Definite Integration · Mathematics · WB JEE

$$\int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^9}x{{\cos }^5}x\,dx} $$ equals

If $$I = \int\limits_{ - \pi }^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} $$, then I equals

If $$h(x) = \int\limits_0^x {{{\sin }^4}t\,dt} $$, then $$h(x + \pi )$$ equals

The value of the integral $$\int\limits_0^2 {|{x^2} - 1|dx} $$ is

The value of $$\int\limits_0^\pi {|\cos x|dx} $$ is

The value of $$\int\limits_{ - 3}^3 {(a{x^5} + b{x^3} + cx + k)dx} $$, where a, b, c, k are constants, depends only on

The value of the integral $$\int\limits_{ - a}^a {{{x{e^{{x^2}}}} \over {1 + {x^2}}}dx} $$ is

The value of the $$\mathop {\lim }\limits_{n \to \infty } \left( {{1 \over {n + 1}} + {1 \over {n + 2}} + ... + {1 \over {6n}}} \right)$$ is

If $$f(x) = f(a - x)$$, then $$\int\limits_0^a {xf(x)dx} $$ is equal to

The value of $$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}}} $$ is

If $${I_1} = \int\limits_0^{\pi /4} {{{\sin }^2}xdx} $$ and $${I_2} = \int\limits_0^{\pi /4} {{{\cos }^2}xdx} $$, then

$$\int\limits_{ - 1}^4 {f(x)dx = 4} $$ and $$\int\limits_2^4 {\{ 3 - f(x)\} dx = 7} $$, then the value of $$\int\limits_{ - 1}^2 {f(x)dx} $$ is

$$\int\limits_0^{1000} {{e^{x - [x]}}dx} $$ is equal to

The value of the integral $$\int\limits_0^{\pi /2} {{{\sin }^5}xdx} $$ is

If $${d \over {dx}}\{ f(x)\} = g(x)$$, then $$\int\limits_a^b {f(x)g(x)dx} $$ is equal to

If $${I_1} = \int\limits_0^{3\pi } {f({{\cos }^2}x)dx} $$ and $${I_2} = \int\limits_0^\pi {f({{\cos }^2}x)dx} $$, then

The value of $$I = \int\limits_{ - \pi /2}^{\pi /2} {|\sin x|dx} $$ is

If $$I = \int\limits_0^1 {{{dx} \over {1 + {x^{\pi /2}}}}} $$, then

The value of $$\int\limits_{ - 2}^2 {(x\cos x + \sin x + 1)dx} $$ is

$$\int\limits_\pi ^{16\pi } {|\sin x|dx = } $$

The value of $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{r^3}} \over {{r^4} + {n^4}}}} $$ is

The value of $$\int\limits_0^\pi {{{\sin }^{50}}x{{\cos }^{49}}x\,dx} $$ is

The value of the integral $\int\limits_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is

$\int_\limits{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} d x$ is equal to

$\int\limits_0^{1 \cdot 5}\left[x^2\right] d x$ is equal to

The value of the integral $\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

Let $f(x)=\max \{x+|x|, x-[x]\}$, where $[x]$ stands for the greatest integer not greater than $x$. Then $\int\limits_{-3}^3 f(x) d x$ has the value

All values of a for which the inequality $$\frac{1}{\sqrt{a}} \int_\limits1^a\left(\frac{3}{2} \sqrt{x}+1-\frac{1}{\sqrt{x}}\right) \mathrm{d} x<4$$ is satisfied, lie in the interval

For any integer $$\mathrm{n}, \int_\limits0^\pi \mathrm{e}^{\cos ^2 x} \cdot \cos ^3(2 n+1) x \mathrm{~d} x$$ has the value :

If $$\mathrm{f}(x)=\frac{\mathrm{e}^x}{1+\mathrm{e}^x}, \mathrm{I}_1=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} x \mathrm{~g}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{g}(x(1-x)) \mathrm{d} x$$, then the value of $$\frac{I_2}{I_1}$$ is

Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a differentiable function and $$f(1)=4$$. Then the value of $$\lim _\limits{x \rightarrow 1} \int_\limits4^{f(x)} \frac{2 t}{x-1} d t$$, if $$f^{\prime}(1)=2$$ is

Let $$\mathrm{I}(\mathrm{R})=\int_\limits0^{\mathrm{R}} \mathrm{e}^{-\mathrm{R} \sin x} \mathrm{~d} x, \mathrm{R}>0$$. then,

$$\lim _\limits{n \rightarrow \infty} \frac{1}{n^{k+1}}[2^k+4^k+6^k+\ldots .+(2 n)^k]=$$

the expression $${{\int\limits_0^n {[x]dx} } \over {\int\limits_0^n {\{ x\} dx} }}$$, where $$[x]$$ and $$\{ x\} $$ are respectively integral and fractional part of $$x$$ and $$n \in N$$, is equal to

The value $$\int\limits_0^{1/2} {{{dx} \over {\sqrt {1 - {x^{2n}}} }}} $$ is $$(n \in N)$$

If $${I_n} = \int\limits_0^{{\pi \over 2}} {{{\cos }^n}x\cos nxdx} $$, then I$$_1$$, I$$_2$$, I$$_3$$ ... are in

$$\int\limits_0^{2\pi } {\theta {{\sin }^6}\theta \cos \theta d\theta } $$ is equal to

The average ordinate of $$y = \sin x$$ over $$[0,\pi ]$$ is :

Let f be derivable in [0, 1], then

The value of $$\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{\sin x}}} \over {{{(\cos x)}^{\sin x}} + {{(\sin x)}^{\cos x}}}}dx} $$ is

Let $$\mathop {\lim }\limits_{ \in \to 0 + } \int\limits_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1,\left( {0 < x < {\pi \over 4}} \right)} $$. Then a and b are given by

Let $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$. Then $$f'\left( {{\pi \over 4}} \right)$$ equals

If I is the greatest of $${I_1} = \int\limits_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} $$, $${I_2} = \int\limits_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} $$, $${I_3} = \int\limits_0^1 {{e^{ - {x^2}}}dx} $$, $${I_4} = \int\limits_0^1 {{e^{ - {x^2}/2}}dx} $$, then

If m, n be integers, then find the value of $$\int\limits_{ - \pi }^\pi {{{(\cos mx - \sin nx)}^2}dx} $$

Evaluate the following integral $$\int\limits_{ - 1}^2 {|x\sin \pi x|dx} $$

Prove that $$I = \int\limits_0^{\pi /2} {{{\sqrt {\sec x} } \over {\sqrt {\cos ecx} + \sqrt {\sec x} }}dx = {\pi \over 4}} $$

If $f(x)=\int_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t$ and $g(x)=\int_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t$, then the value of $f(x)+g(x)$ is

The value of $\int\limits_{-100}^{100} \frac{\left(x+x^3+x^5\right)}{\left(1+x^2+x^4+x^6\right)} d x$ is

$$ \text { The points of extremum of } \int_\limits0^{x^2} \frac{t^2-5 t+4}{2+e^t} d t \text { are } $$

Let f be a non-negative function defined on $$\left[ {0,{\pi \over 2}} \right]$$. If $$\int\limits_0^x {(f'(t) - \sin 2t)dt = \int\limits_x^0 {f(t)\tan t\,dt} } ,f(0) = 1$$ then $$\int\limits_0^{{\pi \over 2}} {f(x)dx} $$ is

Which of the following statements are true?