Mathematics - Integral Calculus
Exam Duration: 45 Mins Total Questions : 30
\(\int { \frac { dx }{ \sqrt { x+1 } +\sqrt { x-1 } } } \); equals
- (a)
\(\frac { 1 }{ 3 } [{ (x+1) }^{ 3/2 }+{ (x-1) }^{ 3/2 }]+c\)
- (b)
\(\frac { 1 }{ 3 } [{ (x+1) }^{ 3/2 }-{ (x-1) }^{ 3/2 }]+c\)
- (c)
\(\frac { 2 }{ 3 } [{ (x-1) }^{ 3/2 }+{ (x+1) }^{ 3/2 }]+c\)
- (d)
\(\frac { 1}{ 3 } [{ (x-1) }^{ 3/2 }-{ (x+1) }^{ 3/2 }]\)
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
The value of integral \(\int _{ 0 }^{ \pi }{ \frac { dx }{ 1+sinx } } \) equals
- (a)
0
- (b)
\(\frac { 1 }{ 2 } \)
- (c)
1
- (d)
\(\frac { 3 }{ 2 } \)
\(\int { \frac { dx }{ sin(x-a)sin(x-b) } } ,\) equals
- (a)
\(\frac { 1 }{ cos(a-b) } log\left| \frac { sin(x-a) }{ sin(x-b) } \right| +c\)
- (b)
\(\frac { 1 }{ sin(a-b) } log\left| \frac { sin(x-a) }{ sin(x-b) } \right| +c\)
- (c)
\(\frac { 1 }{ sin(a-b) } log\left| \frac { sin(x-b) }{ sin(x-a) } \right| +c\)
- (d)
\(\frac { 1 }{ sin(a-b) } log\left| \frac { cos(x-a) }{ cos(x-b) } \right| +c\)
The integral \(\int { \frac { dx }{ x\sqrt { { x }^{ 3 }+1 } } } ,\) equals
- (a)
\(ln\left( \frac { \sqrt { { x }^{ 3 }+1 } +1 }{ \sqrt { { x }^{ 3 }+1 } -1 } \right) \)
- (b)
\(\frac { 2 }{ 3 } ln\left( \frac { \sqrt { { x }^{ 3 }+1 } -1 }{ \sqrt { { x }^{ 3 }+1 } +1 } \right) +c\)
- (c)
\(\frac { 1 }{ 2 } ln\left( \frac { \sqrt { { x }^{ 3 }+1 } -1 }{ \sqrt { { x }^{ 3 }+1 } +1 } \right) +c\)
- (d)
\(\frac { 2 }{ 3 } ln\left( \frac { \sqrt { { x }^{ 3 }+1 }+1 }{ \sqrt { { x }^{ 3 }+1 } -1 } \right) +c\)
The integral \(\int { \frac { dx }{ 9+16{ cos }^{ 2 }x } } ,\) equals
- (a)
\(\frac { 1 }{ 3 } { tan }^{ -1 }\left( \frac { 3tanx }{ 5 } \right) \)
- (b)
\(\frac { 1 }{ 5 } { tan }^{ -1 }\left( \frac { 3tanx }{ 5 } \right) +c\)
- (c)
\( { tan }^{ -1 }\left( \frac { 3tanx }{ 5 } \right) \)
- (d)
\(\frac { 1 }{ 15 } { tan }^{ -1 }\left( \frac { 3tanx }{ 5 } \right) +c\)
\(\int { \frac { (1+x){ e }^{ x } }{ { cos }^{ 2 }(x{ e }^{ x }) } } dx,\) 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)
\(log(x{ e }^{ x })+c\)
The value of \(\int _{ 0 }^{ 1 }{ \sqrt { \frac { 2+x }{ 2-x } } } dx\) is equal to
- (a)
\(\frac { \pi }{ 2 } +1\)
- (b)
\(\pi +\frac { 3 }{ 2 } \)
- (c)
\(\pi +1\)
- (d)
none of these
The value of the integral \(\int _{ 0 }^{ \pi /2 }{ sin2xlogtanxdx } \) equals
- (a)
-1
- (b)
0
- (c)
1
- (d)
none of these
\(\int { \frac { sinx+cosx }{ \sqrt { 1+sin2x } } dx, } \) equals
- (a)
log(sinx+cosx)+c
- (b)
x+c
- (c)
log x+c
- (d)
\(\sqrt { 1+{ sin }^{ 2 }+c } \)
The value of the integral \(\int _{ a }^{ b }{ \frac { |x| }{ x } } dx,a<b\) is
- (a)
a-b
- (b)
a+b
- (c)
b-a
- (d)
\(|b|-|a|\)
\(\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}\int _{ 0 }^{ { x }^{ 2 } }{ \frac { sin\sqrt { x } }{ { x }^{ 3 } } } dx\) equals
- (a)
0
- (b)
\(\frac { 1 }{ 2 } \)
- (c)
1
- (d)
none of these
The integral \(\int { \frac { dx }{ { x }^{ 2 }+2xcos\alpha +1 } } ,\) equals
- (a)
\(\frac { 1 }{ cos\alpha } { tan }^{ -1 }\left( \frac { x+sin\alpha }{ cos\alpha } \right) +c\)
- (b)
\(\frac { 1 }{ cos\alpha } { tan }^{ -1 }\left( \frac { x-sin\alpha }{ cos\alpha } \right) +c\)
- (c)
\(\frac { 1 }{ sin\alpha } { tan }^{ -1 }\left( \frac { x-sin\alpha }{ cos\alpha } \right) +c\)
- (d)
\(\frac { 1 }{ sin\alpha } { tan }^{ -1 }\left( \frac { x+sin\alpha }{ cos\alpha } \right) +c\)
