Definite Integration

\(\int_0^4| | x-2|-x| d x=\)

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

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

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

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

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

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

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 $\t...

If $[x]$ denotes the greatest integer function of $x$ and
\(\int_{-3 / 2}^{3 / 2}[2 x-3] d x=k, \text { then }\left|k+\frac{1}{2}\right|=\)

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

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

$$ \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...

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

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

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

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

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

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

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...

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

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

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

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