Question

Find the area of the shaped region.

Short answer

The shaded area is\(e - \frac{1}{e} + \frac{{10}}{3}\)square units.

Step-by-step solution

Given information

Consider two continuous functions \(f(y)\)and \(g(y)\)such that \(f(y) \ge g(y)\)for \(y \in (c,d).\)

The area bounded between\(x = f(y),x = g(y)\)and the horizontal lines\(y = c,d\)is given by

\(A = \int_c^d f (y) - g(y)dy\)

Finding the area of shaped region

On the interval \(y \in \left( { - 1,1} \right)\), we can observe that \({e^y} \ge {y^2} - 2\), on the graph

As a result, the shaded area is

\(\begin{aligned}{l}\int\limits_{ - 1}^1 {{e^y} - \left( {{y^2} - 2} \right)} dy\\ = \int\limits_{ - 1}^1 {{e^y} - } {y^2} + 2dy\\ = \left( {{e^y} - \frac{{{y^{2 + 1}}}}{{2 + 1}} + 2y} \right)_{ - 1}^1\end{aligned}\)

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

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

\( = e - \frac{1}{e} - \frac{2}{3} + 4\)

The shaded area is\(e - \frac{1}{e} + \frac{{10}}{3}\)square units.