{t}cosec^{99}\left ( t-\frac{1}{t} \right )dt
\displaystyle l=-1
\Rightarrow 2l=0
\Rightarrow l=0
\displaystyle \int_{\tfrac{1}{\sqrt{3}}}^{0}\dfrac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}
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0%
\displaystyle - \tan^{-1} \dfrac{1}{2}
0%
\displaystyle \tan^{-1} 1
0%
\displaystyle - \tan^{-1} \dfrac{1}{3}
0%
\displaystyle \tan^{-1} \dfrac{1}{\sqrt 2}
Explanation
I=\displaystyle \int_{\frac{1}{\sqrt{3}}}^{0}\frac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}
Substitute
x=\displaystyle \frac{1}{z}
\Rightarrow\displaystyle dx=-\frac{1}{z^2}dz
I=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -\dfrac { 1 }{ { z }^{ 2 } } dz }{ \left( \dfrac { 2 }{ { z }^{ 2 } } +1 \right) \sqrt { \dfrac { 1 }{ { z }^{ 2 } } +1 } }
=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -zdz }{ \left( 2+{ z }^{ 2 } \right) \sqrt { 1+{ z }^{ 2 } } }
Substitute
\sqrt { 1+{ z }^{ 2 } } =t
\Rightarrow z^2+1=t^2
\Rightarrow zdz=tdt
I=\displaystyle \int _{ 2 }^{ \infty } \frac { -tdt }{ \left( 1+{ t }^{ 2 } \right) t }
I=[\cot^{-1}t]_{2}^{\infty}
\Rightarrow I=-\cot^{-1}2
\Rightarrow I=-\tan^{-1}{\dfrac{1}{2}}
The value of definite integral
\displaystyle \int_{\infty }^{0}\frac{Ze^{-z}}{\sqrt{1-e^{-2z}}}dz
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0%
\displaystyle -\frac{\pi }{2}ln2
0%
\displaystyle \frac{\pi }{2}ln2
0%
\displaystyle -\pi ln2
0%
\displaystyle \pi ln\frac{1}{\sqrt{2}}
Explanation
\displaystyle l=\int_{\infty }^{0}\frac{ze^{-z}}{\sqrt{1-e^{-2z}}}dz
put
\displaystyle e^{-z}=\sin \theta
\displaystyle l=-\int_{0}^{\pi /2}\frac{ln\left ( \sin \theta \right )\left ( -\cos \theta \right )d\theta }{\sqrt{1-\sin ^{2}\theta }}=\int_{0}^{\pi /2}ln\sin \theta d\theta
\displaystyle \frac{-\pi }{2}ln2
The derivative of
F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}(x >0)
is
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0%
\displaystyle\frac{1}{3log x}
0%
\displaystyle\frac{1}{3log x}-\frac{1}{2log x}
0%
(log x)^{-1}x(x-1)
0%
\displaystyle\frac{3x^2}{log x}
Explanation
Given,
F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}
\Rightarrow F'(x)=\displaystyle\frac{d}{dx}F(x)
=\displaystyle\frac{d}{dx}(x^3)\cdot\frac{1}{log x^3}-\frac{d}{dx}(x^2)\cdot\frac{1}{log x^2}
=\displaystyle\frac{3x^2}{3log x}-\frac{2x}{2log x}=\frac{x^2-x}{log x}
=x(x-1)(log x)^{-1}
If
\displaystyle \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0
, then
Report Question
0%
1<\alpha <2
0%
\alpha <2
0%
0<\alpha<1
0%
\alpha=0
Explanation
\displaystyle\because \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0,
\therefore { e }^{ { x }^{ 2 } }\left( x-\alpha \right)
must be
+ve
and
-ve
both for
x\in (0,1)
i.e.
{ e }^{ x }\left( x-\alpha \right) =0
for one
x\in (0,1)
\therefore \alpha \in(0,1)
Trapezoidal rule for evaluation of
\displaystyle\int _{ a }^{ b }{ f(x)dx }
requires the interval
(a,b)
to be divided into
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0%
2n
sub-intervals of equal width
0%
any number of sub intervals of not equal width
0%
any number of sub intervals of equal width
0%
3n
sub-intervals of equal width
Explanation
Trapezoidal rule for evaluation of
\int _{ a }^{ b }{ f(x)dx }
requires the interval
(a,b)
to be divided into any number of sub-intervals of equal width.
