Mathematics - Define Integrals and Its Application
Exam Duration: 45 Mins Total Questions : 30
If \(\int^{1}_0\ e^{x^2}(x-\alpha)dx=0, \) then
- (a)
\(1 < \alpha < 2\)
- (b)
\(\alpha < 0\)
- (c)
\(0 < \alpha < 1\)
- (d)
\(\alpha = 0\)
The value of \(\int^{\pi}_0\sqrt{1+4sin^2 {x\over 2}-4sin{x\over 2}\ dx}\) is
- (a)
\(4\sqrt3-4\)
- (b)
\(4\sqrt3-4-{\pi \over3}\)
- (c)
\(\pi-4\)
- (d)
\({2\pi\over 3}-4-4\sqrt3\)
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)
\(g(x).g(\pi)\)
\(\int^{\pi}_0\ [cot\ x]dx,\) where [] denotes the greatest integer function, is equal to
- (a)
\(\pi\over 2\)
- (b)
\(1\)
- (c)
\(-1\)
- (d)
\(-{\pi\over 2}\)
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
If \({ I }_{ 1 }=\int _{ 0 }^{ 1 }{ { 2x }^{ 2 } } dx,{ I }_{ 2 }=\int _{ 0 }^{ 1 }{ { 2x }^{ 3 } } dx,{ I }_{ 3 }=\int _{ 1 }^{ 2 }{ { 2 }^{ x^{ 2 } } } dx\) and ,then \({ I }_{ 4 }=\int _{ 1 }^{ 2 }{ { 2 }^{ x^{ 3 } } } dx\)
- (a)
I1>I2
- (b)
I2>I1
- (c)
I3>I4
- (d)
I4>I3
Let f be integrable over [0,a] for any real a. If we define
\({ I }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ cos\theta \quad f(sin\theta +cos^{ 2 }\theta )d\theta } \) and
\({ I }_{ 2 }=\int _{ 0 }^{ \pi /2 }{ sin2\theta \quad f(sin\theta +cos^{ 2 }\theta )d\theta } \), then
- (a)
I1=I2
- (b)
I1=-I2
- (c)
I1=2I2
- (d)
I1=-2I2
If \({ I }_{ n }\int _{ 0 }^{ \pi /4 }{ { tan }^{ n } } x\quad dx\) ,then
- (a)
\({ I }_{ n }+{ I }_{ n-2 }=\frac { 1 }{ n-1 } \)
- (b)
\({ I }_{ n+1 }+{ I }_{ n-1 }=\frac { 1 }{ n } \)
- (c)
\(\frac { 1 }{ n+1 } <2I_{ n }<\frac { 1 }{ n-1 } \) ,where n>1 is a natural number
- (d)
\(\left( { I }_{ n+2 }+{ I }_{ n } \right) (n+1)=1\)
Let \(\int _{ \alpha }^{ \beta }{ \frac { f(\alpha +\beta -x) }{ f(x)+f(\alpha +\beta -x) } } dx=4\) ,then
- (a)
\(\alpha =-1,\beta =7\)
- (b)
\(\alpha =5,\beta =13\)
- (c)
\(\alpha =-2,\beta =6\)
- (d)
\(\alpha =-10,\beta =-2\)
\(\int _{ 0 }^{ \pi /4 }{ sinx } d(x-[x])\) is equal to (where [.] denotes the greatest integer function)
- (a)
1/2
- (b)
\(1-\frac { 1 }{ \sqrt { 2 } } \)
- (c)
1
- (d)
none of these
The value of \(\int _{ -2 }^{ 2 }{ \frac { sin^{ 2 }x }{ [x/\pi ]+[1/2] } } dx\) , where [x] denotes the greatest integer ≤ x, is
- (a)
1
- (b)
0
- (c)
4-sin4
- (d)
none of these
\(\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } } \sum _{ r=n+1 }^{ 2n }{ ln } \left( 1+\frac { r }{ n } \right) \) 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
The value of the integral \(I=\int _{ 1 }^{ \infty }{ \frac { (x^{ 2 }-2) }{ x^{ 3 }\sqrt { (x^{ 2 }-1) } } } \) dx is
- (a)
0
- (b)
2/3
- (c)
1
- (d)
none of these
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