Question
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
✓
Answer
Let $\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
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