Chapter 8, Problem 18RE

To determine

To find : the area of the region enclosed by the lines and curves.

Answer to Problem 18RE

The area of the region is $5.7312$ .

Explanation of Solution

Given information : the bell the region enclosed by the graphs of $y=3^{1-x^2},y=\frac{x^2-3}{10}$ .

Calculation :

The function $y=3^{1-x^2},y=\frac{x^2-3}{10}$ .

The integral is intersects at $x=\pm 1.8933$ .

The integral to find the area bounded by these curve:

  $\begin{array}{l}{\displaystyle \underset{-1.8933}{\overset{1.8933}{\int }}\left(3^{1-x^2}-\frac{x^2-3}{10}\right)dx}\\ \frac{3\sqrt{\pi }\left(\sqrt{\ln \left(3\right)x}\right)}{2\sqrt{\ln \left(3\right)}}-\frac{\frac{x^3}{3}-3x}{10}\\ \frac{3\sqrt{\pi }\left(\frac{18933\sqrt{\ln \left(3\right)}}{10000}\right)+3417662475921\sqrt{\ln \left(3\right)}}{2\sqrt{\ln \left(3\right)}}\\ \Rightarrow 5.7312\end{array}$