Definite Integration · Mathematics · JEE Advanced
Numerical
$$ \int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x $$
is ___________.
The value of $${{16{S_1}} \over \pi }$$ is _____________.
The value of $${{48{S_2}} \over {{\pi ^2}}}$$ is ___________.
If $$F:[0,\pi ] \to R$$ is defined by $$F(x) = \int_0^x {f(t)dt} $$, and if $$\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$$ = 2
then the value of f(0) is ...........
$$\int_0^{1/2} {{{1 + \sqrt 3 } \over {{{({{(x + 1)}^2}{{(1 - x)}^6})}^{1/4}}}}dx} $$ is ........
$$\int\limits_0^x {{{{t^2}} \over {1 + {t^4}}}} dt = 2x - 1$$
$$f\left( x \right) = \left\{ {\matrix{ {\left[ x \right],} & {x \le 2} \cr {0,} & {x > 2} \cr } } \right.$$ where $$\left[ x \right]$$ is the greatest integer less than or equal to $$x$$, if $$I = \int\limits_{ - 1}^2 {{{xf\left( {{x^2}} \right)} \over {2 + f\left( {x + 1} \right)}}dx,} $$ then the value of $$(4I-1)$$ is
Then the value of $${{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f\left( x \right)\cos \,\pi x\,dx} $$ is
Let $$f:R \to R$$ be a continuous function which satisfies $$f(x) = \int\limits_0^x {f(t)dt} $$. Then, the value of $$f(\ln 5)$$ is ____________.
MCQ (Single Correct Answer)
If $$m \le \int\limits_{1/2}^1 {f\left( x \right)dx \le M,} $$ then the possible values of $$m$$ and $$M$$ are
P.$$\,\,\,\,$$ The number of polynomials $$f(x)$$ with non-negative integer coefficients of degree $$ \le 2$$, satisfying $$f(0)=0$$ and $$\int_0^1 {f\left( x \right)dx = 1,} $$ is
Q.$$\,\,\,\,$$ The number of points in the interval $$\left[ { - \sqrt {13} ,\sqrt {13} } \right]$$
at which $$f\left( x \right) = \sin \left( {{x^2}} \right) + \cos \left( {{x^2}} \right)$$ attains its maximum value, is
R.$$\,\,\,\,$$ $$\int\limits_{ - 2}^2 {{{3{x^2}} \over {\left( {1 + {e^x}} \right)}}dx} $$ equals
S.$$\,\,\,\,$$ $${{\left( {\int\limits_{ - {1 \over 2}}^{{1 \over 2}} {\cos 2x\log \left( {{{1 + x} \over {1 - x}}} \right)dx} } \right)} \over {\left( {\int\limits_0^{{1 \over 2}} {\cos 2x\log \left( {{{1 + x} \over {1 - x}}} \right)dx} } \right)}}$$
List $$II$$
1.$$\,\,\,\,$$ $$8$$
2.$$\,\,\,\,$$ $$2$$
3.$$\,\,\,\,$$ $$4$$
4.$$\,\,\,\,$$ $$0$$
The value of $$g\left( {{1 \over 2}} \right)$$ is
The value of $$g'\left( {{1 \over 2}} \right)$$ is
non-constant and differentiable function such that
$$f'\left( x \right) < 2f\left( x \right)$$ and $$f\left( {{1 \over 2}} \right) = 1.$$ Then the value of $$\int\limits_{1/2}^1 {f\left( x \right)} \,dx$$ lies in the interval
$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to
Let $$g\left( x \right) = \int\limits_0^{{e^x}} {{{f'\left( t \right)} \over {1 + {t^2}}}} \,dt.$$
Which of the following is true?
$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$
If $$f''\left( x \right) < 0\,\forall x \in \left( {a,b} \right)$$ and $$c$$ is a point such that $$a < c < b,$$ and
$$\left( {c,f\left( c \right)} \right)$$ is the point lying on the curve for which $$F(c)$$ is
maximum, then $$f'(c)$$ is equal to
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$
If $$\mathop {\lim }\limits_{x \to a} {{\int\limits_a^x {f\left( x \right)dx - \left( {{{x - a} \over 2}} \right)\left( {f\left( x \right) + f\left( a \right)} \right)} } \over {{{\left( {x - a} \right)}^3}}} = 0,\,\,$$ then $$f(x)$$ is
of maximum degree
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$
$$\int\limits_0^{\pi /2} {\sin x\,dx = } $$
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
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
$${1 \over 2} \le f\left( t \right) \le 1,$$ 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
equal to $$y$$, then the value of the integral $$\int\limits_{\pi /2}^{3\pi /2} {\left[ {2\sin x} \right]dx} $$ is
$$\int\limits_0^1 {f\left( x \right)dx = {{2A} \over \pi },} $$ then constants $$A$$ and $$B$$ are
$$\int\limits_{ - \pi /2}^{\pi /2} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx} $$ is
$$\int\limits_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^3}\left( {2n + 1} \right)xdx} $$ has the value
$$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} } $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has
MCQ (More than One Correct Answer)
Consider the equation
$$ \int_{1}^{e} \frac{\left(\log _{\mathrm{e}} x\right)^{1 / 2}}{x\left(a-\left(\log _{\mathrm{e}} x\right)^{3 / 2}\right)^{2}} d x=1, \quad a \in(-\infty, 0) \cup(1, \infty) $$
Which of the following statements is/are TRUE?
for all x$$ \in $$R, then which of the following statements is/are TRUE?
$$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{{n^n}\left( {x + n} \right)\left( {x + {n \over 2}} \right)...\left( {x + {n \over n}} \right)} \over {n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + {{{n^2}} \over 4}} \right)....\left( {{x^2} + {{{n^2}} \over {{n^2}}}} \right)}}} \right)^{{x \over n}}},$$ for
all $$x>0.$$ Then
Then the correct expression(s) is (are)
If $${I_n} = \int\limits_{ - \pi }^\pi {{{\sin nx} \over {(1 + {\pi ^x})\sin x}}dx,n = 0,1,2,} $$ .... then
Subjective
Column $$I$$
(A) $$\int\limits_{ - 1}^1 {{{dx} \over {1 + {x^2}}}} $$
(B) $$\int\limits_0^1 {{{dx} \over {\sqrt {1 - {x^2}} }}} $$
(C) $$\int\limits_2^3 {{{dx} \over {1 - {x^2}}}} $$
(D) $$\int\limits_1^2 {{{dx} \over {x\sqrt {{x^2} - 1} }}} $$
Column $$II$$
(p) $${1 \over 2}\log \left( {{2 \over 3}} \right)$$
(q) $$2\log \left( {{2 \over 3}} \right)$$
(r) $${{\pi \over 3}}$$
(s) $${{\pi \over 2}}$$
$$\int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\cos x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\sin 2x} \right)\cos x\,dx.} $$
$$f\left( x \right) + f\left( {{1 \over x}} \right)$$ and show that $$f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$$
Here, $$\ln t = {\log _e}t$$.
Hence or otherwise, evaluate the integral
$$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $$
$${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$$
Hence prove that $$\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $$
$$f\left( x \right) = f\left( {a - x} \right)$$ and $$g\left( x \right) + g\left( {a - x} \right) = 2,$$
then show that $$\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $$
(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.
Fill in the Blanks
then one of the possible values of $$k$$ is ............
$$\int_1^2 {f\left( x \right)dx} = .......$$
Where [ ] denotes the greatest integer function, equals .............
Then $$\int\limits_0^{\pi /2} {f\left( x \right)dx = .......} $$