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

The value of the definite integral ${\int }_1^{\infty }{\left(e^{x+1}+e^{3-x}\right)}^{-1}dx$ is

The value of the definite integral ${\int }_1^e\left(\left(x+1\right)e^x\ln x\right)dx$ is-

Let $a, b, c$ be non-zero real numbers such that ; $\underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=\underset{0}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$, then the quadratic equation $a{x}^2+bx+c=0$ has -

If $f\left(x\right)=A\sin \left(\frac{\pi x}{2}\right)+B, f^{\text{'}}\left(\frac{1}{2}\right)=\sqrt{2}$ and $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{2A}{\pi }$, then the constant $A$ and $B$ are-

Solve:${\int }_0^{\infty }\frac{x{\tan }^{-1}x}{{\left(1+x^2\right)}^2}dx$.

Suppose $f$, $f^{\text{'}}$ and $f^{\text{'}\text{'}}$ are continuous on $\left[0, e\right]$ and that $f^{\text{'}}\left(e\right)=f\left(e\right)=f\left(1\right)=1$ and ${\int }_1^e\frac{f\left(x\right)}{x^2}dx=\frac{1}{2}$, then the value of ${\int }_1^e{f}^{\text{'}\text{'}}\left(x\right)\ln xdx$ equals -

${\int }_{\frac{1}{2}}^2\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$ has the value equal to -

The value of $\underset{\pi }{\overset{2\pi }{\int }}\left[2\sin x\right]dx$ is equal to (where [.] represents the greatest integer function)