IISER Mathematics - Integral Calculus
Exam Duration: 45 Mins Total Questions : 30
The value of the integral \(\int _{ 0 }^{ \pi /4 }{ \frac { \sin { \theta } +\cos { \theta } }{ 9+16\sin { 2\theta } } } d\theta \), is
- (a)
\(\frac { 1 }{ 20 } log\quad 2\)
- (b)
log 2
- (c)
log 3
- (d)
\(\frac { 1 }{ 20 } log\quad 3\)
Let \(f:R\rightarrow R\) be a continuous function.Then the value of the integral
\(\int _{ -\pi /2 }^{ \pi /2 }{ \{ f(x)+f(-x)\} \{ g(x)-g(-x)\} dx } \), is
- (a)
\(\pi \)
- (b)
1
- (c)
-1
- (d)
0
Let F(x) = (f (x) + f(-x)) (g(x)-g(-x))
F(-x) = (f(-x) + f (x)) (g(-x)-g(x))
=-(f (x) + f(-x)) (g (x)-g(-x)) =-F(x)
Thus, F(-x) is an odd function.
\(\because \quad \int _{ -\pi /2 }^{ \pi /2 }{ F(x)dx=0 } \)
Hence, the correct alternative is (d).
If \(\int { \frac { cos\sqrt { x } }{ \sqrt { x } } } dx=ksin\sqrt { x } +c,\) then k equals
- (a)
1
- (b)
\(\sqrt { 2 } \)
- (c)
2
- (d)
\(\frac { 1 }{ \sqrt { 2 } } \)
Let \(I=\int { \frac { cos\sqrt { x } }{ \sqrt { x } } dx } \)
Put, \(\sqrt { x } =t\)
\(\therefore \frac { 1 }{ 2\sqrt { x } } dx=dt\)
\(or\quad \frac { dx }{ \sqrt { x } } =2dt\)
\(I=2\int { cos\quad t\quad dt=2sin\quad t+c } \)
\(or\quad I=2sin\sqrt { x } +c.\)
\(I=2sin\sqrt { x } +c;k=2\)
Hence, the correct alternative is (c).
\(\int { \frac { (1+x){ e }^{ x } }{ { cos }^{ 2 }(x{ e }^{ x }) } } dx,\) equals
- (a)
\(tan[(x+1){ e }^{ x }]+c\)
- (b)
\(tan\left( \frac { { e }^{ x } }{ x+1 } \right) +c\)
- (c)
\(tan(x{ e }^{ x })+c\)
- (d)
\(log(x{ e }^{ x })+c\)
\(I=\int { \frac { (1+x){ e }^{ x } }{ { cos }^{ 2 }({ xe }^{ x }) } dx } \)
Put xex=t
∴ (xex+ex)dx=dt
or (x+1)exdx=dt
\(I=\int { \frac { dt }{ { cos }^{ 2 }t } } =\int { { sec }^{ 2 }t\quad dt } \)
or I = tan t+c = tan (xex) + c
Hence, the correct alternative is (c).
\(\int { \frac { (x+1){ (x+logx) }^{ 2 } }{ 2x } } dx,\) equals
- (a)
\(\frac { 1 }{ 6 } { (x+logx) }^{ 3 }+c\)
- (b)
\(\frac { 1 }{ 3 } { (x+logx) }^{ 3 }+c\)
- (c)
\(\frac { 1 }{ 2 } { (x+logx) }^{ 3 }+c\)
- (d)
\(\frac { 1 }{ 8 } { (x+logx) }^{ 3 }+c\)
\(I=\frac { 1 }{ 2 } \int { \left( 1+\frac { 1 }{ x } \right) } (x+log\quad x)^{ 2 }dx\)
Put x+log x=t
\(\left( 1+\frac { 1 }{ x } \right) dx=dt\)
\(I=\frac { 1 }{ 2 } \int { { t }^{ 2 } } dt=\frac { 1 }{ 6 } { t }^{ 3 }+c\)
or \(I=\frac { 1 }{ 6 } (x+log\quad x)^{ 3 }+c\)
Hence, the correct alternative is (a).
