IISER Mathematics - Define Integrals and Its Application
Exam Duration: 45 Mins Total Questions : 30
\(\int^{\pi}_0{x\ tan\ x\over sec\ x + tan\ x}dx\) is equal to
- (a)
\({\pi\over 2}(\pi-2)\)
- (b)
\({\pi}(\pi-2)\)
- (c)
\({\pi\over 4}(\pi-2)\)
- (d)
\({\pi\over 4}(\pi+2)\)
The value of \(\int^{\pi}_0{s^2sin\ x\over (2x-\pi)(1+xos^2x)}dx\) is
- (a)
\(\pi^2\over 4\)
- (b)
\(\pi^2\over 2\)
- (c)
\(\pi^2\over 3\)
- (d)
\(\pi^2\over 6\)
\(\int^{\pi}_0{x\over a^2cos^2x+b^2sin^2x}dx\) is equal to
- (a)
\({\pi^2\over ab}\)
- (b)
\({\pi^2\over 2ab}\)
- (c)
\({\pi^2\over 4ab}\)
- (d)
\({\pi^2\over 9ab}\)
Let \(I = \int^3_1\sqrt{3+x^3}\ dx,\) then the value of I will lie in the interval
- (a)
[4,6]
- (b)
[1,3]
- (c)
\([4,2\sqrt{30}]\)
- (d)
[2,3]
If \(\int^{\pi}_{0}{cos\ 4\ t\ dt},\) then \(g(x+\pi)\) equals
- (a)
\(9(x)\over g(\pi)\)
- (b)
\(g(x)+g(\pi)\)
- (c)
\(g(x)-g(\pi)\)
- (d)
\(g(x).g(\pi)\)
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)
\(I_2>I_1\)
If \(f(x)=\int _{ 0 }^{ x }{ \frac { dt }{ \{ f(t)\} ^{ 2 } } } and\quad \int _{ 0 }^{ x }{ \frac { dt }{ \{ f(t)\} ^{ 2 } } } \quad =\sqrt [ 3 ]{ 6 } \), f(9) equals
- (a)
0
- (b)
1
- (c)
2
- (d)
none of these
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
If \({ I }_{ k }=\int _{ -2k\pi }^{ 2k\pi }{ "sinx|[sinx]\quad dx\quad } \forall k\in N\) , where [.] denotes the greatest integer function, then \(\sum _{ k=1 }^{ 10 }{ { I }_{ k } } \) is equal to
- (a)
-110
- (b)
-440
- (c)
-330
- (d)
-220
The values of ∝, which satisfy \(\int _{ \pi /2 }^{ \alpha }{ sinx\quad dx } =sin2\alpha (\alpha \epsilon [0,2\pi ])\) are equal to
- (a)
\(\pi /2\)
- (b)
\(3\pi /2\)
- (c)
\(7\pi /6\)
- (d)
\(11\pi /6\)
\(\int _{ -1/2 }^{ 1/2 }{ \sqrt { \left\{ \left( \frac { x+1 }{ x-1 } \right) ^{ 2 }+\left( \frac { x-1 }{ x+1 } \right) ^{ 2 }-2 \right\} } } dx\) is
- (a)
4 In \(\left( \frac { 4 }{ 3 } \right) \)
- (b)
4 In \(\left( \frac { 3 }{ 4 } \right) \)
- (c)
-In \(\left( \frac { 81 }{ 256 } \right) \)
- (d)
In \(\left( \frac { 256 }{ 81 } \right) \)
Let e be the eccentricity of a hyperbola \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { y^{ 2 } }{ b^{ 2 } } =1\) and f(e) br the eccentricity hyperbola \(-\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { y^{ 2 } }{ b^{ 2 } } =1\) ,then\(\int _{ 1 }^{ 3 }{ \frac { fff...f(e) }{ n\quad times } } de\) is equal to
- (a)
2,if n is even
- (b)
4,if n is even
- (c)
\(2\sqrt { 2 } \) n is odd
- (d)
\(4\sqrt { 2 } \) n is odd
Let \(I=\int _{ 0 }^{ \pi /2 }{ \frac { sinx }{ x } } \) dx, then I lies in the interval
- (a)
(0,1)
- (b)
[0,1]
- (c)
[0,3/2]
- (d)
[1,2]
The value of \(\int _{ 0 }^{ 2\pi }{ \frac { dx }{ e^{ sinx }+1 } } \) is
- (a)
\(\pi \)
- (b)
0
- (c)
\(2\pi \)
- (d)
\(\pi /2\)
If \(I=\int _{ 0 }^{ 50\pi }{ \sqrt { (1-cos2x) } } dx\), then the value of I is
- (a)
\(50\sqrt { 2 } \)
- (b)
\(100\sqrt { 2 } \)
- (c)
\(25\sqrt { 2 } \)
- (d)
none of these
If \(\int _{ 0 }^{ 10 }{ f(x)dx } =5,\) then \(\sum _{ k=1 }^{ 10 }{ \int _{ 0 }^{ 1 }{ f(k-1+x)dx } } \) is
- (a)
50
- (b)
10
- (c)
