JEE Main Mathematics - Define Integrals and Its Application
Exam Duration: 60 Mins Total Questions : 30
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]
Statement I The value of \(\int^{\pi/3}_{\pi/6}{dx\over 1+\sqrt{tan\ x}}\) is equal to \({\pi\over6}\)
Statement II \(\int^b_a f(x)dx=\int^b_a\ f(a+b-x)dx\)
- (a)
Statement I is true, Statement II is true; Statement II is a correct
- (b)
Statement I is true , Statement II is true; Statement II is not a correct explanation for Statement I
- (c)
Statement I is true, Statement II is false
- (d)
Statement I is false, Statement II is true
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)
\(g(x).g(\pi)\)
The value of \(\int^1_0{8log(1+x)\over 1+x^2}dx\) is
- (a)
\({\pi\over 8}\ log\ 2\)
- (b)
\({\pi\over 2}\ log\ 2\)
- (c)
\( log\ 2\)
- (d)
\(\pi\ log 2\)
If f(x) be a real valued function, f(x)+f(x+4)=f(x+2)+f(x+6), \(g(x)=\int _{ x }^{ x+8 }{ f(t) } dt\). Then g'(x) is equal to
- (a)
f(x)
- (b)
f(x+8)
- (c)
8
- (d)
0
The value of \(\int _{ -1 }^{ 15 }{ } \)sgn ({x} dx, where {.} denotes the fractional part function, is
- (a)
8
- (b)
16
- (c)
24
- (d)
0
The value of the integral \(\int _{ 0 }^{ \pi }{ \frac { sin\left( n+\frac { 1 }{ 2 } \right) x }{ sin\quad x/2 } } dx(n\in N)\) is
- (a)
\(\pi \)
- (b)
\(2\pi \)
- (c)
\(3\pi \)
- (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)
\(<\frac { \pi }{ 2 } \)
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
The value of the definite integral \(\int _{ 0 }^{ 1 }{ \frac { x\quad dx }{ ({ x }^{ 2 }+16) } } \) lies in the interval [a,b]. Then smallest such interval is
- (a)
[0,1/17]
- (b)
[0,1]
- (c)
[0,1/27]
- (d)
none of these
Let f(x)=max{x+|x|,x-[x]}, where [x] denotes the greatest integer ≤x. Then \(\int _{ -2 }^{ 2 }{ f(x) } dx\) is equals to
- (a)
3
- (b)
2
- (c)
1
- (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
If [.] stands for the greatest integer function, the value of
\(\int _{ 4 }^{ 10 }{ \frac { \left[ x^{ 2 } \right] }{ \left[ x^{ 2 }-28x+196 \right] +\left[ x^{ 2 } \right] } } \) is
- (a)
0
- (b)
1
- (c)
3
- (d)
none of these
The value of \(\lim _{ n\rightarrow \infty }{ \left( \frac { { 1 }^{ k }+{ 2 }^{ k }+...+{ n }^{ k } }{ { n }^{ k-1 } } \right) } \) is
- (a)
\(\frac { 1 }{ k+2 } \)
- (b)
\(\frac { 1 }{ k+1 } \)
- (c)
\(\frac { 1 }{ k+3 } \)
- (d)
0
The value of \(\int _{ \pi /4 }^{ 3\pi /4 }{ \frac { x }{ 1+sinx } } \) dx is equal to
- (a)
\((\sqrt { 2 } -1)\pi \)
- (b)
\((\sqrt { 2 } +1)\pi \)
- (c)
\(\pi \)
- (d)
none of these
The are bounded by the curves y2=16x and x2=16y is given by
- (a)
\(\frac{64}{3}\) sq.units
- (b)
\(\frac{256}{3}\) sq.units
- (c)
\(\frac{16}{3}\) sq.units
- (d)
none of these