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Found in: Page 369

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the area of the shaped region.

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

See the 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.