Definite Integration
601 Questions
Start JEE Mains Test
2002
Q551
JEE Advanced
MCQ
14 Mar 2026
The integral $\int\limits_{ - 1/2}^{1/2} {\left( {\left[ x \right] + \ell n\left( {{{1 + x} \over {1 - x}}} \right)} \right)dx} $ equal to
A.
$ - {1 \over 2}$
B.
$0$
C.
$1$
D.
$2\ell n\left( {{1 \over 2}} \right)$
2002
Q552
JEE Advanced
MCQ
14 Mar 2026
Let $T>0$ be a fixed real number . Suppose $f$ is a continuous
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
If $I = \int\limits_0^T {f\left( x \right)dx} $ then the value of $\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $ is
A.
$3/2I$
B.
$2I$
C.
$3I$
D.
$6I$
2002
Q553
JEE Advanced
MCQ
14 Mar 2026
Let $T>0$ be a fixed real number . Suppose $f$ is a continuous
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
If $I = \int\limits_0^T {f\left( x \right)dx} $ then the value of $\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $ is
A.
$3/2I$
B.
$2I$
C.
$3I$
D.
$6I$
2001
Q554
JEE Advanced
MCQ
14 Mar 2026
The value of $\int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}dx,\,a > 0,} $ is
A.
$\pi $
B.
$a\pi $
C.
$\pi /2$
D.
$2\pi $
2000
Q555
JEE Advanced
Numerical
14 Mar 2026
For $x>0,$ let $f\left( x \right) = \int\limits_e^x {{{\ln t} \over {1 + t}}dt.} $ Find the function
$f\left( x \right) + f\left( {{1 \over x}} \right)$ and show that $f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$
Here, $\ln t = {\log _e}t$.
$f\left( x \right) + f\left( {{1 \over x}} \right)$ and show that $f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$
Here, $\ln t = {\log _e}t$.
Correct Answer: Solve it.
2000
Q556
JEE Advanced
MCQ
14 Mar 2026
If $f\left( x \right) = \left\{ {\matrix{
{{e^{\cos x}}\sin x,} & {for\,\,\left| x \right| \le 2} \cr
{2,} & {otherwise,} \cr
} } \right.$ then $\int\limits_{ - 2}^3 {f\left( x \right)dx = } $
A.
$0$
B.
$1$
C.
$2$
D.
$3$
2000
Q557
JEE Advanced
MCQ
14 Mar 2026
The value of the integral $\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {{{{{\log }_e}x} \over x}} \right|dx} $ is :
A.
$3/2$
B.
$5/2$
C.
$3$
D.
$5$
2000
Q558
JEE Advanced
MCQ
14 Mar 2026
Let $g\left( x \right) = \int\limits_0^x {f\left( t \right)dt,} $ where f is such that
${1 \over 2} \le f\left( t \right) \le 1,$ for $t \in \left[ {0,1} \right]$ and $\,0 \le f\left( t \right) \le {1 \over 2},$ for $t \in \left[ {1,2} \right]$.
Then $g(2)$ satisfies the inequality
${1 \over 2} \le f\left( t \right) \le 1,$ for $t \in \left[ {0,1} \right]$ and $\,0 \le f\left( t \right) \le {1 \over 2},$ for $t \in \left[ {1,2} \right]$.
Then $g(2)$ satisfies the inequality
A.
$ - {3 \over 2} \le g\left( 2 \right) < {1 \over 2}$
B.
$0 \le g\left( 2 \right) < 2$
C.
${3 \over 2} < g\left( 2 \right) \le {5 \over 2}$
D.
$2 < g\left( 2 \right) < 4$
1999
Q559
JEE Advanced
Numerical
14 Mar 2026
Integrate $\int\limits_0^\pi {{{{e^{\cos x}}} \over {{e^{\cos x}} + {e^{ - \cos x}}}}\,dx.} $
Correct Answer: $${\pi \over 2}$$
1999
Q560
JEE Advanced
MCQ
14 Mar 2026
If for a real number $y$, $\left[ y \right]$ is the greatest integer less than or
equal to $y$, then the value of the integral $\int\limits_{\pi /2}^{3\pi /2} {\left[ {2\sin x} \right]dx} $ is
equal to $y$, then the value of the integral $\int\limits_{\pi /2}^{3\pi /2} {\left[ {2\sin x} \right]dx} $ is
A.
$ - \pi $
B.
$0$
C.
$ - \pi /2$
D.
$ \pi /2$
1999
Q561
JEE Advanced
MCQ
14 Mar 2026
$\int\limits_{\pi /4}^{3\pi /4} {{{dx} \over {1 + \cos x}}} $ is equal to
A.
