Embibe Experts Solutions for Chapter: Definite Integration, Exercise 1: Exercise 1

The tangent to the graph of the function $y=f\left(x\right)$ at the point with abscissa $x=1$ form an angle of $\frac{\pi }{6}$ and at the point $x=2,$ an angle of $\frac{\pi }{3}$ and at the point $x=3,$ an angle of $\frac{\pi }{4}$ with positive $x$-axis. The value of ${\int }_1^{3}f\text{'}\left(x\right)f"\left(x\right)dx+{\int }_2^{3}f"\left(x\right)dx$ ($f"\left(x\right)$ is supposed to be continuous) is

The value of ${\int }_{1/e}^{\tan x}\frac{t}{1+t^2}dt+{\int }_{1/e}^{\cot x}\frac{1}{t\left(1+t^2\right)}dt$, where $x\in \left(\frac{\pi }{6}, \frac{\pi }{3}\right),$ is equal to 

If $f\left(x\right)=x\left|x\right|$, then for any real numbers $a$ and $b$ with $a<b,$  the value of ${\int }_a^{b}f\left(x\right)dx$ equals to

If $I={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\left\{\frac{\pi +4x^5}{1-\sin \left(\left|x\right|+\frac{\pi }{6}\right)}\right\}dx$, then $I$ equals

If $\sum _{i=1}^4\left({\sin }^{-1}{x}_i+{\cos }^{-1}{y}_i\right)=6\pi ,$ then ${\int }_{\sum _{i=1}^4{x}_i}^{\sum _{i=1}^4{y}_i}x\ln \left(1+x^2\right)\left(\frac{e^x}{1+e^{2x}}\right) dx$ is equal to

$\underset{n\to \infty }{\lim } \sum _{r=1}^n\left(\frac{r^3}{r^4+n^4}\right)$ equals to

$\underset{n\to \infty }{Lt} \sum _{r=2n+1}^{3n}\frac{n}{r^2-n^2}$ is equal to

$\underset{n\to \infty }{\lim }{\left(\sin \frac{\pi }{2n}. \sin \frac{2\pi }{2n}. \sin \frac{3\pi }{2n}\dots \dots \sin \frac{\left(n-1\right)\pi }{n}\right)}^{\frac{1}{n}}$ is equal to