$\\int_{0}^{1} x e^{2x} dx$

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = x$ and $\mathrm{d}v=e^{2x}\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$ and $v=\frac{1}{2}e^{2x}$.

Step2: Apply the integration - by - parts formula

$\int_{0}^{1}xe^{2x}\mathrm{d}x=\left[\frac{1}{2}xe^{2x}\right]{0}^{1}-\frac{1}{2}\int{0}^{1}e^{2x}\mathrm{d}x$. First, evaluate $\left[\frac{1}{2}xe^{2x}\right]{0}^{1}=\frac{1}{2}\times1\times e^{2\times1}-\frac{1}{2}\times0\times e^{2\times0}=\frac{1}{2}e^{2}$. Second, evaluate $\frac{1}{2}\int{0}^{1}e^{2x}\mathrm{d}x$. Let $t = 2x$, then $\mathrm{d}t = 2\mathrm{d}x$ and $\frac{1}{2}\int_{0}^{1}e^{2x}\mathrm{d}x=\frac{1}{4}\int_{0}^{2}e^{t}\mathrm{d}t=\frac{1}{4}[e^{t}]_{0}^{2}=\frac{1}{4}(e^{2}-1)$.

Step3: Calculate the final result

$\int_{0}^{1}xe^{2x}\mathrm{d}x=\frac{1}{2}e^{2}-\frac{1}{4}(e^{2}-1)=\frac{1}{2}e^{2}-\frac{1}{4}e^{2}+\frac{1}{4}=\frac{1}{4}e^{2}+\frac{1}{4}$.

Answer:

$\frac{1}{4}e^{2}+\frac{1}{4}$