Definite Integration · Mathematics · TS EAMCET

$$ \lim _{n \rightarrow \infty} \frac{\left(2 n(2 n-1) \ldots .(n+2)(n+1)^{1 / n}\right.}{n}= $$

If $\int_0^{\frac{\pi}{2}} \tan ^{14}\left(\frac{x}{2}\right) d x=2\left[\sum_{n=1}^7 f(n)-\frac{\pi}{4}\right]$, then $f(n)=$

$$ \int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} d x= $$

$$ \int_0^{\frac{\pi}{2}} \frac{d x}{\cos x-\sqrt{3} \sin x}= $$

$$ \int_0^{\frac{\pi}{2}} \sqrt{\tan x d x}= $$

$$ \int_{-1}^1 \frac{\log 2-\log (1+x)}{\sqrt{1-x^2}} d x= $$

$$ \int_0^{\frac{\pi}{4}} \frac{\sec x}{3 \cos x+4 \sin x} d x= $$

$$ \int_{-2}^4\left|2-x^2\right| d x= $$

$$ \int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x= $$

$$ \int_8^{18} \frac{1}{(x+2) \sqrt{x-3}} d x= $$

If [.] denotes the greatest integer function, then $\int_1^2\left[x^2\right] d x=$

$$ \mathop {\lim }\limits_{n \to \infty } \frac{1}{n^2}\left[e^{1 / n}+2 e^{2 / n}+3 e^{3 / n}+\ldots+2 n e^2\right]= $$

Let $m, n, p, q$ be four positive integers. If

$$ \begin{aligned} & \int_0^{2 \pi} \sin ^m x \cos ^n x d x=4 \int_0^{\pi / 2} \sin ^m x \cos ^n x d x \int_0^{2 \pi} \sin ^p x \cos ^n x d x=0 \\ & \int_0^\pi \sin ^p x \cos ^q x d x=0, a=m+n+p \text { and } b=m+n+q, \text { then } \end{aligned} $$

$$ \int_0^2 \sqrt{(x+3)(2-x)} d x= $$

$$ \int_0^{\pi / 4} x^2 \sin 2 x d x $$

$$ \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x d x= $$

$ \int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) d x= $

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

$$ \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x= $$

$$ \lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots(2)\right]^{1 / n}= $$

$$ \int_{-1}^1 x|x| d x= $$

$$ \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x= $$

If $\int_0^3\left(3 x^2-4 x+2\right) d x=k$, then an integer root of $3 x^2-4 x+2=3 k / 5$ is

$$ \int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x= $$

If $[x]$ represents greatest integer function, then

$$ \int_{-2}^2[2-x] d x= $$

$$ \int_0^2 \frac{x}{(2-x)^{\frac{3}{4}}} d x= $$

$$ \int_0^2 x^3(2-x)^4 d x= $$

$$ \int_0^3\left|x^2-3 x+2\right| d x= $$

$$ \int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x= $$

$$ \int_0^2 x^{\frac{5}{2}} \sqrt{2-x} d x= $$

$$ \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x= $$

$$ \lim\limits_{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots \ldots+\frac{1}{n} \sec ^2 1\right]= $$

$$ \int\limits_2^5 \sqrt{\frac{5-x}{x-2}} d x= $$

$$ \int\limits_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x d x= $$

[.] is the greatest integer function, then

$$ \int_0^{2 \pi}[|\sin x|+|\cos x|] d x= $$

$$ \int_0^4| | x-2|-x| d x= $$

If $\int_{-a}^a f(x) d x=\int_0^a f(x) d x+\int_0^a g(x) d x$, then $g(x)=$

76. Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: R \rightarrow R$ is defined by $f(x)=x^2+2$, then

$$ \lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]= $$

If $f(x)=\left|\begin{array}{ccc}2 \cos ^2 x & \sin 2 x & \sin x \\ \sin 2 x & 2 \sin ^2 x & -\cos x \\ \sin x & -\cos x & 0\end{array}\right|$, then

$$ \left.\int_0^{\pi / 4}|2| f(x) \mid+5 f^{\prime}(x)\right) d x= $$

$\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x=$

$$ \int_0^{\pi / 2} \sin ^4 \theta \cos ^3 \theta d \theta= $$

It is given that $\frac{d}{d t}(t \log t-t)=\log t$, then $\exp \left(\int_0^1 2 x \log \left(1+x^2\right) d x\right)=$

$$ \int_0^{2 a} f(x) d x= $$

$$ \int_1^2 x \sqrt{4-x^2} d x= $$

If $[x]$ denotes the greatest integer function of $x$ and

$$ \int_{-3 / 2}^{3 / 2}[2 x-3] d x=k, \text { then }\left|k+\frac{1}{2}\right|= $$

$$ \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} d x= $$

Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$

If $f(x)=\frac{1}{x^3} \int_5^x\left(2 u^2-u f^{\prime}(u) d u\right.$, then $f^{\prime}(5)=$

Assertion (A) $\int\limits_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$

Reason (R) $\int\limits_a^b f(x) d x=\int_{g(a)}^{g(b)} f(g(u)) g^{\prime}(u) d u$

The correct option among the following is

If $\cos x+\cos 2 x+\ldots+\cos n x=\frac{A(x)}{2 \sin x / 2}$, then $\int\limits_0^\pi A(x) d x=$

$$ \begin{array}{r}\mathop {\lim }\limits_{n \to \infty }\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}\right. \\ +\frac{n^{3 / 2}}{(n+4)^{5 / 2}}-\frac{n^{1 / 2}}{(n+6)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}} \\ \left.-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]= \end{array} $$

$$ \lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]= $$

If

$$ f(x)=\left|\begin{array}{ccc} 1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{array}\right| $$

then $\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=$

$$ \lim\limits_{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{n} \sin ^{-1} \frac{1}{n}+\frac{2}{n} \sin ^{-1} \frac{2}{n}+\ldots+\frac{\pi}{2}\right]= $$

The positive integer $n \leq 5$ for which $\int_0^1 e^x(x-1)^n d x=16-6 e$ is

If $f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x$, then $\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x=$

$$ \int_3^5(x-3)^3(5-x)^5 d x= $$

$$\mathop {\lim }\limits_{x \to \infty } \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]= $$

$$ \int_0^{\pi / 2} \frac{d x}{4+5 \sin x} $$

$$ \mathop {\lim }\limits_{x \to \infty }\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}= $$

$$ \int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}= $$