Eamcet Mathematics - Define Integrals and Its Application Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
\(\int^{\pi/2}{tan^2\ x\over 1+tan^2\ x}dx\) is equal to
- (a)
\(\infty\)
- (b)
0
- (c)
\(\pi\over4\)
- (d)
\(\pi\over2\)
\(\int^{\pi\over 2}_0{tan^2x\over 1+tan^2\ x}=\int^{\pi\ 2}_0\ tan^2cos^2x\ dx\)
\(=\int^{\pi\ 2}_0\ sin^2xdx=\int^{\pi\over2}_0{1-cos2x\over 2}dx\)
\(=[{{1\over2}\times-{sin2x\over 4}}]^{\pi\over2}_0\quad={\pi\over4}\)
\(\int^{\pi/4}_0\ {sin\ x+cos\ x\over 9+16\ sin2x }dx\) is equal to
- (a)
\(in\ 3\over 20\)
- (b)
\(In\ 3\over 40\)
- (c)
\(In\ 3\over 60\)
- (d)
\(In\ 3\over 100\)
I = \(\int^{\pi/4}_0{sin\ x+cosx\over 9+16\ sin2x}dx=\int^{\pi/4}_0{\sqrt2cos({\pi\over4}-x)\over 9+16\ sin\ 2x}dx\)
\(= \int^{\pi/4}_0{\sqrt2\ cos[{{\pi\over4}-{({\pi\over4}-x})}]\over 9+16\ sin[2({\pi\over4}-4)]}dx\)
\(=\int^{\pi/4}_0{\sqrt2cosx\over 9+16\ cos2x}dx=\int^{\pi/4}_0{\sqrt2\ cos\ x\ dx\over 25-32\ sin^2x}\)
Put sin x = t
\(\Rightarrow\) cos x dx = dt
\(\therefore\) \(I = \int^{1\over\sqrt2}_0{\sqrt2\over 25-32\ t^2}dt\)
After solving \(I = {In3\over20}\)
\(\int^1_{-1}{x^3+|x|+1\over x^2+2|x|+1}dx\) is equal to
- (a)
ln 2
- (b)
2 ln 2
- (c)
\({1\over2}ln 2\)
- (d)
4 ln 2
\(\int^1_{-1}{x^3+|x|+1\over x^2+2|x|+1}dx\)
\(=\int^1_{-1}{x^3\over x^2+2|x|+1}dx+\int^1_{-1}{|x|+1\over x^2+2|x|+1}dx\)
\(= 0+2\int^1_0\ {|x|+1\over (|x|+1)^2}dx\)
\(= 2\int^1_0{x+1\over (x+1)^2}dx\)
= 2 ln 2 \([\because x>0, from 0\ to\ 1]\)
\(\int^2_{-2}|x\ cos\ \pi x|dx\) is equal to
- (a)
\(8\over \pi\)
- (b)
\(4\over \pi\)
- (c)
\(2\over 7\)
- (d)
\(1\over \pi\)
\(\int^2_{-2}|x\ cos\ \pi x|dx=2\int^2_0|x\ cos\ \pi x|dx\)
\(=2\left\{ \int^{1/2}_0|x\ cos\ \pi x| dx+\int^{3/2}_{1/2} |xcos\ \pi x| dx +\int^{3}_{3/2} |x\ cos\ \ x|dx \right\}\)
\(={8\over \pi}\)
If \(\int^{1}_0\ e^{x^2}(x-\alpha)dx=0, \) then
- (a)
\(1 < \alpha < 2\)
- (b)
\(\alpha < 0\)
- (c)
\(0 < \alpha < 1\)
- (d)
\(\alpha = 0\)
\(\int^1_0e^{x^2}(x-\alpha)dx=0\)
\(f(x)=e^{x^2}(x-\alpha)\) must cross X-axis between x=0 and x=1. Now, f(x) = 0 at \(x=\alpha\),
0 < x < 1
\(\therefore\) \(0<\alpha<1\)
\(\int^{\pi/2}_0{dx\over sin({x-{\pi\over 3}})sin({x-\pi\over 6})}\) is equal to
- (a)
\(4\ log\sqrt3\)
- (b)
\(-4\ log\sqrt3\)
- (c)
\(2\ log\sqrt3\)
- (d)
\(-2\ log\sqrt3\)
Let I = \(\int_{0}^{\pi/2}{{{dx}\over{sin\left( x-{{{\pi}\over{3}}} \right)sin\left( x-{{{\pi}\over{6}}} \right)}}}\)
= \(2\int _{ 2 }^{ \pi /2 }{ \frac { sin\left( \frac { \pi }{ 3 } -\frac { \pi }{ 6 } \right) }{ sin\left( x-\frac { \pi }{ 3 } \right) sin\left( x-\frac { \pi }{ 6 } \right) } dx } \)
= \(2\int _{ 2 }^{ \pi /2 }{ \frac { sin\left[ \left( x-\frac { \pi }{ 6 } \right) -\left( x-\frac { \pi }{ 3 } \right) \right] }{ sin\left( x-\frac { \pi }{ 3 } \right) sin\left( x-\frac { \pi }{ 6 } \right) } dx } \)
= \(2\int _{ 0 }^{ \pi /2 }{ \left[ \frac { sin\left( x-\frac { \pi }{ 6 } \right) cos\left( x-\frac { \pi }{ 3 } \right) -cos\left( x-\frac { \pi }{ 6 } \right) sin\left( x-\frac { \pi }{ 3 } \right) }{ sin\left( x-\frac { \pi }{ 3 } \right) sin\left( x-\frac { \pi }{ 6 } \right) } \right] dx } \)
= \(2\int _{ 0 }^{ \pi /2 }{ \left[ cot\left( x-\frac { \pi }{ 3 } \right) -cot\left( x-\frac { \pi }{ 6 } \right) \right] } dx\)
= \({ 2\left[ log\frac { sin\left( x-\frac { \pi }{ 3 } \right) }{ sin\left( x-\frac { \pi }{ 6 } \right) } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
= - 4 log \(\sqrt{3}\)
If \(f(2-x)=f(2+x)\) and \(f(4-x)=f(4+x)\) for all x and f(x) is a function for which \(\int^2_0f(dx)=5,\) then \(\int^{50}_0\ f(x)\) dx is equal to
- (a)
125
- (b)
\(\int^{46}_{-4}f(x)\ dx\)
- (c)
\(\int^{51}_{1}f(x)\ dx\)
- (d)
\(\int^{52}_{2}f(x)\ dx\)
Given, \(f(2-x)=f(2+x),f(4-x)=f(4+x)\) Now, \(f(4+x)=f(4-x)=f(2+2-x) = \left\{ 2-(2-x) \right\}=f(x)\)
\(\therefore\) \(\int^{50}_0f(x)\ dx=\int^{48}_0f(x)\ dx+\int^{50}_{48}f(x)\ dx\)
\(=12\int^4_0f(x)\ dx+\int^2_0f(x)\ dx\)
\(=12[\int^2_0 f(x)dx+\int^2_0f(4-x)dx]+5\)
\(=24\int^2_0f(x)\ dx+5=24(5)+5=125\)
The value of \(\int^1_0{8log(1+x)\over 1+x^2}dx\) is
- (a)
\({\pi\over 8}\ log\ 2\)
- (b)
\({\pi\over 2}\ log\ 2\)
