# Find the area bounded by the curve y=e^(x/3) , the tangent line which is y=1/3 e^(3)x - 2e^3 and the x-axis?

## Do I need integration by part to do this?

Jun 4, 2018

$\frac{3}{2} {e}^{3} - 3$

#### Explanation: The straight line $y = \frac{1}{2} {e}^{3} x - 2 {e}^{3}$ is tangent to $y = {e}^{\frac{x}{3}}$ at the point $\left(9 , {e}^{3}\right)$. It is also easy to see that it cuts the $X$ axis at $\left(6 , 0\right)$

The area under the curve $y = {e}^{\frac{x}{3}}$ from $x = 0$ to $x = 9$ is given by

${\int}_{0}^{9} {e}^{\frac{x}{3}} \mathrm{dx} = {\left[3 {e}^{\frac{x}{3}}\right]}_{0}^{9} = 3 {e}^{3} - 3$

The area that we are looking for is the difference between the area under the curve $y = {e}^{\frac{x}{3}}$ from $x = 0$ to $x = 9$ and the area of the triangle with vertices at $\left(6 , 0\right)$, $\left(9 , 0\right)$ and (9,e^3). The area of the triangle is $\frac{1}{2} \times 3 \times {e}^{3} = \frac{3}{2} {e}^{3}$

Thus the area that we need is

$3 {e}^{3} - 3 - \frac{3}{2} {e}^{3} = \frac{3}{2} {e}^{3} - 3$