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)
\({\pi\over 4}(\pi+2)\)
\(\int^4_1|(1-x)(3-x)log\ x| dx\) is equal to
- (a)
\({4\over3}log4+{7\over9}\)
- (b)
\({4\over3}log4-{7\over9}\)
- (c)
\({7\over9}-{4\over3}log4\)
- (d)
\({7\over9}+{4\over 6}log 3\)
Find the value of \(\int^2_{-2}max{(1-x),(1+x),2}dx\)
- (a)
8
- (b)
-8
- (c)
9
- (d)
-9
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 \(\int _{ 0 }^{ 1 }{ \frac { e^{ t }dt }{ t+1 } =a } \), 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
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
Let \(f(x)=\frac { 1 }{ 2 } a_{ 0 }+\sum _{ i=1 }^{ n }{ a_{ i }cos(ix) } +\sum _{ j=1 }^{ n }{ b_{ j } } sin(jx)\), then \(\int _{ -\pi }^{ \pi }{ fx) } \) cos kx dx is equal to
- (a)
ak
- (b)
bk
- (c)
\(\pi \)ak
- (d)
\(\pi \)bk
Let f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g be the function satisfying f(x)+g(x)=x2 the value of the integral \(\int _{ 0 }^{ 1 }{ f(x)g(x) } dx\) is
- (a)
1/4(e-7)
- (b)
1/4(e-2)
- (c)
1/2(e-3)
- (d)
none of these
If function f(x) is continuous in the interval (a, b) and having same definition between a and b, then we can find \(\int _{ a }^{ b }{ f(x) } \) dx and if f(x) is discontinuous and not having same definition between a and b, then we must break the interval such that f(x) becomes continuous and having same defintion in the breaking intervals.
Now, if f (x) is discontinuous at \(x=c(a<c<b),\) ,then \(\int _{ a }^{ b }{ f(x) } dx=\int _{ a }^{ b }{ f(x) } dx+\int _{ a }^{ b }{ f(x) } dx\) and also if f(x) is discontinuous at x=a in (0,2a),then we can write \(\int _{ a }^{ 2a }{ f(x) } dx=\int _{ 0 }^{ a }{ \{ f(a-x)+ } f(a=x)\} \quad dx\)
\(\int _{ -1 }^{ 1 }{ [|x|]d\left( \frac { q }{ 1+e^{ -1/x } } \right) } \) (where [.] denotes greatest integer function) is equal to
- (a)
-3
- (b)
-2
- (c)
-1
- (d)
none of these
\(\int _{ a }^{ b }{ f(x)d\alpha (x)+ } \int _{ a }^{ b }{ \alpha (x)+ } df(x)=\alpha (b)f(b)-\alpha (a)
f(a)\)
\(\int _{ 0 }^{ 3 }{ ({ x }^{ 2 }+1)d[x] } \) (where [.] denotes the greatest integer function) is equal to
- (a)
3
- (b)
\(\frac { 9 }{ 2 } \)
- (c)
17
- (d)
\(\frac { 27 }{ 2 } \)
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
f(x) is continuous periodic function with period T, then the integral \(I=\int _{ a }^{ a+T }{ f(x)\quad dx } \) is
- (a)
equal to 2a
- (b)
equal to 3a
- (c)
independent of a
- (d)
none of these
\(\int _{ a/4 }^{ 3a/4 }{ \frac { \sqrt { x } }{ \sqrt { (a-x)+\sqrt { x } } } } \) dx is equal to
- (a)
a/2
- (b)
a
- (c)
-a
- (d)
none of these