- (c)
\( log\ 2\)
- (d)
\(\pi\ log 2\)
Let I = \(\int _{ 0 }^{ 1 }{ {{8log(1+x)}\over{(1+{x}^{2})}} } dx\)
Put x = tan \(\theta\) \(\Rightarrow\) dx = sec2 \(\theta\)d\(\theta\)
When x = 0, then tan \(\theta\) = 0
\(\therefore\) \(\theta\) = 0
When x =1, then tan \(\theta\) = 1
\(\Rightarrow\) \(\theta={{\pi}\over{4}}\)
\(\therefore\) I = \(\int _{ 0 }^{ \pi/4 }{ {{8 \quad log[1+tan \theta]}\over{1+{tan}^{2} \theta}} } .{sec}^{2}\theta d\theta\)
\(\Rightarrow\) I = 8 \(\int _{ 0 }^{ \pi/4 }{ log(1+tan \theta) } d \theta\) ..(i)
Using \(\int _{ 0 }^{ a }{ f(x) }dx=\int _{ 0 }^{ a }{ f(a-x) } dx\)
\(\therefore\) I = \(8\int _{0 }^{ \pi/4 }{ log\left\{ 1+tan\left( {{\pi}\over{4}}-\theta \right) \right\} } d\theta\)
= \(8 \int _{ 0 }^{ \pi/4 }{ log} \left\{ 1+{{1-tan \theta}\over{1+tan \theta}} \right\} d\theta\)
\(\Rightarrow\) I = \(8\int _{ 0 }^{ \pi/4 }{ log \left\{ {{2}\over{1+tan\theta}} \right\} } d\theta\) ...(ii)
On adding Eqs. (i) and (ii), we get
2I = \(8\int _{ 0 }^{ \pi/4 }{ \left[ log\left\{ 1+tan \theta \right\} + log\left\{ {{2}\over{1+tan\theta}} \right\} \right] } d\theta\)
\(\Rightarrow\) I = \(4\int_{0}^{\pi/4}{log2d\theta =log2[\theta]^{\pi/4}_{0}}\)
=4 log2.\(\left( {{\pi}\over{4}}-0 \right)=\pi\) log 2
The values of \(\int^a_1\ [x]\ f'(x) dx, a > 1,\) where [x] denotes the greatest integer not exceeding x, is
- (a)
\([a]f(a)- \left\{ {f(x)+f(2)+....+f(|a|)} \right\}\)
- (b)
\([a]f([a])- \left\{ {f(1)+f(2)+....+f(a)} \right\}\)
- (c)
\(af([a])- \left\{ {f(1)+f(2)+....+f(a)} \right\}\)
- (d)
\([a]f(a)- \left\{ {f(1)+f(2)+....+f([a])} \right\}\)
Now, \(\int^a_1[x]f'(x)=dx=\int^2_1f'(x)dx+\int^3_2\ 2f'(x) dx+...+\int^a_[a]f'(x)\ dx\)
\(=[f(x)]^2_1+2[f(x)]^3_2+...+[a][f(x)]^a_{[a]}\)
\(=f(2)-f(1)+2f(3)-2f(2)+...+[a]f(a)-[a]f([a])\)
\(=[a]f(a)- \left\{ {f(1)+f(2)+....+f([a])} \right\}\)
If \(I_1 = \int^1_0\ 2^{x^2} dx,\) \(I_2=\int^1_0\ 2^{x^3}\ dx,\) \(I_3\int^{2}_1{2^{x^2}}\ dx\) and \(I_4=\int^2_1\ 2^{x^3}dx,\) then
- (a)
\(I_3>I_4\)
- (b)
\(I_3>I_4\)
- (c)
\(I_I>I_2\)
- (d)
\(I_2>I_1\)
Given, \(I_1=\int^1_0\ 2^{x^2}\ dx, I_2=\int^1_0\ 2^{x^2}dx\)
\(I_3=\int^2_1\ 2^{x^2}dx\) and \(I_4=\int^2_12^{x^2}dx\)
Since, \(2^{x^3}>2^{x^2}\) for \(0
\(\therefore\) \(\int^1_0\ 2^{x^3}\ dx< \int^1_0 2^{x^2}dx\) and \(\int^2_12^{x^3}dx>\int^2_1 2^{x^2}\ dx\)
\(\Rightarrow\) \(I_2
The value of \(\int _{ x }^{ 11 }{ \left[ x \right] ^{ 3 } } \) .dx, where [.] denotes the greatest integer function, is
- (a)
0
- (b)
14400
- (c)
2200
- (d)
3025
\(\int _{ 0 }^{ 11 }{ \left[ x \right] } ^{ 3 }dx=\sum _{ r=0 }^{ 10 }{ \int _{ r }^{ r+1 }{ \left[ x \right] ^{ 3 }dx } = } \sum _{ r=0 }^{ 10 }{ \int _{ r }^{ r+1 }{ r^{ 3 }dx } } \)
\(=\sum _{ r=0 }^{ 10 }{ (r^{ 3 }) } ={ 0 }^{ 3 }+{ 1 }^{ 3 }+{ 2 }^{ 3 }+....{ 10 }^{ 3 }\)
\(=\left( \frac { 10.11 }{ 2 } \right) ^{ 2 }=({ 55) }^{ 2 }\)
=3025
If \(I_{ 1 }=\int _{ 0 }^{ \pi /2 }{ cos(sinx)dx; } \quad I_{ 2 }\int _{ 0 }^{ \pi /2 }{ (sin(cosx)dx } \quad and\quad I_{ 3 }\int _{ 0 }^{ \pi /2 }{ cosxdx } \), then
- (a)
I1>I2>I3
- (b)
I2>I3>I1
- (c)
I3>I1>I2
- (d)
I1>I3>I2
∵ x>0
∴ sinx<x ⇒ cos(sinx)>cosx ....(i)
Also ∵ 0<x<\(\frac { \pi }{ 2 } \)
∴ 1>cosx>0
Now, sinx<x for x\(\in \)\(\left( 0,\frac { \pi }{ 2 } \right) \)
∴ sin(cos x)<cos x ....(ii)
∴ from Eqs. (i) and (ii), we get
cos(sinx)>cosx>sin(cosx)
or \(\int _{ 0 }^{ \pi /2 }{ cos(sinx)dx> } \int _{ 0 }^{ \pi /2 }{ cosx\quad dx } >\int _{ 0 }^{ \pi /2 }{ sin(cosx)dx } \)
⇒ I1>I3>I2
The value of the integral \(\int _{ a }^{ b }{ \frac { |x| }{ x } dx } ,a
- (a)
b-a
- (b)
a-b
- (c)
b+a
- (d)
none of the above
Case I : 0
then \(\int _{ a }^{ b }{ \frac { \left| x \right| }{ x } } dx=b-a\)
Case II: a<0
\(=\int _{ a }^{ 0 }{ (-1)dx+\int _{ 0 }^{ b }{ (1)dx=a+b } } \)
Case III: a
\(=a-b\)
The value of the integral
\(\int _{ -10 }^{ 0 }{ \frac { \left| \frac { 2[x] }{ 3x-[x] } \right| }{ \left( \frac { 2[x] }{ 3x-[x] } \right) } } \)dx, where [.] denotes the greatest integer function, is
- (a)
0
- (b)
-10
- (c)
10
- (d)
none of these
Let \(f(x)=\frac { 2[x] }{ 3x-[x] } \)
It is clear that f(x) is not defined if x=0 and if 3x[x]
