Answer
Integrate with respect to x.
$$area = e - \frac{1}{e} + \frac{4}{3}$$
Work Step by Step
$y = e^x, y = x^2 - 1, x = -1, x=1$
Before solving for the area, the aspects of the curves need to be identified.
$\underline{y = e^x}$:
y-intercept:
$e^0 = y$
$y = 1$
$(0, 1)$
$\underline{x^2 - 1}$:
x-intercepts:
$x^2 - 1 = 0$
$x = \pm1$
$(-1,0),(1,0)$
y-intercept:
$y = 0^2 -1$
$y = -1 \rightarrow (0,-1)$
vertex (when $y\prime$ = 0):
$y\prime = 2x$
$2x = 0$
$x = 0$
$(0,-1)$
With both graphs' aspects calculated, the graphs can be drawn.
To find the area, integrate with respect to $x$.
Based on graphs, $f(x) = e^x$ and $g(x) = x^2 - 1$
Limits of integration: $a = -1$ and $b = 1$
$$\int_a^b f(x) - g(x) dx$$
$$\int^1_{-1} e^x - (x^2 - 1) dx$$
Solve the definite integral and simplify completely.
$$= \int^1_{-1} (e^x - x^2 + 1)dx$$
$$= (e^x - \frac{x^3}{3} + x)|^1_{-1}$$
$$= [e^1 - \frac{1^3}{3} + 1] - [e^{-1} - \frac{(-1)^3}{3} + (-1)]$$
$$= e - \frac{1}{3} + 1 - \frac{1}{e} - \frac{1}{3} + 1$$
$$area = e - \frac{1}{e} + \frac{4}{3}$$
