Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

\(y = {e^x},y = {x^2} - 1,x = - 1,x = 1\)

Short answer

The area of the region is\(e - \frac{1}{e} + \frac{4}{3}\)

Step-by-step solution

Given information

The given equation is

\(y = {e^x},y = {x^2} - 1,x = - 1,x = 1\)

The area bounded by the curve

Choose whether to integrate in terms of x or y.

\(y = {e^x},y = {x^2} - 1,x = - 1,x = 1\)

Because integrating along the y-axis would involve sophisticated splitting up, it will be easier to integrate along the x.

The width of the approximate rectangle is\(\Delta x\), and the height is the top function minus the bottom function.

\(f - g = {e^x} - \left( {{x^2} - 1} \right) = {e^x} - {x^2} + 1\)

Determining the area of shaped region

\(A = \int\limits_{ - 1}^1 {\left( {{e^x} - {x^2} + 1} \right)} dx = \left[ {{e^x} - \frac{1}{3}{x^3} + x} \right]_{ - 1}^1\)

\(\begin{aligned} &= e - \frac{1}{3} + 1 - \left( {{e^{ - 1}} + \frac{1}{3} - 1} \right)\\ &= e - \frac{1}{3} + 1 - \frac{1}{e} - \frac{1}{3} + 1\end{aligned}\)

\( = e - \frac{2}{3} + 2 - \frac{1}{e}\)

\( = e - \frac{1}{e} + \frac{4}{2}\)

\( \approx 3.684\)

The area of the region is \(e - \frac{1}{e} + \frac{4}{3}\)