Answer
$$\int^{\sqrt2}_{1}x2^{(x^2)}dx=\frac{1}{\ln2}$$
Work Step by Step
$$A=\int^{\sqrt2}_{1}x2^{(x^2)}dx$$
We set $x^2=u$, which means $$2xdx=du$$ $$xdx=\frac{1}{2}du$$
- For $x=\sqrt2$, we have $u=2$ and for $x=1$, we have $u=1$
Therefore, $$A=\frac{1}{2}\int^{2}_{1}2^udu$$ $$A=\frac{1}{2}\times\frac{2^u}{\ln2}\Big]^{2}_{1}$$ $$A=\frac{1}{2\ln2}(2^2-2^1)$$ $$A=\frac{2}{2\ln2}=\frac{1}{\ln2}$$
