Definite Integration · Mathematics · MHT CET

If $[x]$ denotes the greatest integer function, then $$\int_\limits0^5 x^2[x] d x=$$

The value of $\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} d x$ is

$$\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=$$

The value of the integral $\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} \mathrm{dx}$ is

$$\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{(1+\tan x)(2+\tan x)} d x=$$

After $t$ seconds, the acceleration of a particle, which starts from rest and moves in a straight line is $\left(8-\frac{\mathrm{t}}{5}\right) \mathrm{cm} / \mathrm{s}^2$, then velocity of the particle at the instant, when the acceleration is zero, is

The value of integral $\int_\limits{-2}^0\left(x^3+3 x^2+3 x+5+(x+1) \cos (x+1)\right) d x$ is equal to

If $\mathrm{I}=\int_0^{\frac{\pi}{4}} \log (1+\tan x) \mathrm{d} x$, then value of $\mathrm{I}$ is

$$\int_\limits{0.2}^{3.5}[x] \mathrm{d} x=$$ (where $[x]=$ greatest integer not greater than $x$ )

$$\int_\limits0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$$

$$\int_\limits0^a \frac{x-a}{x+a} d x=$$

$\int_\limits{\frac{-\pi}{4}}^{\frac{\pi}{4}}(\sin x)^{-4} \mathrm{~d} x$ has the value

$\int_\limits0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ has the value

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)}$ is equal to

The value of $\mathrm{I}=\mathrm{I}=\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+\mathrm{e}^{-x}} \mathrm{~d} x$ is equal to

The value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)} d x$ is

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right)\right) \mathrm{d} x$, where $[x]$ represent greatest integer function, equals

The value of the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \frac{\pi-x}{\pi+x}\right) \cos x d x$ is equal to

If $\int_\limits0^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=1-\frac{1}{\sqrt{2}},(k>0)$, then the value of $k$ is

The value of $\mathrm{I}=\int_\limits{\sqrt{\log _{\mathrm{e}}}}^{\sqrt{\log _{\mathrm{e}} 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log _{\mathrm{e}} 6-x^2\right)} d x$ is

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$, then $\int_0^a f(x) g(x) d x$ is equal to

If $$I_n=\int_\limits0^{\frac{\pi}{4}} \tan ^n \theta d \theta$$, then $$I_{12}+I_{10}=$$

The integral $$\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x$$ is equal to

The integral $$\int_\limits{\pi / 6}^{\pi / 3} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x$$ is equal to

If $$\mathrm{f}(x)=\left\{\begin{array}{ll}\mathrm{e}^{\cos x} \sin x & , \text { for }|x| \leq 2 \\ 2, & \text { otherwise }\end{array}\right.$$, then $$\int_\limits{-2}^3 \mathrm{f}(x) \mathrm{d} x$$ is equal to

The value of $$\int_\limits0^\pi\left|\sin x-\frac{2 x}{\pi}\right| \mathrm{d} x$$ is

$$\int_\limits0^4|2 x-5| d x=$$

Let $$f:[-1,2] \rightarrow[0, \infty)$$ be a continuous function such that $$\mathrm{f}(x)=\mathrm{f}(1-x), \forall x \in[-1,2]$$

Let $$\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$$ and $$\mathrm{R}_2$$ be the area of the region bounded by $$y=\mathrm{f}(x), x=-1, x=2$$ and the $$\mathrm{X}$$-axis, then $$\mathrm{R}_2$$ is

$$\int\limits_0^\pi \frac{d x}{4+3 \cos x}=$$

Let $$\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$$, then $$\mathrm{f}(3)-\mathrm{f}(1)$$ is equal to

Let $$\mathrm{f}(x)$$ be positive for all real $$x$$. If $$\mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x$$, where $$(2 h-1)>0$$, then $$\frac{I_1}{I_2}$$ is

$$\int_\limits{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=$$

Let $$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$$ and $$\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$$ be continuous functions. Then the value of the integral $$\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \mathrm{d} x$$ is

$$\int_\limits 0^\pi \frac{x \tan x}{\sec x+\cos x} d x= $$

$$\int_\limits0^1 \cos ^{-1} x d x=$$

If $$\int_\limits0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6}$$, then the value of $$\mathrm{k}$$ is

$$\int_\limits1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5} \text {, then } \mathrm{k} \text { has the value }$$