So, in (-10,0), f is not defined at x=-1/3
Case I: \(x\in \left( -10,-\frac { 1 }{ 3 } \right) \)
[x]<0 and 3x-[x]<0
so \(\frac { [x] }{ 3x-[x] } >0\)
\(\therefore \quad \left| \frac { 2[x] }{ 3x-[x] } \right| =\frac { 2[x] }{ 3x-[x] } \)
\(\Rightarrow \quad \int _{ -10 }^{ -1/3 }{ 1\quad dx } =-\frac { 1 }{ 3 } +10=\frac { 29 }{ 3 } \)
Case II: \(x\in \left( -\frac { 1 }{ 3 } ,0 \right) \)
[x]<0 and 3x-[x]>0
so \(\frac { [x] }{ 3x-[x] } <0\)
\(\therefore \quad \left| \frac { 2[x] }{ 3x-[x] } \right| =-\frac { 2[x] }{ 3x-[x] } \)
Hence, \(\int _{ -1/3 }^{ 0 }{ \frac { |f(x)| }{ f(x) } dx } =\int _{ -10 }^{ -1/3 }{ \frac { |f(x)| }{ f(x) } dx } +\int _{ -1/3 }^{ 0 }{ \frac { |f(x)| }{ f(x) } } \)
\(=\frac { 29 }{ 3 } -\frac { 1 }{ 3 } =\frac { 28 }{ 3 } \)
The value of \(\int _{ 0 }^{ 4 }{ \{ \sqrt { x } \} } \)dx, where {.} denotes the fractional part function, is
- (a)
1/3
- (b)
1
- (c)
5/3
- (d)
7/3
\(\int _{ 0 }^{ 4 }{ \{ \sqrt { x\} dx } =\int _{ 0 }^{ 4 }{ (\sqrt { x } -[\sqrt { x } ]) } } dx\)
\(=\int _{ 0 }^{ 4 }{ (\sqrt { x } dx } -\int _{ 0 }^{ 4 }{ [\sqrt { x } ]dx } \)
\(=\frac { 2 }{ 3 } \{ x^{ 3/2 }\} _{ 0 }^{ 4 }-\left( \int _{ 0 }^{ 1^{ 2 } }{ [\sqrt { x } ]dx } +\int _{ 1^{ 2 } }^{ 2^{ 2 } }{ [\sqrt { x } dx } \right) \)
=\(\frac { 2 }{ 3 } \).8-[0+1.(22-12)]
=\(\frac { 16 }{ 3 } -3\)
=\(\frac { 7 }{ 3 } \)
If \({ I }_{ 1 }=\int _{ x }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } \) and \({ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \frac { dt }{ 1+{ t }^{ 2 } } } \) for x>0 , then
- (a)
I1=I2
- (b)
I1>I2
- (c)
I2>I1
- (d)
I2=cot -1 x-π/4
\(\because \quad \quad { I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \frac { dt }{ 1+{ t }^{ 2 } } } \)
\(=[{ tan }^{ -1 }]_{ 1 }^{ 1/x }\)
\(={ tan }^{ -1 }\left( \frac { 1 }{ x } \right) -{ tan }^{ -1 }\) (1)
\({ I }_{ 2 }={ cot }^{ -1 }(x)-\pi /4\)
and \({ I }_{ 1 }=\int _{ x }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } \)
\(=[{ tan }^{ -1 }t]_{ x }^{ 1 }={ tan }^{ -1 }1-{ tan }^{ -1 }x\)
\(=\frac { \pi }{ 4 } -\left( \frac { \pi }{ 2 } -{ cot }^{ -1 }x \right) \)
\(={ cot }^{ -1 }x-\pi /4\)
\(={ I }_{ 2 }\)
Hence \({ I }_{ 1 }={ I }_{ 2 }\quad and\quad { I }_{ 2 }={ cot }^{ -1 }x-\pi /4\)
If f(x) be a real valued function, f(x)+f(x+4)=f(x+2)+f(x+6), \(g(x)=\int _{ x }^{ x+8 }{ f(t) } dt\). Then g'(x) is equal to
- (a)
f(x)
- (b)
f(x+8)
- (c)
8
- (d)
0
∵ f(x)+f(x+4)=f(x+2)+f(x+6) ....(i)
Replacing x by x+2 in Eq. (i), then
f(x+2)+f(x+6)=f(x+4)+f(x+8), .....(ii)
From Eqs. (i) and (ii), we ger
f(x+8)=f(x)
⇒ f(x+8)-f(x)=0 ....(iii)
Now, \(g(x)=\int _{ x }^{ x+8 }{ f(t) } dt\)
∴ g'(x)=f(x+8)-f(x)
=0 [from Eq. (iii)]
If \(I=\int _{ 0 }^{ 1 }{ \sqrt { (1+{ x }^{ 3 }) } } dx\) then
- (a)
\(I<1\)
- (b)
\(I\neq \sqrt { 5 } /2\)
- (c)
\(I<\sqrt { 7 } /2\)
- (d)
none of these
Since \(I=\int _{ 0 }^{ 1 }{ \sqrt { (1+{ x }^{ 3 }) } } dx\)
\(\because \quad \quad \quad 0<x<1\)
\(\therefore \quad \quad \quad { x }^{ 3 }<x\)
or \(\sqrt { (1+{ x }^{ 3 }) } <\sqrt { (x+1) } \)
\(\int _{ 0 }^{ 1 }{ \sqrt { (1+{ x }^{ 3 }) } } dx<\int _{ 0 }^{ 1 }{ \sqrt { (1+{ x }) } } dx\)
\(\Rightarrow \quad I<\frac { 2 }{ 3 } [(1+x)^{ 3/2 }]_{ 0 }^{ 1 }\)
\(\Rightarrow \quad I<\frac { 2 }{ 3 } (2\sqrt { 2 } -1)\)
Hence \(I<1\quad and\quad I<\sqrt { 7 } /2\)
The value of \(\frac { \int _{ 0 }^{ n }{ [x]dx } }{ \int _{ 0 }^{ n }{ \{ n\} dx } } \) is, (where [x} and {x} denotes the integral part and fractional part functions of x and n\(\in \)N)
- (a)
n+2
- (b)
n+1
- (c)
n
- (d)
n-1
\(\because \int _{ 0 }^{ n }{ [x]dx } =\int _{ 0 }^{ n }{ (x-\{ x\} )dx= } \int _{ 0 }^{ n }{ x } dx-\int _{ 0 }^{ n }{ \{ x\} } dx\)
\(=\int _{ 0 }^{ n }{ x } dx-n\int _{ 0 }^{ n }{ \{ x\} } dx\)
\(=\int _{ 0 }^{ n }{ x } dx-n\int _{ 0 }^{ n }{ x } dx\)
\(=\left( \frac { { n }^{ 2 } }{ 2 } -\frac { n }{ 2 } \right) \)
\(\therefore \frac { \int _{ 0 }^{ n }{ [x]dx } }{ \int _{ 0 }^{ n }{ \{ x\} } dx } =\frac { \left( \frac { { n }^{ 2 } }{ 2 } -\frac { n }{ 2 } \right) }{ \frac { n }{ 2 } } =n-1\)
The value of \(\int _{ -1 }^{ 15 }{ } \)sgn ({x} dx, where {.} denotes the fractional part function, is
