Answer
$$\frac{e-1}{2}$$
Work Step by Step
Area = $\int^b_a f(x) - g(x) dx$ Where f(x) is above g(x) on the coordinate system.
In this question $f(x) = e^x$ and $g(x) = xe^{x^2}$. The limits of integration will be $a$ = line (x = 0) and $b$ = the x-coord of the right interception (where $f(x)$ = $g(x)$).
$$\int^b_a f(x) - g(x) dx$$
$$= \int^1_0 e^x - xe^{x^2}$$
Next step is to solve the definite integral. Split the integral into two and use the substitution rule for the integral $xe^{x^2}$.
$$\int^1_0e^x dx - \int^1_0xe^{x^2}$$
Substitution:
$u = x^2$
$du = 2x dx$
$\frac{1}{2}du = x dx$
$a = 0 \rightarrow u = 0$ ||| $b = 1 \rightarrow u = 1$
$$= \int^1_0 e^x dx - \frac{1}{2}\int^1_0 e^u du$$
$$= (e^x)|^1_0 - \frac{1}{2}(e^u)|^1_0$$
$$= (e^1 - e^0)-\frac{1}{2}(e^1 - e^0)$$
$$= e - 1 -\frac{e-1}{2}$$
$$= \frac{e-1}{2}$$