The point of maxima or minima of the function
\(f(x)=\int _{ 0 }^{ x }{ \frac { sint }{ t } } dx\) are given by
- (a)
\(n\pi \)
- (b)
\((2n+1)\frac { \pi }{ 2 } \)
- (c)
\(2n\pi \)
- (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
\(\int { \frac { { e }^{ x }(xlogx+1) }{ x } } dx,\) equals
- (a)
\(\frac { { e }^{ x } }{ logx } +c\)
- (b)
\({ e }^{ x }{ logx }^{ }+c\)
- (c)
\({ e }^{ x }{ (logx) }^{ 2 }+c\)
- (d)
NONE OF THESE
\(\int { \frac { d\theta }{ { sin }^{ 4 }\theta +{ cos }^{ 4 }\theta } } ,\)equals
- (a)
\({ tan }^{ -1 }\left( \frac { { tan }^{ 2 }\theta -1 }{ \sqrt { 2 } tan\theta } \right) +c\)
- (b)
\({ tan }^{ -1 }\left( \frac { { tan }^{ 2 }\theta -1 }{ \sqrt { 2 } } \right) +c\)
- (c)
\(\frac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\left( \frac { { tan }^{ 2 }\theta -1 }{ \sqrt { 2 } tan\theta } \right) +c\)
- (d)
NONE OF THESE
\(\int _{ 2 }^{ 3 }{ \frac { \sqrt { x } }{ \sqrt { 5-x } +\sqrt { x } } } dx\),equals
- (a)
\(\frac { 1 }{ 2 } \)
- (b)
2
- (c)
3
- (d)
4
for \(0\le x<\pi \), the area bounded by y=x and y=x+sin x, is
- (a)
2 sq. units
- (b)
4 sq. units
- (c)
\(2\pi \) sq. units
- (d)
\(4\pi \) sq. units
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 integral \(\int { \frac { { x }^{ 3 }+3x+2 }{ { ({ x }^{ 2 }+1) }^{ 2 }(x+1) } dx, } \) equals
- (a)
\(\frac { 3 }{ 2 } { tan }^{ -1 }x+\frac { 1 }{ 4 } log({ x }^{ 2 }+1)-\frac { 1 }{ 2 } log(x+1)+\frac { x }{ 1+{ x }^{ 2 } } +c\)
- (b)
\(\frac { 1 }{ 2 } { tan }^{ -1 }x-\frac { 3 }{ 2 } log({ x }^{ 2 }+1)+\frac { 1 }{ 2 } log(x+1)-\frac { x }{ 1+{ x }^{ 2 } } +c\)
- (c)
\(\frac { 1 }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 4 } log({ x }^{ 2 }+1)-\frac { 1 }{ 2 } log(x+1)-\frac { x }{ 1+{ x }^{ 2 } } +c\)
- (d)
NONE OF THESE
The integral
\(-\int { \frac { { sin }^{ -1 }x-{ cos }^{ -1 }x }{ { sin }^{ -1 }x+{ cos }^{ -1 }x } dx, } \) equals
- (a)
\(\frac { 4 }{ \pi } \left( x{ sin }^{ -1 }-\sqrt { 1-{ x }^{ 2 } } \right) -x+c\)
- (b)
\(\frac { 4 }{ \pi } \left( x{ sin }^{ -1 }+\sqrt { 1-{ x }^{ 2 } } \right) -x+c\)
- (c)
\(x{ sin }^{ -1 }+\sqrt { 1-{ x }^{ 2 } } -x+c\)
- (d)
NONE OF THESE
\(\int { \frac { \sqrt { tanx } }{ secxcosx } } dx,\) equals
- (a)
\(\frac { 1 }{ 2 } \sqrt { tanx } +c\)
- (b)
\(\sqrt { 2tanx+c } \)
- (c)
\(2\sqrt { tanx } +c\)
- (d)
\(\sqrt { 2 } cotx+c\)
The value of the integral \(\int _{ 0 }^{ 3/2 }{ [{ x }^{ 2 }] } dx\) where [ ], is the greastest integer function, equals
- (a)
\(2+\sqrt { 2 } \)
- (b)
\(2-\sqrt { 2 } \)
- (c)
\(2+\frac { 1 }{ \sqrt { 2 } } \)
- (d)
none of these
The value of integral \(\int _{ { e }^{ -1 } }^{ { e }^{ 2 } }{ \left| \frac { { log }_{ e }x }{ x } \right| } dx\), is
- (a)
3/2
- (b)
5/2
- (c)
3
- (d)
5
The value of \(\int _{ \pi /2 }^{ 3\pi /2 }{ [2sinx]dx } \), where [ ] represents the greatest integer function, is
- (a)
\(-\pi \)
- (b)
0
- (c)
\(-\frac { \pi }{ 2 } \)
- (d)
\(\frac { \pi }{ 2 } \)
The points of extremum of \(\int _{ 0 }^{ { x }^{ 2 } }{ \frac { { t }^{ 2 }-5t+6 }{ 2+{ e }^{ t } } dt } \) in the [1,2] are
- (a)
x=1,x=2
- (b)
\(x=\sqrt { 2 } ,x=\sqrt { 3 } \)
- (c)
x=1,5,x=1.9
- (d)
none of these
If f(y)=ey,g(y)=y; y>o and
\(F(t)=\int _{ 0 }^{ t }{ f(t-y)g(y)dy } \), then F(t) equals
- (a)
1-e-t(1+t)
- (b)
e-t-(1+t)
- (c)
tet
- (d)
te-t