- (a)
8
- (b)
16
- (c)
24
- (d)
0
\(\int _{ -1 }^{ 15 }{ sgn(\{ x\} )dx } =\int _{ 0 }^{ 16 }{ sgn(\{ x-1\} )dx } \)
\(=\int _{ 0 }^{ 16 }{ sgn(\{ x\} ) } dx=16\int _{ 0 }^{ 1 }{ sgn } (\{ x\} )dx=16\int _{ 0 }^{ 1 }{ sgn(x) } dx\)
=\(=16\int _{ 0 }^{ 1 }{ 1.dx } \)=16
If \(\int _{ 0 }^{ 1 }{ \frac { e^{ t }dt }{ t+1 } =a } \), then \(\int _{ b-1 }^{ b }{ \frac { e^{ -t }dt }{ t-b-1 } =a } \) is equal to
- (a)
ae-b
- (b)
-ae-b
- (c)
-be-a
- (d)
aeb
\(\because \quad a=\int _{ 0 }^{ 1 }{ \frac { { e }^{ t }dt }{ t+1 } } \quad \quad ...(i)\)
\(\therefore \quad \int _{ b-1 }^{ b }{ \frac { e^{ -t }dt }{ t-b-1 } } \)
∴ dt=-dy
Then, \(=-e^{ -b }\int _{ 0 }^{ 1 }{ \frac { e^{ y }dy }{ y+1 } =-ae^{ -b } } \) [from Eq(i)]
If \(G(x,t)=\begin{cases} x(t-1)\quad where\quad x\le t \\ t(x-1)\quad where\quad t<x \end{cases}\) and if t is continuous function of x in[0,1] let g(x) ,then
- (a)
g(0)=1
- (b)
g(0)=0
- (c)
g(1)=1
- (d)
g'' (x)=f(x)
For x=0
\(G(0,t)=0,t\ge 0\)
\(\therefore \quad g(0)=\int _{ 0 }^{ 1 }{ f(t)1G(0,t)dt } =\int _{ 0 }^{ 1 }{ 0 } .dt=0\)
and for x=1
G(1,t)=0,t<1
g(1) =\(\int _{ 0 }^{ 1 }{ f(t) } G(1,t)dt=0\)
Also, \(g(x)=\int _{ 0 }^{ x }{ f(t) } G(x,t)dt+\int _{ x }^{ 1 }{ f(t) } G(x,t)dt\)
\(\int _{ 0 }^{ x }{ f(t).t(x-1) } dt+\int _{ 0 }^{ x }{ f(t).x(t-1) } dt\)
\({ g }^{ ' }(x)=(x-1).xf(x)+\int _{ 0 }^{ x }{ t\quad f(t)dt } .1+x\{ 0-(x-1)f(x)\} +\int _{ x }^{ 1 }{ (t-1) } f(t)dt.1\)
\(=\int _{ 0 }^{ x }{ t } f(t)dt+\int _{ x }^{ 1 }{ (t-1) } f(t)dt\)
\(\therefore \quad { g }^{ '\quad ' }(x)=xf(x)+\{ 0-(x-1)f(x)\} \)
\(=f(x)\)
For any t\(\in \)R and f be a continuous function,
Let \({ I }_{ 1 }=\int _{ sin^{ 2 }t }^{ 1+cos^{ 2 }t }{ xf(x2-x))dx } \) and \({ I }_{ 2 }=\int _{ sin^{ 2 }t }^{ 1+cos^{ 2 }t }{ f(x(2-x))dx } \quad \) then \(\frac { { I }_{ 1 } }{ { I }_{ 2 } } \)is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Using property \(\int _{ x }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx } \)
Then, \({ I }_{ 1 }=\int _{ sin^{ 2 }t }^{ 1+cos^{ 2 }t }{ (2-x)f\{ 2-x)(2-(2-x)\} dx } \)
\(=\int _{ sin^{ 2 }t }^{ 1+cos^{ 2 }t }{ (2-x)f((2-x)xdx } \)
=2I2-I1
⇒ 2I1=2I2
∴ \(\frac { { I }_{ 1 } }{ I_{ 2 } } \)=1
\(\int _{ 0 }^{ \pi /4 }{ sinx } d(x-[x])\) is equal to (where [.] denotes the greatest integer function)
- (a)
1/2
- (b)
\(1-\frac { 1 }{ \sqrt { 2 } } \)
- (c)
1
- (d)
none of these
\(\because \quad 0\le x<\pi /4\)
∴ [x]=0
Then, \(\int _{ 0 }^{ \pi /4 }{ sinx } d(x-[x]=\int _{ 0 }^{ \pi /4 }{ sinx } dx\)
\(=\{ cosx\} _{ 0 }^{ \pi /4 }\)
\(=-\left( \frac { 1 }{ \sqrt { 2 } } -1 \right) \)
\(=1-\frac { 1 }{ \sqrt { 2 } } \)
The value of \(\int _{ 1/e }^{ tan\quad x }{ \frac { t\quad dt }{ 1+{ t }^{ 2 } } } +\int _{ 1/e }^{ cot\quad x }{ \frac { dt }{ t(1+{ t }^{ 2 }) } } \) is
- (a)
\(\frac { 1 }{ 2+{ tan }^{ 2 }x } \)
- (b)
1
- (c)
\(\frac { \pi }{ 4 } \)
- (d)
\(\frac { 2 }{ \pi } \int _{ -1 }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } \)
Let \(I=\int _{ 1/e }^{ cot\quad x }{ \frac { dt }{ t(1+{ t }^{ 2 }) } } \)
Put \(t=\frac { 1 }{ z } \)
\(\therefore \quad dt=-\frac { 1 }{ { z }^{ 2 } } dz\)
\(\therefore \quad I=\int _{ e }^{ tanx }{ \frac { -\frac { 1 }{ { z }^{ 2 } } dz }{ \frac { 1 }{ z } \left( \frac { 1 }{ { z }^{ 2 } } \right) } } =\int _{ tan\quad x }^{ e }{ \frac { z\quad dz }{ ({ z }^{ 2 }+1) } } \)
\(=\int _{ tan\quad x }^{ e }{ \frac { t\quad dt }{ (1+t^{ 2 }) } } \) (By property)
\(\therefore \quad \int _{ 1/e }^{ tan\quad x }{ \frac { t\quad dt }{ (1+{ t }^{ 2 }) } } +\int _{ 1/e }^{ cot\quad x }{ \frac { dt }{ t(1+{ t }^{ 2 }) } } \)
\(=\int _{ 1/e }^{ tan\quad x }{ \frac { t\quad }{ (1+{ t }^{ 2 }) } } dt+\int _{ tan\quad x }^{ e }{ \frac { t\quad }{ (1+{ t }^{ 2 }) } } dt\)
\(=\int _{ 1/e }^{ e }{ \frac { t\quad dt }{ (1+{ t }^{ 2 }) } } =\frac { 1 }{ 2 } [In\quad (1+{ t }^{ 2 })]_{ 1/e }^{ e }\)
\(=\frac { 1 }{ 2 } \left\{ In\quad (1+{ e }^{ 2 })-In\left( 1+\frac { 1 }{ { e }^{ 2 } } \right) \right\} \)
\(=\frac { 1 }{ 2 } (In\quad { e }^{ 2 })=1\)
