Question

What is the net area between `f(x) = e^(3x)-4x` and the x-axis over `x in [1, 2 ]`?

Answer 1

`1/3e^6-1/3e^3-6~=121.78`

Explanation:

We can represent this area as the integral of `f(x)` on the interval `[1,2]`:
`int_1^2e^(3x)-4x\ dx`

We start by computing the antiderivative:
`int\ e^(3x)-4x\ dx=int\ e^(3x)\ dx-int\ 4x\ dx`

The left one can be solved using a u-substitution with `u=3x` and `(du)/dx=3`:
`int\ e^(3x)\ dx=int\ e^u/3\ du=e^u/3=e^(3x)/3`

So now we know:
`int\ e^(3x)-4x\ dx=e^(3x)/3-2x^2`

We can then evaluate the definite integral:
`int_1^2e^(3x)-4x\ dx=[e^(3x)/3-2x^2]_1^2=(1/3e^6-8)-(1/3e^3-2)`

`=1/3e^6-1/3e^3-8+2=1/3e^6-1/3e^3-6~=121.78`