$2$
B.
$-2$
C.
$1/2$
D.
$-1/2$
1998
Q562
JEE Advanced
Numerical
14 Mar 2026
Prove that $\int_0^1 {{{\tan }^{ - 1}}} \,\left( {{1 \over {1 - x + {x^2}}}} \right)dx = 2\int_0^1 {{{\tan }^{ - 1}}} \,x\,dx.$
Hence or otherwise, evaluate the integral
$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $
Hence or otherwise, evaluate the integral
$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $
Correct Answer: $$\log 2$$
1998
Q563
JEE Advanced
MCQ
14 Mar 2026
Let $f\left( x \right) = x - \left[ x \right],$ for every real number $x$, where $\left[ x \right]$ is the integral part of $x$. Then $\int_{ - 1}^1 {f\left( x \right)\,dx} $ is
A.
$1$
B.
$2$
C.
$0$
D.
$1/2$
1998
Q564
JEE Advanced
MCQ
14 Mar 2026
If $\int_0^x {f\left( t \right)dt = x + \int_x^1 {t\,\,f\left( t \right)\,\,dt,} } $ then the value of $f(1)$ is
A.
$1/2$
B.
$0$
C.
$1$
D.
$-1/2$
1997
Q565
JEE Advanced
Numerical
14 Mar 2026
Determine the value of $\int_\pi ^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} \,dx.$
Correct Answer: $${\pi ^2}$$
1997
Q566
JEE Advanced
Numerical
14 Mar 2026
The value of $\int_1^{{e^{37}}} {{{\pi \sin \left( {\pi In\,x} \right)} \over x}\,dx} $ is ...............
Correct Answer: $$2$$
1997
Q567
JEE Advanced
Numerical
14 Mar 2026
Let ${d \over {dx}}\,F\left( x \right) = {{{e^{\sin x}}} \over x},\,x > 0.$ If $\int_1^4 {{{2{e^{\sin {x^2}}}} \over x}} \,\,dx = F\left( k \right) - F\left( 1 \right)$
then one of the possible values of $k$ is ............
then one of the possible values of $k$ is ............
Correct Answer: $$16$$
1996
Q568
JEE Advanced
Numerical
14 Mar 2026
For $n>0,$ $\int_0^{2\pi } {{{x{{\sin }^{2n}}x} \over {{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx = $
Correct Answer: $${\pi ^2}$$
1996
Q569
JEE Advanced
Numerical
14 Mar 2026
If for nonzero $x$, $af(x)+$ $bf\left( {{1 \over x}} \right) = {1 \over x} - 5$ where $a \ne b,$ then
$\int_1^2 {f\left( x \right)dx} = .......$
$\int_1^2 {f\left( x \right)dx} = .......$
Correct Answer: $${1 \over {{a^2} - {b^2}}}\left[ {a\left( {\log 2 - 5} \right) + {{7b} \over 2}} \right]$$
1995
Q570
JEE Advanced
Numerical
14 Mar 2026
Let ${I_m} = \int\limits_0^\pi {{{1 - \cos mx} \over {1 - \cos x}}} dx.$ Use mathematical induction to prove that ${I_m} = m\,\pi ,m = 0,1,2,........$
Correct Answer: Solve it.
1995
Q571
JEE Advanced
Numerical
14 Mar 2026
Evaluate the definite integral :
$$\int\limits_{ - 1/\sqrt 3 }^{1/\sqrt 3 } {\left( {{{{x^4}} \over {1 - {x^4}}}} \right){{\cos }^{ - 1}}\left( {{{2x} \over {1 + {x^2}}}} \right)} dx$$
Correct Answer: $${\pi \over {12}}\left[ {\pi + 3{{\log }_e}\left( {2 + \sqrt 3 } \right) - 4\sqrt 3 } \right]$$
1995
Q572
JEE Advanced
MCQ
14 Mar 2026
If $f\left( x \right)\,\,\, = \,\,\,A\sin \left( {{{\pi x} \over 2}} \right)\,\,\, + \,\,\,B,\,\,\,f'\left( {{1 \over 2}} \right) = \sqrt 2 $ and
$\int\limits_0^1 {f\left( x \right)dx = {{2A} \over \pi },} $ then constants $A$ and $B$ are
$\int\limits_0^1 {f\left( x \right)dx = {{2A} \over \pi },} $ then constants $A$ and $B$ are
A.
${\pi \over 2}$ and ${\pi \over 2}$
B.
${2 \over \pi }$ and ${3 \over \pi }$
C.
$0$ and ${-4 \over \pi }$
D.