The value of \(\int _{ 0 }^{ 1 }{ x(1-x)^{ 2/3 } } dx\) equals
- (a)
\(\frac { 9 }{ 20 } \)
- (b)
\(\frac { 9 }{ 40 } \)
- (c)
\(\frac { -9 }{ 20 } \)
- (d)
none of these
\(I=\int _{ 0 }^{ 1 }{ x(1-x)^{ 2/3 }dx } \)
\(=\int _{ 0 }^{ 1 }{ (1-x)\left\{ 1-(1-x) \right\} dx } \)
\(=\int _{ 0 }^{ 0 }{ { x }^{ 2/3 } } (1-x)dx=\int _{ 0 }^{ 1 }{ \left( { x }^{ 2/3 }-{ x }^{ 5/3 } \right) } dx\)
\(=\left( \frac { 3 }{ 5 } { x }^{ 5/3 }-\frac { 3 }{ 8 } { x }^{ 8/3 } \right) _{ 0 }^{ 1 }=\frac { 3 }{ 5 } -\frac { 3 }{ 8 } =\frac { 9 }{ 40 } \)
The integral \(\int _{ 0 }^{ \pi }{ xf(sinx)dx } \) equals
- (a)
\(\pi \int _{ 0 }^{ \pi }{ f(sinx)dx } \)
- (b)
\(\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ f(sinx)dx } \)
- (c)
\(\frac { \pi }{ 2 } \int _{ 0 }^{ \pi /2 }{ f(sinx)dx } \)
- (d)
none of these
\(I=\int _{ 0 }^{ \pi }{ (\pi -x) } f\left[ sin(\pi -x) \right] dx\)
\(I=\pi \int _{ 0 }^{ \pi }{ (sin\quad x) } dx-I\)
\(\Rightarrow I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ f(sin\quad x)dx. } \)
If \(\int { f(x)dx=F(x),\quad } \) then \(\int { { x }^{ 3 } } f({ x }^{ 2 })dx,\) equals
- (a)
\(\frac { 1 }{ 2 } \left[ { x }^{ 2 }F({ x }^{ 2 })-\int { F({ x }^{ 2 })dx } \right] \)
- (b)
\(\frac { 1 }{ 2 } \left[ { x }^{ 2 }F({ x }^{ 2 })-\ { F({ x }^{ 2 }) } \right] \)
- (c)
\(\frac { 1 }{ 2 } \left[ { x }^{ 2 }F({ x }^{ 2 })-\int { F({ x }^{ 2 })d({ x }^{ 2) } } \right] \)
- (d)
NONE OF THESE
Let \(I=\int { { x }^{ 3 }f\left( { x }^{ 2 } \right) dx } \)
Put x2=t
2x dx=dt
or xdx=\(\frac { 1 }{ 2 } \)dt
∴ \(I=\frac { 1 }{ 2 } \int { tf(t)dt } \)
\(=\frac { 1 }{ 2 } \left[ t\int { f(t)dt } -\int { 1 } \left( \int { f(t)dt } \right) dt \right] \)
\(=\frac { 1 }{ 2 } \left[ { x }^{ 2 }F\left( { x }^{ 2 } \right) -\int { F(t)dt } \right] \)
\(\left( \because \int { f(x)dx=F(x)or\int { f(t)dt=F(t) } } \right) \)
\(=\frac { 1 }{ 2 } \left[ { x }^{ 2 }F\left( { x }^{ 2 } \right) -\int { F } \left( { x }^{ 2 } \right) d\left( { x }^{ 2 } \right) \right] \)
Hence, the correct alternative is (c).