\displaystyle \int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } }
is equal to
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0%
\displaystyle \frac { \pi }{ 2 } +1
0%
\displaystyle \frac { \pi }{ 2 } -1
0%
\displaystyle 1-\frac { \pi }{ 2 }
0%
none of these
Explanation
Let
\displaystyle I=\int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\int _{ 0 }^{ \pi /2 }{ \frac { a\cos { \theta } d\theta }{ a+a\cos { \theta } } }
By putting
x=a\sin { \theta } \Rightarrow dx=a\cos { \theta } d\theta
\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \cos { \theta } }{ 1+\cos { \theta } } d\theta } =\int _{ 0 }^{ \pi /2 }{ \left( 1-\frac { 1 }{ 1+\cos { \theta } } \right) d\theta }
\displaystyle =\int _{ 0 }^{ \pi /2 }{ d\theta } -\frac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \sec ^{ 2 }{ \frac { \theta }{ 2 } } d\theta }
\displaystyle ={ \left[ \theta -\tan { \frac { \theta }{ 2 } } \right] }_{ 0 }^{ \pi /2 }=\frac { \pi }{ 2 } -\tan { \frac { \pi }{ 4 } } =\left( \frac { \pi }{ 2 } -1 \right)
The value of the integral
\int_0^{\dfrac{\pi}{2}}\,sin^5x\,dx
is
Report Question
0%
\dfrac{4}{15}
0%
\dfrac{8}{5}
0%
\dfrac{8}{15}
0%
\dfrac{4}{5}
Explanation
I=\int_0^{\dfrac{\pi}{2}}sin^4x.sin\,x\,dx
=\int_0^{\dfrac{\pi}{2}}(1-cos^2\,x)^2\,sin\,x\,dx
Put
cos\,x=t
\Rightarrow\;-sin\,x\,dx=dt
=-\int_1^0(1-t^2)^2dt=\int_0^(t^4-2t^2+1)dt
[\because\,-\int_a^bf(x)dx=\int_a^bf(x)dx]
=\dfrac{1}{5}(t^5)_0^1-\dfrac{2}{3}(t^3)_0^1+(t)_0^1
=\dfrac{1}{5}-\dfrac{2}{3}+1=\dfrac{3-10+15}{15}=\dfrac{8}{15}
Using Trapezoidal rule and following table
\int_0^8 {f(x)}dx
is equal to
x
0
2
4
6
8
f(x)
2
5
10
17
26
Report Question
0%
184
0%
92
0%
46
0%
-36
\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}
equals
Report Question
0%
\dfrac {\pi}{3}
0%
\dfrac {2\pi}{3}
0%
\dfrac {\pi}{6}
0%
\dfrac {\pi}{12}
Explanation
\displaystyle \int \frac {d}{1+x^2}=\tan^{-1}x=F(x)
By second fundamental theorem of calculus, we obtain
\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}=F(\sqrt 3)-F(1)
=\tan^{-1}\sqrt 3-\tan^{-1}1
\displaystyle =\frac {\pi}{3}-\frac {\pi}{4}=\frac {\pi}{12}
\displaystyle\int _{ 1 }^{ 3 }{ \dfrac { \cos { \left( \log { x } \right) } }{ x } dx }
is equal to
Report Question
0%
1
0%
\cos { \left( \log { 3 } \right) }
0%
\sin { \left( \log { 3 } \right) }
0%
\dfrac { \pi }{ 4 }
Explanation
\displaystyle \int_{1}^{3} \frac{cos(logx)}{x}dx
Substituting
log\,x = t
\displaystyle dt = \frac{1}{x}dx
\displaystyle \therefore \int_{t_{1}}^{t_{2}} cos\,t\,dt
Lim
: t = log\,1 = 0
to
t = log\,3
\displaystyle \therefore \int_{0}^{log3} cos\,t\,dt \Rightarrow [sint]_{0}^{log3} = sin\,log3-0
\displaystyle \therefore \int_{1}^{3} \frac{cos(log\,x)}{x}dx = sin\,log3
The value of
I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \frac { x }{ 2 } \right) } \right) dx }
is
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0%
\cfrac{1}{n}
0%
\cfrac{n+2}{2n+1}
0%
\cfrac{2n-1}{n}
0%
\cfrac{2n-3}{3n-2}
Explanation
Given
I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\cfrac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \cfrac { x }{ 2 } \right) } \right) dx }