Also \(\frac { 2 }{ \pi } \int _{ -1 }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } =\frac { 4 }{ \pi } \int _{ 0 }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } =\frac { 4 }{ \pi } .{ tan }^{ -`1 }1\)
\(=\frac { 4 }{ \pi } .\frac { \pi }{ 4 } =1\)
If \(\int _{ 1/2 }^{ 2 }{ \frac { 1 }{ 2 } } cosec^{ 101 }\left( x-\frac { 1 }{ x } \right) \)dx=k, then the value of k is
- (a)
1
- (b)
1/2
- (c)
0
- (d)
1/101
\(\int _{ 1/2 }^{ 2 }{ \frac { 1 }{ 2 } } cosec^{ 101 }\left( x-\frac { 1 }{ x } \right) dx=k\)
Put \(x=\frac { 1 }{ t } ,\quad \therefore \quad dx=-\frac { 1 }{ { t }^{ 2 } } dt\)
\(\int _{ 2 }^{ 1/2 }{ t. } cosec^{ 101 }\left( \frac { 1 }{ t } -t \right) .\left( -\frac { 1 }{ t^{ 2 } } \right) dt=k\)
-k=k
∴ 2k=0
⇒ k=0
If \(\int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } }dx } =\frac { \sqrt { \pi } }{ 2 } \) and \(\int _{ 0 }^{ \infty }{ e^{ { -ax }^{ 2 } } } dx\), a>0 is
- (a)
\(\frac { \sqrt { \pi } }{ 2 } \)
- (b)
\(\frac { \sqrt { \pi } }{ 2a } \)
- (c)
\(2\frac { \sqrt { \pi } }{ a } \)
- (d)
\(\frac { 1 }{ 2 } \sqrt { \frac { \pi }{ a } } \)
\(I=\int _{ 0 }^{ \infty }{ e^{ { -ax }^{ 2 } } } dx\)
Put \(\sqrt { a } x=t\)
\(dx=\frac { dt }{ \sqrt { a } } \)
Then \(I=\frac { 1 }{ \sqrt { a } } \int _{ 0 }^{ \infty }{ e^{ { -t }^{ 2 } } } dt\)
\(=\frac { 1 }{ \sqrt { a } } \int _{ 0 }^{ \infty }{ e^{ { -x }^{ 2 } } } dt\) (by property)
\(=\frac { 1 }{ \sqrt { a } } .\frac { \sqrt { \pi } }{ 2 } \) (given)
\(=\frac { 1 }{ 2 } \sqrt { \left( \frac { \pi }{ a } \right) } \)
Let \({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { c0s }^{ n }x\quad dx } ,n\epsilon N\) ,then
- (a)
\({ I }_{ n-2 }>{ I }_{ n }\)
- (b)
\(n({ I }_{ n-2 }-{ I }_{ n })={ I }_{ n-2 }\)
- (c)
\({ I }_{ n }:{ I }_{ n-1 }=n"(n-1)\)
- (d)
none of these
\(\because \quad { I }_{ n }=\int _{ 0 }^{ n/2 }{ cos^{ n-1 } } x.cos\quad x\quad dx\)
\(=[{ cos }^{ n-1 }x.sin\quad x]_{ 0 }^{ \pi /2 }-\int _{ 0 }^{ \pi /2 }{ (n-1) } { cos }^{ n-2 }\quad x(-sin\quad x)(sin\quad x)dx\)
\(=0+(n-1)\int _{ 0 }^{ \pi /2 }{ cos^{ n-2 } } x(1-cos^{ 2 }\quad x)dx\)
\(=(n-1)({ I }_{ n-2 }-{ I }_{ n })\)
\(\therefore \quad n({ I }_{ n-2 }-{ I }_{ n })={ I }_{ n-2 }\)
Also \(\because \quad 0<x<\pi /2\)
\(\therefore \quad 1>cos\quad x>0\)
\(\therefore \quad { cos }^{ n }x<{ cos }^{ n-2 }x\)
\(\Rightarrow \quad \int _{ 0 }^{ \pi /2 }{ cos^{ n }x\quad dx } <\int _{ 0 }^{ \pi /2 }{ cos^{ n-2 }x\quad dx } \)
or \({ I }_{ n }<{ I }_{ n-2 }\)
\(\Rightarrow \quad { I }_{ n-2 }>{ I }_{ n }\)
The value of integral \(\int _{ 0 }^{ \pi /4 }{ \frac { dx }{ (a^{ 2 }cos^{ 2 }x+b^{ 2 }sin^{ 2 }x) } } \) is
- (a)
\(\frac { 1 }{ ab } \tan ^{ -1 }{ \left( \frac { b }{ a } \right) } (a>0,b>0)\)
- (b)
\(rac { 1 }{ ab } \tan ^{ -1 }{ \left( \frac { b }{ a } \right) } (a<0,b<0)\)
- (c)
\(\frac { \pi }{ 4 } (a=1,b=1)\)
- (d)
none of the above
Let \(I=\int _{ 0 }^{ \pi /4 }{ \frac { { sec }^{ 2 }x\quad dx }{ { a }^{ 2 }+b^{ 2 }tan^{ 2 }x } } \)
Put b tan \(x=t\Rightarrow \sec ^{ 2 }{ x } dx=\frac { dt }{ b } \)
\(\therefore \quad I=\frac { 1 }{ b } \int _{ 0 }^{ b }{ \frac { dt }{ { a }^{ 2 }+t^{ 2 } } } =\frac { 1 }{ ab } \left\{ { tan }^{ -1 }\left( \frac { t }{ a } \right) \right\} _{ 0 }^{ b }\)
\(=\frac { 1 }{ ab } \tan ^{ -1 }{ \left( \frac { b }{ a } \right) } \)
For \(a=b=1\)
\(I={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)
If f(x) satisfies the requirements of Rolle's theorem in [1,2] and f'(x) is continuous in [1,2], then \(\int _{ 1 }^{ 2 }{ f^{ ' }(x) } \)dx is equal to
- (a)
0
- (b)
1
- (c)
3
- (d)
-1
f(1)=f(2)=0
\(\therefore \quad \int _{ 1 }^{ 2 }{ f^{ ' }(x)dx } =\left[ f(x) \right] _{ 1 }^{ 2 }\)
=f(2)-f(1)=0
Let f(x) be a continuous function defined on the closed interval [a,b] ,then
\(\lim _{ n\rightarrow \infty }{ \sum _{ r=0 }^{ n-1 }{ \frac { 1 }{ n } } f\left( \frac { r }{ n } \right) } =\int _{ 0 }^{ 1 }{ f(x)\quad dx } \)
The value of \(\lim _{ x\rightarrow \infty }{ \left\{ \frac { 1 }{ n } +\frac { { n }^{ 2 } }{ (n+1)^{ 3 } } +\frac { { n }^{ 2 } }{ (n+2)^{ 3 } } +..+\frac { 1 }{ 8n } \right\} } \) is
- (a)
5/4
- (b)
3/4
- (c)
5/6
- (d)
3/8