The value of \(\int _{ 0 }^{ \pi /2 }{ \frac { 1+2cosx }{ (2+cosx)^{ 2 } } } dx\) is
- (a)
\(\frac { \pi }{ 2 } \)
- (b)
\(\pi\)
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
none of these
Put \(\frac { sin\quad x }{ 2+cos\quad x } =t\)
\(\int { \frac { (x-1){ e }^{ x } }{ { (x+1) }^{ 3 } } dx, } \) equals
- (a)
\(\frac { { e }^{ x } }{ (x+1) } +c\)
- (b)
\(\frac { { e }^{ x } }{ { (x+1) }^{ 2 } } +c\)
- (c)
\(\frac { { xe }^{ x } }{ { (x+1) }^{ } } +c\)
- (d)
\(\frac { { x.e }^{ x } }{ { (x+1) }^{ 2 } } +c\)
\(I=\int { \frac { \left( x-1 \right) { e }^{ x } }{ \left( x+1 \right) ^{ 3 } } dx } \)
\(=\int { \frac { \left( x+1-2 \right) { e }^{ x } }{ \left( x+1 \right) } dx } \)
\(=\int { \frac { 1 }{ \left( x+1 \right) ^{ 2 } } } { e }^{ x }dx-\int { \frac { 2 }{ \left( x+1 \right) ^{ 3 } } } { e }^{ x }dx\)
\(=\int { \left( x+1 \right) ^{ -2 } } { e }^{ x }dx-\int { \frac { 2 }{ \left( x+1 \right) ^{ 3 } } } { e }^{ x }dx\)
\(=\left( x+1 \right) ^{ -2 }{ e }^{ x }-\int { (-2)(x+1)^{ -3 } } { e }^{ x }dx-\int { \frac { 2 }{ \left( x+1 \right) ^{ 3 } } } { e }^{ x }dx\)
\(=\frac { { e }^{ x } }{ \left( x+1 \right) ^{ 2 } } +\int { \frac { { 2 } }{ \left( x+1 \right) ^{ 2 } } } { e }^{ x }dx-\int { \frac { 2 }{ \left( x+1 \right) ^{ 2 } } } { e }^{ x }dx\)
\(I=\frac { { e }^{ x } }{ \left( x+1 \right) ^{ 2 } } +c\)
Hence, the correct alternative is (b).
f \(\int { \frac { 4{ e }^{ x }+6{ e }^{ -x } }{ 9{ e }^{ x }-4{ e }^{ -x } } } dx=Ax+Blo{ g }_{ e }(9{ e }^{ 2x }-4)+c,\) then values of A,B,C are given by
- (a)
\(A=\frac { -3 }{ 2 } ,B=\frac { 35 }{ 36 } ,C\) is arbitrary constant
- (b)
\(A=\frac { 35}{ 36 } ,B=\frac { -3 }{ 2 } ,C=\frac { 1 }{ 2 } \)
- (c)
\(A=\frac { -35 }{ 36 } ,B=\frac { 3 }{ 2 } ,C=1\)
- (d)
\(A=\frac { 3 }{ 2 } ,B=\frac { 35 }{ 36 } ,C=2 \)
If 0<a<1, then the integral
\(\int { \frac { dx }{ 1-2acosx+{ a }^{ 2 } } } ,\) equals
- (a)
\(\frac { 2 }{ 1+{ a }^{ 2 } } { tan }^{ -1 }\left( \frac { 1+a }{ 1-a } tan(x/2) \right) +c\)
- (b)
\(\frac { 2 }{ 1-{ a }^{ 2 } } { tan }^{ -1 }\left( \frac { 1+a }{ 1-a } tan(x/2) \right) +c\)
- (c)
\(\frac { 2 }{ 1-{ a }^{ 2 } } { tan }^{ -1 }\left( \frac { 1-a }{ 1+a } tan(x/2) \right) +c\)
- (d)
NONE OF THESE
The integral \(\int _{ 0 }^{ \pi /2 }{ \varphi (cos2x)cosxdx } \) equals
- (a)
\(\sqrt { 2 } \int _{ -\pi /4 }^{ \pi /4 }{ \varphi (cos2x)cosxdx } \)