${4 \over \pi }$ and $0$
1995
Q573
JEE Advanced
MCQ
14 Mar 2026
The value of $\int\limits_\pi ^{2\pi } {\left[ {2\,\sin x} \right]\,dx} $ where [ . ] represents the greatest integer function is
A.
${{ - 5\pi } \over 3}$
B.
$\pi $
C.
${{ 5\pi } \over 3}$
D.
$ - 2\pi $
1994
Q574
JEE Advanced
Numerical
14 Mar 2026
Show that $\int\limits_0^{n\pi + v} {\left| {\sin x} \right|dx = 2n + 1 - \cos \,v} $ where $n$ is a positive integer and $\,0 \le v < \pi .$
Correct Answer: $$2n + 1 - \cos \gamma $$
1994
Q575
JEE Advanced
Numerical
14 Mar 2026
The value of $\int\limits_2^3 {{{\sqrt x } \over {\sqrt {3 - x} + \sqrt x }}} dx$ is ...........
Correct Answer: $${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$$
1993
Q576
JEE Advanced
Numerical
14 Mar 2026
Evaluate $\int_2^3 {{{2{x^5} + {x^4} - 2{x^3} + 2{x^2} + 1} \over {\left( {{x^2} + 1} \right)\left( {{x^4} - 1} \right)}}} dx.$
Correct Answer: $${1 \over 6}\log 6 - {1 \over {10}}$$
1993
Q577
JEE Advanced
MCQ
14 Mar 2026
The value of $\int\limits_0^{\pi /2} {{{dx} \over {1 + {{\tan }^3}\,x}}} $ is
A.
$0$
B.
$1$
C.
$\pi /2$
D.
$\pi /4$
1993
Q578
JEE Advanced
Numerical
14 Mar 2026
The value of $\int\limits_{\pi /4}^{3\pi /4} {{\phi \over {1 + \sin \phi }}d\phi } $ is ..............
Correct Answer: $$\pi \left( {\sqrt 2 - 1} \right)$$
1992
Q579
JEE Advanced
Numerical
14 Mar 2026
Determine a positive integer $n \le 5,$ such that
$$\int\limits_0^1 {{e^x}{{\left( {x - 1} \right)}^n}dx = 16 - 6e} $$
Correct Answer: $$n = 3$$
1991
Q580
JEE Advanced
Numerical
14 Mar 2026
Evaluate $\,\int\limits_0^\pi {{{x\,\sin \,2x\,\sin \left( {{\pi \over 2}\cos x} \right)} \over {2x - \pi }}dx} $
Correct Answer: $${8 \over {{\pi ^2}}}$$
1990
Q581
JEE Advanced
Numerical
14 Mar 2026
Prove that for any positive integer $k$,
${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$
Hence prove that $\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $
${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$
Hence prove that $\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $
Correct Answer: Solve it.
1990
Q582
JEE Advanced
Numerical
14 Mar 2026
Show that $\int\limits_0^{\pi /2} {f\left( {\sin 2x} \right)\sin x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\cos 2x} \right)\cos x\,dx} $
Correct Answer: Solve it.
1990
Q583
JEE Advanced
MCQ
14 Mar 2026
Let $f:R \to R$ and $\,\,g:R \to R$ be continuous functions. Then the value of the integral
$\int\limits_{ - \pi /2}^{\pi /2} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx} $ is
$\int\limits_{ - \pi /2}^{\pi /2} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx} $ is
A.
$\pi $
B.
$1$
C.
$-1$
D.
$0$
1989
Q584
JEE Advanced
Numerical
14 Mar 2026
If $f$ and $g$ are continuous function on $\left[ {0,a} \right]$ satisfying
$f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 2,$
then show that $\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $
$f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 2,$
then show that $\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $
Correct Answer: Solve it.
1989
Q585
JEE Advanced
Numerical
14 Mar 2026
The value of $\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|dx} $ is ...............
Correct Answer: $$4$$
1988
Q586
JEE Advanced
Numerical
14 Mar 2026
Evaluate $\int\limits_0^1 {\log \left[ {\sqrt {1 - x} + \sqrt {1 + x} } \right]dx} $
Correct Answer: $${1 \over 2}\left[ {\log 2 + {\pi \over 2} - 1} \right]$$
1988
Q587
JEE Advanced
Numerical
14 Mar 2026
The integral $\int\limits_0^{1.5} {\left[ {{x^2}} \right]dx,} $
Where [ ] denotes the greatest integer function, equals .............
Correct Answer: $$2 - \sqrt 2 $$
1988
Q588
JEE Advanced
MCQ
14 Mar 2026
The value of the integral $\int\limits_0^{2a} {[{{f\left( x \right)} \over {\left\{ {f\left( x \right) + f\left( {2a - x} \right)} \right\}}}]\,dx} $ is equal to $a$.
