Question

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?

Answer 1

`3/2e^3-3`

Explanation:

The straight line `y=1/2e^{3}x-2e^3` is tangent to `y=e^{x/3}` at the point `(9,e^3)`. It is also easy to see that it cuts the `X` axis at `(6,0)`

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

`int_0^9 e^{x/3}dx = [3 e^{x/3}]_0^9 = 3e^3-3`

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

Thus the area that we need is

`3e^3-3-3/2e^3 = 3/2e^3-3`