\(\therefore\lim _{ x\rightarrow \infty }{ \left\{ \frac { 1 }{ n } +\frac { { n }^{ 2 } }{ (n+1)^{ 3 } } +\frac { { n }^{ 2 } }{ (n+2)^{ 3 } } +..+\frac { 1 }{ 8n } \right\} } \)
\(=\lim _{ x\rightarrow \infty }{ \left\{ \frac { { n }^{ 2 } }{ (n+0)^{ 3 } } +\frac { { n }^{ 2 } }{ (n+1)^{ 3 } } +\frac { { n }^{ 2 } }{ (n+2)^{ 3 } } +..+\frac { { n }^{ 2 } }{ (n+n)^{ 3 } } \right\} } \)
\(=\lim _{ n\rightarrow \infty }{ \sum _{ r=0 }^{ n }{ \frac { { n }^{ 2 } }{ (n+r)^{ 3 } } } } \)
\(=\lim _{ n\rightarrow \infty }{ \sum _{ r=0 }^{ n }{ \frac { 1 }{ n\left( 1+\frac { r }{ n } \right) ^{ 3 } } } } \)
\(=\int _{ 0 }^{ 1 }{ \frac { dx }{ (1+x)^{ 3 } } } \)
\(=-\left\{ \frac { 1 }{ 8 } -\frac { 1 }{ 2 } \right\} =\frac { 1 }{ 2 } -\frac { 1 }{ 8 } =\frac { 3 }{ 8 } \)
The value of \(\int _{ 0 }^{ 2 }{ } \)[x2-1]dx, where [x] denotes the greatest integer function is given by
- (a)
\(3-\sqrt { 3 } -\sqrt { 2 } \)
- (b)
2
- (c)
1
- (d)
none of these
Let \(I=\int _{ 0 }^{ 2 }{ [x^{ 2 }-1] } dx\)
At x=2 the value of x2-1 is 3
and at x=0 the value of x2-1 is -1
integers in between -1 to 3 are 0,1,2
∴ x2-1=0, x2-1=1 and x2-1=0
Or, x=1, \(x=\sqrt { 2 } ,\)\(x=\sqrt { 3 } \) (∵ x>0)
\(\therefore \quad I=\int _{ 0 }^{ 1 }{ [x^{ 2 }-1]dx } +1\int _{ 1 }^{ \sqrt { 2 } }{ [x^{ 2 }-1]dx } +\int _{ \sqrt { 2 } }^{ \sqrt { 3 } }{ [x^{ 2 }-1]dx } +\int _{ \sqrt { } 3 }^{ 2 }{ [x^{ 2 }-1]dx } \)
\(=(-1)\int _{ 0 }^{ 1 }{ dx } +0+1\int _{ \sqrt { 2 } }^{ \sqrt { 3 } }{ dx+2 } \int _{ \sqrt { 3 } }^{ 2 }{ dx } \)
\(=-1+0+\sqrt { 3 } -\sqrt { 2 } +2(2-\sqrt { 3 } )\)
\(=3-\sqrt { 3 } -\sqrt { 2 } \)
If \(f(x)=cosx-\int _{ 0 }^{ x }{ (x-t) } \)f(t)dt, then f''(x)+f(x) equals
- (a)
-cos x
- (b)
0
- (c)
\(\int _{ 0 }^{ x }{ (x-t) } f(t)dt\)
- (d)
\(-\int _{ 0 }^{ x }{ (x-t) } f(t)dt\)
\(f(x)=cosx-\int _{ 0 }^{ x }{ (x-t } )f(t)dt\)
\(f(x)=cosx-x\int _{ 0 }^{ x }{ f(t) } dt+\int _{ 0 }^{ x }{ t } f(t)dt\)
\(\therefore \quad f^{ ' }(x)=-sinx-\left\{ x\quad f(x)+\int _{ 0 }^{ x }{ f(t) } dt \right\} x\quad f(x)\)
\(=-sinx-\int _{ 0 }^{ x }{ f(t) } dt\)
∴ f''(x)=-cosx-f(x)
⇒ f''(x)+f(x)=-cosx
Let f: R⟶R such that f(x+2y)=f(x)+f(2y)+4xy ∀x,y\(\in \)R and f'(0)=0. If \(I_{ 1 }=\int _{ 0 }^{ 1 }{ f(x) } dx,\quad I_{ 2 }=\int _{ -1 }^{ 0 }{ f(x) } dx\) and \({ I }_{ 3 }=\int _{ 1/2 }^{ 2 }{ f(x) } dx\), then
- (a)
I1=I2>I3
- (b)
I1>I2>I3
- (c)
I1=I2<I3
- (d)
I1<I2<I3
\(\because \quad f^{ ' }(x)=\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(x) }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { f\left( x+2\left( \frac { h }{ 2 } \right) \right) -f(x+2.0) }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { f(x)+f(h)+2xh-f(x)-f(0)-0 }{ h } } \)
=f'(0)+2x=0+2x
∴ =f'(x)=x2+c ⇒ f(0)=0+c=0
∴ c=0
Then f(x)=x2
∴ I1=1/3, I2=1/3 and I3=21/8
Hence, I1=I2<I3
The value of \(\int _{ 0 }^{ \pi /2 }{ \frac { 1+2cosx }{ (2+cosx)^{ 2 } } } \)dx is
- (a)
\(-\frac { 1 }{ 2 } \)
- (b)
2
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
none of these
\( I=\int _{ 0 }^{ \pi /2 }{ \frac { 1+2cosx }{ (2+cosx)^{ 2 } } dx } \)
\(=\int _{ 0 }^{ \pi /2 }{ \frac { cosec^{ 2 }x(1+2\quad cosx) }{ cosec^{ 2 }x(2+cosx)^{ 2 } } } dx\)
\(=\int _{ 0 }^{ \pi /2 }{ \frac { (cosec^{ 2 }x+2\quad cosecx\quad cotx) }{ (2\quad cosecx+cotx)^{ 2 } } } \)
\(=\left[ \frac { 1 }{ 2\quad cosecx+cotx } \right] _{ 0 }^{ \pi /2 }=\frac { 1 }{ 2 } \)
If \(f(x)=\int _{ 2 }^{ x^{ 2 } }{ \frac { (sin^{ -1 }\sqrt { t } )^{ 2 } }{ \sqrt { t } } } dt\) , then the value of (1-x2){f''(x)}2-2f'(x) at x=\(\frac { 1 }{ \sqrt { t } } \) is
- (a)
2-\(\pi \)
- (b)
3+\(\pi \)
- (c)
4-\(\pi \)
- (d)
none of these
\(\because \quad f(x)=\int _{ 2 }^{ x^{ 2 } }{ \frac { (sin^{ -1 }\sqrt { t } )^{ 2 } }{ \sqrt { t } } } dt\quad \)
\(\therefore \quad f^{ ' }(x)=\frac { (sin^{ -1 }x)^{ 2 } }{ x } .(2x-0)\)
f'(x)=2(sin-1x)2 .......(i)
\(f^{ '' }(x)=\frac { 4\quad sin^{ -1 }x }{ \sqrt { (1-x^{ 2 }) } } \)
\(\therefore \quad [f^{ '' }(x)]^{ 2 }=\frac { 16(sin^{ -1 }x)^{ 2 } }{ (1-x^{ 2 }) } \quad \quad \quad ...(ii)\)
From Eqs (i) and (ii), we have
(1-x2)[f''(x)]2-2f'(x)=12(sin-1x)2
\(=12\left( sin^{ -1 }\frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }\)