- (b)
\(\sqrt { 2 } \int _{ -\pi /4 }^{ \pi /4 }{ \varphi (cos2x)sinxdx } \)
- (c)
\(\sqrt { 2 } \int _{ -\pi /4 }^{ \pi /4 }{ \varphi (sin2x)sinxdx } \)
- (d)
none of these
If f(a+b-x)=f(x), then \(\int _{ a }^{ b }{ xf(x)dx } \) is equal to
- (a)
\(\frac { a+b }{ 2 } \int _{ a }^{ b }{ f(b-x)dx } \)
- (b)
\(\frac { a+b }{ 2 } \int _{ a }^{ b }{ f(x)dx } \)
- (c)
\(\frac { b-a }{ 2 } \int _{ a }^{ b }{ f(x)dx } \)
- (d)
none of these
Let \(I=\int _{ a }^{ b }{ xf } (x)dx.\)
Put x=a+b-y;dx=-dy
\(I=-\int _{ b }^{ a }{ (a+b-y)f(a+b-y)dy } \)
\(=-\int _{ a }^{ b }{ (a+b-x)f(a+b-x)dx } \)
\(I=-\int _{ a }^{ b }{ (a+b-x)f(x)dx } \)
[∵ f(a+b-x)=f(x)]
or \(I=(a+b)\int _{ a }^{ b }{ f(x)dx } -\int _{ a }^{ b }{ f(x)dx } \)
or \(I=(a+b)\int _{ a }^{ b }{ f(x)dx } -I\)
or \(I=\frac { a+b }{ 2 } \int _{ a }^{ b }{ f(x)dx } .\)
\(\int _{ 0 }^{ \pi /2 }{ { sin }^{ 6 }x } { cos }^{ 4 }xdx,\) equals
- (a)
\(\frac { 3\pi }{ 256 } \)
- (b)
\(\frac { \pi }{ 128 } \)
- (c)
\(\frac { \pi }{ 64 } \)
- (d)
none of these
Let \(I=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 6 } } x\quad { cos }^{ 4 }x\quad dx\)
\(I=\frac { \left( 5\times 3\times 1 \right) \times \left( 3\times 1 \right) }{ 10\times 8\times 6\times 4\times 2 } \times \frac { \pi }{ 2 } =\frac { 3\pi }{ 256 } \)
Hence, the correct alternative is (a).
\(\int { \frac { x{ ({ tan }^{ -1 }x) }^{ 2 } }{ { (1+{ x }^{ 2) } }^{ 3/2 } } dx, } \) equals
- (a)
\(\frac { [2-{ ({ tan }^{ -1 }x) }^{ 2 }]+2x{ tan }^{ -1 }x+2 }{ \sqrt { 1+{ x }^{ 2 } } } +c\)
- (b)
\(\frac { [2-{ ({ tan }^{ -1 }x) }^{ 2 }]+2x{ tan }^{ -1 }x }{ \sqrt { 1+{ x }^{ } } } +c\)
- (c)
\(\frac { 2-{ ({ tan }^{ -1 }x) }^{ 2 }}{ \sqrt { 1+{ x }^{ 2 } } } \)
- (d)
NONE OF THESE
\(\int { \frac { logx-1 }{ { (logx) }^{ 2 } } dx, } \) equals
- (a)
xlogx+c
- (b)
\(\frac { x }{ { (logx) }^{ 2 } } +c\)
- (c)
\(\frac { x }{ logx } +c\)
- (d)
NONE OF THESE
Let \(I=\int { \frac { log\quad x-1 }{ \left( log\quad x \right) ^{ 2 } } } dx\)
Put log x=t
or x=et
dx=etdt
\(I=\int { \left( \frac { t-1 }{ { t }^{ 2 } } \right) } { e }^{ t }dt\)
\(=\int { { e }^{ t } } \left( \frac { 1 }{ t } -\frac { 1 }{ { t }^{ 2 } } \right) dt\)
\(I=\frac { { e }^{ t } }{ t } +c\)
\(\left( \because \int { { e }^{ x }(f(x))+f'(x)dx={ e }^{ x } } f(x) \right) \)
\(I=\frac { x }{ log\quad x } +c\)
Hence, the correct alternative is (c).