If $$\mathrm{f}(x)$$ is a function satisfying $$\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$$ with $$\mathrm{f}(0)=1$$ and $$\mathrm{g}(x)$$ is a function that satisfies $$\mathrm{f}(x)+\mathrm{g}(x)=x^2$$. Then the value of the integral $$\int_\limits0^1 f(x) g(x) d x$$ is

$$\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}$$ is equal to

The value of the integral $$\int_\limits0^1 \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x$$ is

$$\int_\limits0^2[x] \mathrm{d} x+\int_\limits0^2|x-1| \mathrm{d} x=$$

(where $$[x]$$ denotes the greatest integer function.)

$$\int_\limits2^5 2[\mathrm{x}] \mathrm{dx}=\{\text { where }[\mathrm{x}] \text { denotes the greatest integer function } \leq \mathrm{x}\}$$

$$\int_\limits0^\pi x \sin x \cos ^4 x d x=$$

$$\int_\limits0^4 x[x] d x=$$ (where $$[\mathrm{x}]$$ denotes greatest integer function not greater than $$\mathrm{x}]$$

$$\int_\limits0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x=$$

$$\int_\limits0^\pi \frac{1}{4+3 \cos x} d x=$$

$$\int_\limits1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x=$$

$$\int_\limits0^1|5 x-3| d x=$$

$$\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=$$

$$\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$$

$$\int_\limits0^2|2 x-3| d x=$$

If $$\int_\limits0^a \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}$$, then $$k=$$

$$\int\limits_{ - \pi }^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx = } $$

The value of $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{2x - 1} \over {1 + x - {x^2}}}} \right)dx} $$ is

$$\int\limits_5^{10} \frac{d x}{(x-1)(x-2)}=$$

$$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{\cos x} \over {1 + {e^x}}}dx = } $$

$$\int_\limits0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=$$

If $$f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$$, then $$\int_\limits1^4 f(x) d x=$$

If $$2 f(x)-3 f\left(\frac{1}{x}\right)=x$$, then $$\int_\limits1^e f(x) d x=$$

If $$\int_\limits2^e\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}$$, then

$$\int_\limits0^{\pi / 4} \log (1+\tan x) d x=$$

If $$\int_\limits0^{\frac{\pi}{2}} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B$$, then A + B =

$\int_\limits0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x=$

The c.d.f, $F(x)$ associated with p.d.f. $f(x)=3\left(1-2 x^2\right)$. If $0< x<1$ is $k\left(x-\frac{2 x^3}{k}\right)$, then value of $k$ is

$$\int_\limits0^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} d x=$$

$\int_\limits0^1\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots\right.$ upto $\left.\infty\right) e^{2 x} d x=$

$$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x=$$

$$\int_\limits0^1\left(\frac{x^2-2}{x^2+1}\right) d x=$$

$$\int_\limits{-5}^5 \log \left(\frac{7-x}{7+x}\right) d x=$$

$$\int_0^a \sqrt{\frac{x}{a-x}} d x=$$

$$\int_\limits2^3 \frac{x}{x^2-1} d x=$$

$$\int_\limits0^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=$$

$$\int_0^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta=$$ ............

$$\int_{\frac{\pi}{18}}^{\frac{4 \pi}{9}} \frac{2 \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\ldots \ldots$$

$$\int_\limits a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\ldots \ldots$$

$$\int_0^1 x(1-x)^5 d x=\ldots \ldots$$

If $\int_0^a \sqrt{\frac{a-x}{x}} d x=\frac{K}{2}$, then $K=\ldots .$.

The value of $\int_{-3}^3\left(a x^5+b x^3+c x+k\right) d x$, where $a, b, c, k$ are constants, depends only on

$$\int_0^4 \frac{1}{1+\sqrt{x}} d x=\ldots \ldots$$