\(=12\left( \frac { \pi }{ 4 } \right) ^{ 2 }\)
\(=\frac { 3 }{ 4 } { \pi }^{ 2 }\)
The equation
\(\int _{ -\pi /4 }^{ \pi /4 }{ \left( a|sinx|+\frac { b\quad sinx }{ 1+cosx } +c \right) } \)dx=0, where a,b,c are constants, gives a relation between
- (a)
a,b and c
- (b)
and c
- (c)
a and b
- (d)
b and c
\(\int _{ -\pi /4 }^{ \pi /4 }{ a|sin } x|dx+b\int _{ -\pi /4 }^{ \pi /4 }{ \frac { sinx }{ 1+cosx } } dx+\int _{ -\pi /4 }^{ \pi /4 }{ c } dx=0\)
⇒ 2a\(\int _{ 0 }^{ \pi /4 }{ sinx\quad dx } +0+c\left( \frac { \pi }{ 2 } \right) =0\)
⇒ 2a\(\left( 1-\frac { 1 }{ \sqrt { 2 } } \right) +0+\frac { \pi c }{ 2 } =0\) (by property)
The value of \(\int _{ 0 }^{ 1000 }{ e^{ x-\left[ x \right] } } dx\) is ([.] denotes the greatest integer function)
- (a)
1000e
- (b)
1000(e-1)
- (c)
1001(e-1)
- (d)
none of these
∵ ex-[x] is periodic with period 1
\(\therefore \quad =\int _{ 0 }^{ 1000 }{ { e }^{ x-\left[ x \right] } } dx\)=1000\(\int _{ 0 }^{ 1 }{ { e }^{ x-\left[ x \right] } } \)dx (by property)
=1000\(\int _{ 0 }^{ 1 }{ e^{ x }dx } \)
=1000(e-1)
The value of \(\lim _{ n\rightarrow \infty }{ \left( \frac { { 1 }^{ k }+{ 2 }^{ k }+...+{ n }^{ k } }{ { n }^{ k-1 } } \right) } \) is
- (a)
\(\frac { 1 }{ k+2 } \)
- (b)
\(\frac { 1 }{ k+1 } \)
- (c)
\(\frac { 1 }{ k+3 } \)
- (d)
0
\(P=\lim _{ n\rightarrow \infty }{ \left( \frac { { 1 }^{ k }+{ 2 }^{ k }+...+{ n }^{ k } }{ { n }^{ k-1 } } \right) } \)
\(=\lim _{ n\rightarrow \infty }{ \left( \left( \frac { 1 }{ n } \right) ^{ k }+\left( \frac { 2 }{ n } \right) ^{ k }+\left( \frac { 3 }{ n } \right) ^{ k }+....+\left( \frac { n }{ n } \right) ^{ k } \right) } \)
\(=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } } r=1\sum _{ r=1 }^{ n }{ \left( \frac { r }{ n } \right) ^{ k } } \)
\(=\int _{ 0 }^{ 1 }{ x^{ k }dx } \)
\(=\left[ \frac { x^{ k+1 } }{ k+1 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 1 }{ k+1 } \)
The value of the integral \(\int _{ 1/e }^{ e }{ |ln\quad x| } dx\)
- (a)
1-1/e
- (b)
2(1-1/e)
- (c)
e-1-1
- (d)
none of these
\(\int _{ 1/e }^{ e }{ |ln\quad x| } dx=\int _{ 1/e }^{ e }{ |ln\quad x| } dx+\int _{ 1 }^{ e }{ |ln\quad x| } dx\)
\(=\int _{ 1/e }^{ 1 }{ (-ln\quad x) } dx+\int _{ 1 }^{ e }{ (ln\quad x) } dx\)
\(=\{ x(ln\quad x-1)\} _{ 1/e }^{ 1 }+\{ x(ln\quad x-1)\} _{ 1 }^{ e }\)
\(=-\left( -1+\frac { 2 }{ e } \right) +1\)
\(=2-\frac { 2 }{ e } \)
=2(1-1/e)
\(\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } } \sum _{ r=n+1 }^{ 2n }{ ln } \left( 1+\frac { r }{ n } \right) \) equals
- (a)
\(ln\left( \frac { 27 }{ 4e } \right) \)
- (b)
\(ln\left( \frac { 27 }{ e^{ 2 } } \right) \)
- (c)
\(ln\left( \frac { 4 }{ e } \right) \)
- (d)
none of these
Let \(P=\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } } \sum _{ r=n+1 }^{ 2n }{ ln } \left( 1+\frac { r }{ n } \right) \)
\(=\int _{ 1 }^{ 2 }{ log_{ e }(1+x) } dx\)
Put 1+x=t
∴ dx=dt
Then \(P=\int _{ 2 }^{ 3 }{ log_{ e }t } dt\)
\(=\left[ log_{ e }t.t-t \right] _{ 2 }^{ 3 }\)
=(3loge3-3)-(2loge2-2)
\(=log_{ e }\left( \frac { 27 }{ 4 } \right) -1\quad \)
\(={ log }_{ e }\left( \frac { 27 }{ 4 } \right) -{ log }_{ e }e\)
\(={ log }_{ e }\left( \frac { 27 }{ 4e } \right) \)
\(\int _{ a/4 }^{ 3a/4 }{ \frac { \sqrt { x } }{ \sqrt { (a-x)+\sqrt { x } } } } \) dx is equal to
- (a)
a/2
- (b)
a
- (c)
-a
- (d)
none of these
Let \(I=\int _{ a/4 }^{ 3a/4 }{ \frac { \sqrt { x } }{ \sqrt { (a-x)+\sqrt { x } } } } \quad \quad .....(i)\)
\(=\int _{ a/4 }^{ 3a/4 }{ \frac { \sqrt { (a-x) } }{ \sqrt { x } +\sqrt { (a-x) } } } dx\quad \quad .....(ii)\)
(by property)
Adding Eqs. (i) and (ii), we get
\(2I=\int _{ a/4 }^{ 3a/4 }{ 1.dx } =\frac { 3a }{ 4 } -\frac { a }{ 4 } =\frac { 2a }{ 4 } \)
The value of \(\int _{ 0 }^{ \pi /4 }{ (tan^{ n }x } +tan^{ n-2 }x)\)d (x-[x]), (where [.] denotes the greatest integer function) is
- (a)
\(\frac { 1 }{ n-1 } \)
- (b)
\(\frac { 1 }{ n+1 } \)
- (c)
\(\frac { 2 }{ n-1 } \)
- (d)
none of these
\(\int _{ 0 }^{ \pi /4 }{ (tan^{ n }x+tan^{ n-2 }x } )dx=\int _{ 0 }^{ \pi /4 }{ tan^{ n-2 } } x(tan^{ 2 }x+1)\)