The integral
\(\int { \frac { d\theta }{ 5+4cos\theta } , } \) equals
- (a)
\(\frac { 1 }{ 3 } { tan }^{ -1 }\left( \frac { tan(\theta /2) }{ 3 } \right) +c\)
- (b)
\(\frac { }{ } { tan }^{ -1 }\left( \frac { tan(\theta /2) }{ 3 } \right) +c\)
- (c)
\(\frac { 1 }{ 6 } { tan }^{ -1 }\left( \frac { tan(\theta /2) }{ 3 } \right) +c\)
- (d)
\(\frac { 2 }{ 3 } { tan }^{ -1 }\left( \frac { tan(\theta /2) }{ 3 } \right) +c\)
If \(\int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } (x-\alpha )dx=0\) then
- (a)
\(0<\alpha <1\)
- (b)
\(\alpha <0\)
- (c)
\(1<\alpha <2\)
- (d)
\(\alpha =0\)
\(\int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } (x-\alpha )dx=0\)
\(\int _{ 0 }^{ 1 }{ { xe }^{ { x }^{ 2 } } } dx-\alpha \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx=0\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ { 2x\quad e }^{ { x }^{ 2 } } } dx=\alpha \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx\)
\(\frac { 1 }{ 2 } \left( { e }^{ { x }^{ 2 } } \right) _{ 0 }^{ 1 }=\alpha \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx\)
\(\frac { 1 }{ 2 } (e-1)=\alpha \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx\quad \quad \quad ...(i)\)
Now, \({ e }^{ { x }^{ 2 } }\)- is an increasing function of x is [0, 1]
∴ \(1(1-0)\le \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx\le e(1-0)\)
\(1\le \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx\le e\quad \quad \quad \quad \quad \quad ...(ii)\)
From (i) and (ii); we find a is a positive real number. and also
\(\alpha =\frac { \frac { 1 }{ 2 } (e-1) }{ \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } } } dx } <1\)
∴ 0<\(\alpha \)<1
Hence, the correct alternative is (a).
\(\begin{matrix} lim \\ n\rightarrow \infty \end{matrix}\frac { \pi }{ n } \left( sin\frac { \pi }{ n } +sin\frac { 2\pi }{ n } +......sin(n-1)\frac { \pi }{ n } \right) \) equals
- (a)
0
- (b)
\(\pi \)
- (c)
2
- (d)
none of these
Adding, \(\frac { sin\quad n\pi }{ n } =0,\) to the given expression, we get
Given expression
\(=\underset { n\rightarrow \infty }{ lim } \frac { \pi }{ n } \left( sin\frac { \pi }{ n } +sin\frac { 2\pi }{ n } +...+\frac { sin\quad n\pi }{ n } \right) \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { \pi }{ n } \sum _{ r=1 }^{ n }{ sin } \left( \frac { rn }{ n } \right) \)
\(=\int _{ 0 }^{ \pi }{ sin } x\quad dx=-\left[ cos\quad x \right] _{ 0 }^{ \pi }=2.\)
The area of the positive yriangle formed by the positive x-axis and the normal and tangant to the circle x2+y2=4 at \(\left( 1,\sqrt { 3 } \right) \) is
- (a)
\(\sqrt { 3 } \)
- (b)
\(\frac { 1 }{ \sqrt { 3 } } \)
- (c)
\(2\sqrt { 3 } \)
- (d)
none of these
Area bounded by the curves, \(y=|x|-1\) \(y=-|x|+1\), in square units, is
- (a)
1
- (b)
2
- (c)
\(2\sqrt { 2 } \)
- (d)
4
The area bounded by the curves
\(y=tanx;-\frac { \pi }{ 3 } \le x\le \frac { \pi }{ 3 } \)
\(y=cotx;-\frac { \pi }{ 6 } \le x\le \frac { 3\pi }{ 2 } \)
- (a)
\(log\left( \frac { 2 }{ 3 } \right) \)
- (b)
\(log\left( \frac { 3 }{ 2 } \right) \)
- (c)
log 2
- (d)
none of these
The area bounded by the a-axis part of the curve \(y=\left( 1+\frac { 8 }{ { x }^{ 2 } } \right) \) and the ordinate x=2 and x=4; is divided into the ordinate x=a; then value of a is