In second integral, put
t=\cfrac{x}{2} \Rightarrow
dx=2dt
\Rightarrow
Also, when
x=0
then
t=0
When
x=\dfrac{\pi}2
then
t=\dfrac{\pi}4
Then
I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ t } \right) } dt
I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) } dx\quad (\because \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(y)dy } )
\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } +\tan ^{ n-1 }{ x } \right) } dx
\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \tan ^{ 2 }{ x+1 } \right) } dx
\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \sec ^{ 2 }{ x } \right) } dx
Put
t=\tan{x}
\Rightarrow dt=\sec ^{ 2 }{ x } dx
Also when
x=0
then
t=0
when
x=\pi /4
, then
t=1
I=\displaystyle \int _{ 0 }^{ 1 }{ { t }^{ n-1 }dt }
={ \left[ \cfrac { { t }^{ n } }{ n } \right] }_{ 0 }^{ 1 }=\cfrac { 1 }{ n }
What is
\displaystyle \int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } } } dx
equal to?
Report Question
0%
\cfrac { { \pi }^{ 2 } }{ 8 }
0%
\cfrac { { \pi }^{ 2 } }{ 32 }
0%
\cfrac { \pi }{ 4 }
0%
\cfrac { \pi }{ 8 }
Explanation
I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { \tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } } } dx
Put
\tan ^{ -1 }{ x } =p\Rightarrow \dfrac { dx }{ 1+{ x }^{ 2 } } =dp
When
x=0, p=0
and
x=1,p=\dfrac{\pi}{4}
\Rightarrow I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ pdp }=\dfrac{p^{2}}{2} =\dfrac { { \pi }^{ 2 } }{ 32}
Hence, B is correct.
What is
\displaystyle\int _{ 1 }^{ 2 }{ \ln { x } dx }
equal to?
Report Question
0%
\ln { 2 }
0%
1
0%
\ln { \left( \dfrac { 4 }{ e } \right) }
0%
\ln { \left( \dfrac { e }{ 4 } \right) }
Explanation
Let,
y=\ln x
then,
x={ e^{ y } }
Now at,
x=1,y=0
x=2,y=\ln 2
\Rightarrow dx={ e }^{ y }.dy
Putting the values, we get
\displaystyle \int_{1}^{2} \ln x dx=\int _{ 0 }^{ \ln2 }{ { e }^{ y }.y.dy }
=\displaystyle { \left[ y.{ e }^{ y } \right] }_{ 0 }^{\ln2 }-\int _{ 0 }^{\ln2 }{ { e }^{ y }dy }
=2\ln 2-1
=2\ln 2-\ln(e)
=\ln\left(\dfrac { 4 }{ e }\right)
Hence, C is correct.
If
\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}
, then
b =
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0%
\tan ^{-1}\, \left (\displaystyle \frac{1}{3} \right )
0%
\displaystyle \frac{\sqrt 3}{2}
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\sqrt 3
0%
1
Explanation
\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}
\Rightarrow [\tan^{-1}x]_0^b=[\tan^{-1}x]_b^{\infty}
\Rightarrow \tan^{-1}b-0=\dfrac{\pi}{2}-\tan^{-1}b
\Rightarrow \tan^{-1}b=\dfrac{\pi}{4}
\Rightarrow b=\tan\dfrac{\pi}{4}=1
Select and write the most appropriate answer from the given alternatives for question :
If
\displaystyle \int^k_0 4x^3dx=16
, then the value of
k
is _____.
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0%
1
0%
2
0%
3
0%
4
Explanation
Given:
\displaystyle \int^K_0 4x^3dx=16
4\left(\dfrac{x^4}{4}\right)^K_0=16
(x^4)^k_0=16
k^4=16
k^4=2^4
\boxed{k=2}
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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