\(\begin{pmatrix} \therefore \quad 0<x< & \frac { \pi }{ 4 } \\ \therefore \quad [x]= & 0 \end{pmatrix}\)
\(=\int _{ 0 }^{ \pi /4 }{ tan^{ n-2 }x } .sec^{ 2 }xdx\)
\(=\left\{ \frac { tan^{ n-1 }x }{ n-1 } \right\} _{ 0 }^{ \pi /4 }\)
\(=\frac { 1 }{ (n-1) } \)
If \(I_{ 1 }=\int _{ 0 }^{ x }{ e^{ zx }.e^{ -z^{ 2 } } } dz\) and \(I_{ 2 }=\int _{ 0 }^{ x }{ e^{ -z^{ 2 }/4 } } dz\) then
- (a)
I1=exI2
- (b)
I1=\(e^{ x^{ 2 } }\)I2
- (c)
I1=\(e^{ x^{ 2 }/2 }\)I2
- (d)
none of these
\(\because \quad I_{ 1 }=\int _{ 0 }^{ x }{ e^{ zx }.e^{ -z^{ 2 } } } dz=\int _{ 0 }^{ x }{ e^{ -(z^{ 2 }-zx) } } dz\)
\(\int _{ 0 }^{ x }{ e^{ -\left\{ \left( z-\frac { x }{ 2 } \right) ^{ 2 }-\frac { x^{ 2 } }{ 4 } \right\} } } dz=e^{ x^{ 2 }/4 }\int _{ 0 }^{ x }{ e^{ -\left( z-\frac { x }{ 2 } \right) ^{ 2 } }dz } \)
\(={ e }^{ x^{ 2 }/4 }\int _{ -x/2 }^{ x/2 }{ e^{ -z^{ 2 } }dz } =\frac { 1 }{ 2 } e^{ x^{ 2 }/4 }\int _{ -x }^{ x }{ e^{ -z^{ 2 }/4 } } dz\)
Now, \(={ e }^{ x^{ 2 }/4 }\int _{ 0 }^{ x }{ e^{ -z^{ 2 }/4 } } dz\)
If z=x+3i, then the value of \(\int _{ 2 }^{ 4 }{ \left[ arg\left| \frac { z-i }{ z+i } \right| \right] dx } \) (where [.] denotes the greatest integer function and i=\(\sqrt { -1 } \) , is
- (a)
\(3\sqrt { 2 } \)
- (b)
\(6\sqrt { 3 } \)
- (c)
\(\sqrt { 6 } \)
- (d)
none of these
\(\because \quad \left| \frac { z-i }{ z+i } \right| =\left| \frac { x+2i }{ x+4i } \right| =\frac { \sqrt { (x^{ 2 }+4) } }{ \sqrt { (x^{ 2 }+16) } } \)=purely real
\(\therefore \quad arg\left| \frac { z-i }{ z+i } \right| =0\) (∵ imaginary part=0)
\(\therefore \quad \int _{ 2 }^{ 4 }{ \left[ arg\left| \frac { z-i }{ z+i } \right| \right] dx } =\int _{ 2 }^{ 4 }{ [0] } dx=0\)
The area between the curvey y=4+3x-x2 and x-axis is
- (a)
\(\frac{125}{6}\) sq.units
- (b)
\(\frac{125}{3}\) sq.units
- (c)
\(\frac{125}{2}\) sq.units
- (d)
none of these
We have, y=4+3x-x2, a parabola
putting y=0, we get x2-3x-4=0
⇒ (x-4)(x+1)=0 ⇒ x=-1 or x=4
∴ Required area =\(\int _{ -1 }^{ 4 }{ (4+3x-x^{ 2 })dx } \)
=\(\left[ 4x+\frac { 3x^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 4 }=\frac { 125 }{ 6 } \) sq.units.
The area of the ellipse \(\frac { x^{ 2 } }{ { 4 }^{ 2 } } +\frac { { y }^{ 2 } }{ { 9 }^{ 2 } } \)=1 is
- (a)
6\(\pi\) sq.units
- (b)
\(\frac { \pi ({ a }^{ 2 }+{ b }^{ 2 }) }{ 4 } \)
- (c)
\(\pi\)(a+b) sq.units
- (d)
none of these
Area of the ellipse \(\frac { x^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { y }^{ 2 } } \)=1 between the limits x=4 abd x=b is \(\pi\)ab
∴ Area of the given ellipse =\(\pi\) x 4 x 9 =36\(\pi\) sq.units.
The are abounded by the curve y=f(x), the x-axis and x=1 and x=b is (b-1) is sin (3b+4). Then, f(x) is
- (a)
(x-1)cos(3x+4)
- (b)
sin(3x+4)
- (c)
sin(3x+4)+3(x-1)cos(3x+4)
- (d)
none of these
\(\int _{ 1 }^{ b }{ f(x) } \)dx=(b-1) sin(3b+4)
Area of function=\(\int _{ 1 }^{ x }{ f(x) } \)=(x-1) sin(3x+4)
On differentiating, we ger
f(x)=sin(3x+4)+3(x-1).cos(3x+4)
The area enclosed between the curves y=sin2x and y=cos2x in the interval 0 ≤ x ≤\(\pi\).
- (a)
3 sq.units
- (b)
\(\frac{1}{2}\) sq.units
- (c)
5 sq.units
- (d)
none of these
The area enclosed by the curve y=\(\sqrt { x } \) and x = -\(\sqrt { y } \) , the circle x2 + y2 = 2 above the x-axis, is
- (a)
\(\frac { \pi }{ 4 } sq.units\)
- (b)
\(\frac { 3\pi }{ 2 } sq.units\)
- (c)
\(\pi\) sq. units
- (d)
\(\frac { \pi }{ 2 } sq.units\)
We have, \(y=\sqrt { x } \) ......(i)
\(x=-\sqrt { y } \) ......(ii)
x2 + y2 = 2 ......(iii)
Solving (i) and (iii), we get B(1, 1)
Solving (ii) and (iii), we get A( -1, 1)
Now required area,
\(=\int _{ -1 }^{ 0 }{ \left[ \sqrt { 2-{ x }^{ 2 } } -{ x }^{ 2 } \right] } dx+\int _{ 0 }^{ 1 }{ \left[ \sqrt { 2-{ x }^{ 2 } } -\sqrt { x } \right] } dx\)
\(={ \left[ \frac { x }{ 2 } \sqrt { 2-{ x }^{ 2 } } +{ sin }^{ -1 }\left( \frac { x }{ \sqrt { 2 } } \right) -\frac { { x }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 0 }\)\(+{ \left[ \frac { x }{ 2 } \sqrt { 2-{ x }^{ 2 } } +{ sin }^{ -1 }\left( \frac { x }{ \sqrt { 2 } } \right) -\frac { 2 }{ 3 } { x }^{ 3/2 } \right] }_{ 0 }^{ 1 }=\frac { \pi }{ 2 } sq.units\)
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