- (a)
\(2\sqrt { 2 } \)
- (b)
\(\pm 2\sqrt { 2 } \)
- (c)
\(\pm 2\sqrt { 2 } \)
- (d)
\(\pm 2\)
The value of \(\int { \frac { dx }{ (1+{ x }^{ 2 })\sqrt { 1-{ x }^{ 2 } } } } ,\) is
- (a)
\(\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { \sqrt { 2x } }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
- (b)
\(\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { \sqrt { 2x } }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
- (c)
\(\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { \sqrt { 2x } }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
- (d)
NONE OF THESE
Put x=sin \(\theta \), dx=cos \(\theta \) d\(\theta \)
\(I=\int { \frac { cos\quad \theta \quad d\theta }{ \left( 1+{ sin }^{ 2 }\theta \right) cos\theta } } =\int { \frac { d\theta }{ 1+{ sin }^{ 2 }\theta } } \)
\(=\int { \frac { { sec }^{ 2 }\theta d\theta }{ { sec }^{ 2 }\theta +{ tan }^{ 2 }\theta } } =\int { \frac { { sec }^{ 2 }\theta d\theta }{ 1+2{ tan }^{ 2 }\theta } } \)
\(I=\frac { 1 }{ 2 } \int { \frac { { sec }^{ 2 }\theta d\theta }{ \left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }+{ tan }^{ 2 }\theta } } =\frac { \sqrt { 2 } }{ 2 } { tan }^{ -1 }\left( \frac { tan\theta }{ \frac { 1 }{ \sqrt { 2 } } } \right) +c\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \sqrt { 2 } tan\theta \right) +c\)
\(=\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { x\sqrt { 2 } }{ \sqrt { 1-{ x }^{ 2 } } } \right) +c.\)
Hence, the correct alternative is (c).
\(\int { \frac { dx }{ (x+2)\sqrt { x+3 } } , } \) equals
- (a)
\(log\left( \frac { \sqrt { x+3 } +1 }{ \sqrt { x-3 } -1 } \right) +c\)
- (b)
\(log\left( \frac { \sqrt { x+3 } -1 }{ \sqrt { x-3 } +1 } \right) +c\)
- (c)
\(\frac { 1 }{ 2 } log\left( \frac { \sqrt { x+3 } -1 }{ \sqrt { x-3 } +1 } \right) +c\)
- (d)
\(\frac { 1 }{ 3 } log\left( \frac { \sqrt { x+3 } -1 }{ \sqrt { x-3 } +1 } \right) +c\)
Let \(I=\int { \frac { dx }{ (x+2)\sqrt { x+3 } } } \)
(The integrand is of the form \(\frac { 1 }{ X\sqrt { Y } } ,\) where X and Y both are linear).
Put \(\sqrt { x+3 } =t,\)
or x+3=t2
⇒ x+2=t2-1
dx=2t dt
∴ \(I=\int { \frac { 2t }{ \left( { t }^{ 2 }-1 \right) .t } } dt\)
\(=2\int { \frac { 1 }{ { t }^{ 2 }-1 } } dt=2.\frac { 1 }{ 2 } log\frac { t-1 }{ t+1 } +c\)
\(or\quad I=log\left( \frac { \sqrt { x+3 } -1 }{ \sqrt { x+3 } +1 } \right) +c\) \(\left( \because t=\sqrt { x+3 } \right) \)
Hence, the correct alternative is (b).
The value of the integral
\(I=\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } \), is
- (a)
\(\frac { 1 }{ n+1 } \)
- (b)
\(\frac { 1 }{ n+2 } \)
- (c)
\(\frac { 1 }{ n+1 } \)-\(\frac { 1 }{ n+2 } \)
- (d)
\(\frac { 1 }{ n+1 } \)+\(\frac { 1 }{ n+2 } \)
Let \(I=\int _{ 0 }^{ 1 }{ (1-x)(1-(1-x))^{ n } } dx\)
\(\left( \because \int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x)dx } \right) \)
\(=\int _{ 0 }^{ 1 }{ { x }^{ n } } (1-x)dx=\int _{ 0 }^{ 1 }{ \left( { x }^{ n }-{ x }^{ n+1 } \right) } dx\)
\(I=\left( \frac { { x }^{ n+1 } }{ n+1 } -\frac { { x }^{ n+2 } }{ n+2 } \right) _{ 0 }^{ 1 }=\frac { 1 }{ n+1 } -\frac { 1 }{ n+2 } \)
Hence, the correct alternative is (c).
