Embibe Experts Solutions for Chapter: Definite Integration, Exercise 2: Exercise-2

The value of ${\int }_0^1\left(\left\{2x\right\}-1\right)\left(\left\{3x\right\}-1\right)dx$ (where, $\left\{·\right\}$ denotes fractional part of $x$) is equal to:

Let $A_1={\int }_0^x\left({\int }_0^{u}f\left(t\right)dt\right)du$ and $A_2={\int }_0^{x}f\left(u\right)\left(x-u\right)du$, then $\frac{A_1}{A_2}$ is equal to:

Let $f\left(x\right)$ be a strictly increasing, non-negative function such that $f^{"}\left(x\right)<0\forall x\in R \& I={\int }_{\alpha }^{\beta }f\left(x\right)dx \left(\beta >\alpha \right)$, then

$I_1={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx, I_2={\int }_0^{\pi }\frac{x^3\sin x}{\left({\pi }^2-3\pi x+3x^2\right)\left(1+{\cos }^{2}x\right)}dx,$ then

Let $L_1=\underset{x\to 0^{+}}{\lim }\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x-\sin x}, L_2=\underset{x\to 0^{-}}{\lim }\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x-\sin x},$ then identify the correct option(s).

$\underset{n\to \infty }{\lim }\frac{\left(1^k+2^k+3^k+\dots ..+n^k\right)}{\left(1^2+2^2+\dots ..+n^2\right)\left(1^3+2^3+\dots \dots .+n^3\right)}=F\left(k\right),$ then $\left(k\in N\right)$

Let $T_n=\sum _{r=1}^n\frac{n}{r^2-2r.n+2n^2}, S_n=\sum _{r=0}^{n-1}\frac{n}{r^2-2r.n+2n^2},$ then

$f\left(x\right)={\int }_0^{1}f\left(tx\right)dt,$ where $f^{\text{'}}\left(x\right)$ is a continuous function such that $f\left(1\right)=2,$ then