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

Let $f\left(x\right)=\left|x-2\right|, g\left(x\right)=x^2-5 \text{and} h\left(x\right)=x^3$. Then ${\int }_0^{3}f\left(x\right)dx=$

Let $f\left(x\right)=\left|x-2\right|, g\left(x\right)=x^2-5 \text{and} h\left(x\right)=x^3$. Then ${\int }_0^3\left(g\left(x\right)+h\left(x\right)\right)dx=$

Let $f\left(x\right)=\left|x-2\right|, g\left(x\right)=x^2-5 \text{and} h\left(x\right)=x^3$. Then ${\int }_2^{4}g\left(x\right)dx+{\int }_1^{2}4h\left(x\right)dx=$

Let $f\left(x\right)=\left|x-2\right|, g\left(x\right)=x^2-5 \text{and} h\left(x\right)=x^3$. Then ${\int }_1^4\left(\frac{g\left(x\right)}{h\left(x\right)}\right)dx=$

Let $f\left(x\right)=\left|x-2\right|, g\left(x\right)=x^2-5 \text{and} h\left(x\right)=x^3$. Then ${\int }_0^{3}f\left(x\right)dx+{\int }_2^{3}h\left(x\right)dx=$

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left({\sec }^{2}x\right) dx$ is equal to 

The value of the integral ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left[\sin \right(\mathrm{x}\left)\right]\mathrm{dx}$ is

The area of the region bounded by the curve $y^2=2y-x$ and the $y-$axis is