\(\int _{ 0 }^{ \pi /4 }{ log(1+tan\theta ) } d\theta \), equals
- (a)
\(\frac { \pi }{ 4 } log2\)
- (b)
\(\frac { \pi }{ 8 } log2\)
- (c)
log2
- (d)
\(\frac { \pi }{ 8 } \)
Let \(I=\int _{ 0 }^{ \pi /4 }{ log } (1+tan\theta )d\theta \)
\(=\int _{ 0 }^{ \pi /4 }{ log } \left( 1+tan\left( \frac { \pi }{ 4 } -\theta \right) \right) d\theta \)
\(\left( \because \int _{ 0 }^{ a }{ f(a)dx=f(a-x)dx } \right) \)
\(=\int _{ 0 }^{ \pi /4 }{ log } \left( 1+\frac { tan\frac { \pi }{ 4 } -tan\theta }{ 1+tan\frac { \pi }{ 4 } tan\theta } \right) d\theta \)
\(=\int _{ 0 }^{ \pi /4 }{ log } \left( 1+\frac { 1-tan\theta }{ 1+tan\theta } \right) d\theta \)
\(=\int _{ 0 }^{ \pi /4 }{ log } \left( \frac { 1+tan\theta +1-tan\theta }{ 1+tan\theta } \right) d\theta \)
\(=\int _{ 0 }^{ \pi /4 }{ log } \left( \frac { 2 }{ 1+tan\theta } \right) d\theta \)
\(=\int _{ 0 }^{ \pi /4 }{ log } 2d\theta -\int _{ 0 }^{ \pi /4 }{ log } (1+tan\theta )d\theta \)
\(or\quad I=\frac { \pi }{ 4 } log2-I\)
\(or\quad 2I=\frac { \pi }{ 4 } log2\)
\(\Rightarrow I=\frac { \pi }{ 8 } log2\)
Hence, the correct alternative is (b).
\(\int _{ 0 }^{ 1 }{ |sin2\pi x|dx } \), is equal to
- (a)
0
- (b)
\(-\frac { 1 }{ \pi } \)
- (c)
\(\frac { 1 }{ \pi } \)
- (d)
\(\frac { 2 }{ \pi } \)
Let \(I=\int _{ 0 }^{ 1 }{ \left| sin\quad 2\pi x \right| } dx\)
Now, when 0≤x≤\(\frac { 1 }{ 2 } ,\) sin2 \(\pi x\) ≥0
and \(\frac { 1 }{ 2 } ,\) <x≤1 sin2 \(\pi x\) ≤0
∴ \(I=\int _{ 0 }^{ 1/2 }{ sin\quad 2\pi x } dx+\int _{ 1/2 }^{ 1 }{ -(sin\quad 2\pi x)dx } \)
\(=-\left( \frac { cos\quad 2\pi x }{ 2\pi } \right) _{ 0 }^{ 1/2 }+\left( \frac { cos\quad 2\pi x }{ 2\pi } \right) _{ 1/2 }^{ 1 }\)
\(=\frac { 1 }{ 2\pi } \left[ -(cos\quad \pi -cos\quad 0)+(cos\quad 2\pi -cos\quad \pi ) \right] \)
\(=\frac { 1 }{ 2\pi } \left[ -(-1-1)+(1+1) \right] =\frac { 2 }{ \pi } \)
Hence, the correct alternative is (d).
The value of the integral \(\int _{ \alpha }^{ \beta }{ \frac { dx }{ (x-\alpha )(\beta -x) } } for\alpha <\beta \), is
- (a)
\(\pi\)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
\({ sin }^{ -1 }(\alpha /\beta )\)
- (d)
\({ sin }^{ -1 }\left( \frac { \beta }{ 2\alpha } \right) \)