Answer
$$\int^{\sqrt3}_{0}\frac{4x}{\sqrt{x^2+1}}dx=4$$
$$\int^{\sqrt3}_{-\sqrt3}\frac{4x}{\sqrt{x^2+1}}dx=0$$
Work Step by Step
Part a) $$A=\int^{\sqrt3}_{0}\frac{4x}{\sqrt{x^2+1}}dx$$
Set $u=\sqrt{x^2+1}$, we then have $$du=\frac{(x^2+1)'}{2\sqrt{x^2+1}}dx=\frac{2xdx}{2\sqrt{x^2+1}}=\frac{xdx}{\sqrt{x^2+1}}$$
For $x=0$, we have $u=1$ and for $x=\sqrt3$, we have $u=2$.
Therefore, $$A=\int^2_14du=4u\Big]^2_1$$ $$A=4(2-1)=4$$
Part b) $$A=\int^{\sqrt3}_{-\sqrt3}\frac{4x}{\sqrt{x^2+1}}dx$$
We carry out a similar substitution as in part a).
For $x=-\sqrt3$, we have $u=2$ and for $x=\sqrt3$, we have $u=2$.
Therefore, $$A=\int^2_24du=0$$