A.
TRUE
B.
FALSE
1987
Q589
JEE Advanced
Numerical
14 Mar 2026
$f\left( x \right) = \left| {\matrix{
{\sec x} & {\cos x} & {{{\sec }^2}x + \cot x\cos ec\,x} \cr
{{{\cos }^2}x} & {{{\cos }^2}x} & {\cos e{c^2}x} \cr
1 & {{{\cos }^2}x} & {{{\cos }^2}x} \cr
} } \right|.$
Then $\int\limits_0^{\pi /2} {f\left( x \right)dx = .......} $
Then $\int\limits_0^{\pi /2} {f\left( x \right)dx = .......} $
Correct Answer: $$ - \left( {{{15\pi + {{32}^ \circ }} \over {60}}} \right)$$
1986
Q590
JEE Advanced
Numerical
14 Mar 2026
Evaluate : $\int\limits_0^\pi {{{x\,dx} \over {1 + \cos \,\alpha \,\sin x}},0 < \alpha < \pi } $
Correct Answer: $${{\pi \alpha } \over {\sin x}}$$
1985
Q591
JEE Advanced
Numerical
14 Mar 2026
Evaluate the following : $\,\,\int\limits_0^{\pi /2} {{{x\sin x\cos x} \over {{{\cos }^4}x + {{\sin }^4}x}}} dx$
Correct Answer: $${{{\pi ^2}} \over {16}}$$
1985
Q592
JEE Advanced
MCQ
14 Mar 2026
For any integer $n$ the integral ...........
$\int\limits_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} $ has the value
$\int\limits_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} $ has the value
A.
$\pi $
B.
$1$
C.
$0$
D.
none of these
1984
Q593
JEE Advanced
Numerical
14 Mar 2026
Given a function $f(x)$ such that
(i) it is integrable over every interval on the real line and
(ii) $f(t+x)=f(x),$ for every $x$ and a real $t$, then show that
the integral $\int\limits_a^{a + 1} {f\,\,\left( x \right)} \,dx$ is independent of a.
(i) it is integrable over every interval on the real line and
(ii) $f(t+x)=f(x),$ for every $x$ and a real $t$, then show that
the integral $\int\limits_a^{a + 1} {f\,\,\left( x \right)} \,dx$ is independent of a.
Correct Answer: Solve it.
1984
Q594
JEE Advanced
Numerical
14 Mar 2026
Evaluate the following $\int\limits_0^{{1 \over 2}} {{{x{{\sin }^{ - 1}}x} \over {\sqrt {1 - {x^2}} }}dx} $
Correct Answer: $${{6 - \pi \sqrt 3 } \over {12}}$$
1983
Q595
JEE Advanced
Numerical
14 Mar 2026
Evaluate : $\int\limits_0^{\pi /4} {{{\sin x + \cos x} \over {9 + 16\sin 2x}}dx} $
Correct Answer: $${1 \over {20}}\log 3$$
1983
Q596
JEE Advanced
MCQ
14 Mar 2026
The value of the integral $\int\limits_0^{\pi /2} {{{\sqrt {\cot x} } \over {\sqrt {\cot x} + \sqrt {\tan x} }}dx} $ is
A.
$\pi /4$
B.
$\pi /2$
C.
$\pi $
D.
none of these
1982
Q597
JEE Advanced
Numerical
14 Mar 2026
Show that $\int\limits_0^\pi {xf\left( {\sin x} \right)dx} = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx.} $
Correct Answer: Solve it.
1982
Q598
JEE Advanced
Numerical
14 Mar 2026
Find the value of $\int\limits_{ - 1}^{3/2} {\left| {x\sin \,\pi \,x} \right|\,dx} $
Correct Answer: $${3 \over \pi } + {1 \over {{\pi ^2}}}$$
1981
Q599
JEE Advanced
Numerical
14 Mar 2026
Show that : $\mathop {\lim }\limits_{n \to \infty } \left( {{1 \over {n + 1}} + {1 \over {n + 2}} + .... + {1 \over {6n}}} \right) = \log 6$
Correct Answer: Solve it.
1981
Q600
JEE Advanced
MCQ
14 Mar 2026
Let $a, b, c$ be non-zero real numbers such that
$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} } $
Then the quadratic equation $a{x^2} + bx + c = 0$ has
$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} } $
Then the quadratic equation $a{x^2} + bx + c = 0$ has
A.
no root in $(0, 2)$
B.
at least one root in $(0, 2)$
C.
a double root in $(0, 2)$
D.
two imaginary roots