5
- (d)
none of these
If \(\frac { 2x }{ \pi } <sin\quad x<x\) for \(0<x<\frac { \pi }{ 2 } \) ,then the value of the integral \(\int _{ 0 }^{ \pi /2 }{ \frac { sin\quad x }{ x } } dx\) is
- (a)
>1
- (b)
<1
- (c)
\(>\frac { \pi }{ 2 } \)
- (d)
\(<\frac { \pi }{ 2 } \)
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)
\(\frac { 1 }{ 2 } \sqrt { \frac { \pi }{ a } } \)
If \(\int _{ 0 }^{ 1 }{ e^{ { e }^{ 2 } } } (x-\alpha )dx=0\) , then
- (a)
1<∝<2
- (b)
∝<0
- (c)
0<∝<1
- (d)
∝=0
If [x] denotes the greatest integer less than or equal to x, then \(\int _{ 0 }^{ \infty }{ \left[ \frac { 2 }{ e^{ x } } \right] } \)dx is equal to
- (a)
loge2
- (b)
e2
- (c)
0
- (d)
2/e
If f(x) and g (x) be two function,such that f(a)=g(a)=0 and f and g are both differentiable at every where in some neighbourhood of point a except possibly 'a'
Then \(\lim _{ x\rightarrow a }{ \frac { f(x) }{ g(x) } } =\lim _{ x\rightarrow 0 }{ \frac { f^{ ' }(x) }{ g^{ ' }(x) } } \) provided f'(a) and g' (a) are not both zero
The value of \(\lim _{ x\rightarrow 0 }{ \frac { \int _{ 0 }^{ { x }^{ 2 } }{ sin\sqrt { t } } dt }{ { x }^{ 3 } } } \) is
- (a)
0
- (b)
2/9
- (c)
1/3
- (d)
2/3
Which of the following is/are true
- (a)
\(\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 2 } } } dx>\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 5 } } } dx\)
- (b)
\(\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 3 } } } dx>\int _{ 0 }^{ 1 }{ { 2 }^{ { x }^{ 5 } } } dx\)
- (c)
\(\int _{ 0 }^{ 1 }{ In\quad x } dx>\int _{ 0 }^{ 1 }{ (In\quad x)^{ 2 } } dx\)
- (d)
\(\int _{ 3 }^{ 4 }{ In\quad x } dx>\int _{ 3 }^{ 4 }{ (In\quad x)^{ 2 } } dx\)
Suppose for every integer n,\(\int _{ n }^{ n+1 }{ f(x) } \)dx=n2, the value of \(\int _{ -2 }^{ 4 }{ f(x) } dx\) is
- (a)
16
- (b)
4
- (c)
19
- (d)
none of these
The value of \(\int _{ 0 }^{ \pi }{ \left( \sum _{ r=0 }^{ 3 }{ a_{ r }cos^{ 3-r }x } sin^{ r }x \right) } \)dx depends on
- (a)
a0 and a2
- (b)
a1 and a2
- (c)
a0 and a3
- (d)
a1 and a3
If \(\int _{ a }^{ b }{ \frac { x^{ n } }{ x^{ n }+(1+-x)^{ n } } } dx\)=6, then
- (a)
a=4,b=12,n\(\in \)R
- (b)
a=2,b=14,n\(\in \)R
- (c)
a=-4,b=20,n\(\in \)R
- (d)
a=2,b=8,n\(\in \)R
\(\int _{ a }^{ b }{ f(x)d\alpha (x)+ } \int _{ a }^{ b }{ \alpha (x)+ } df(x)=\alpha (b)f(b)-\alpha (a)
f(a)\)
\(\int _{ 0 }^{ 3 }{ ({ x }^{ 2 }+1)d[x] } \) (where [.] denotes the greatest integer function) is equal to
- (a)
3
- (b)
\(\frac { 9 }{ 2 } \)
- (c)
17
- (d)
\(\frac { 27 }{ 2 } \)
Let f(x) be a continuous function such that f(a-x)+f(x)=0 for all x\(\in \)[0,a], then \(\int _{ 0 }^{ a }{ \frac { dx }{ 1+e^{ f(x) } } } \) is equal to
- (a)
a
- (b)
\(\frac { a }{ 2 } \)
- (c)
f(a)
- (d)
\(\frac { 1 }{ 2 } \)f(a)
The area enclosed by the curve \(\frac { x^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ { 16 } } \) =1 is
- (a)
10\(\pi\) sq.units
- (b)
10\(\pi\) sq.units
- (c)
5\(\pi\) sq.units
- (d)
4\(\pi\) 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
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)
\(\frac { \pi }{ 2 } sq.units\)