Question

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

Answer 1

The area is `(e^6-e^3-6)/3approx125.7811`.

Explanation:

The area between a curve and the `x`-axis can be found through integrating the function.

Here, we have

`"area"=int_1^2(e^(3x)-2x+1)dx`

We can integrate term by term and then evaluate from `1` to `2`.

`=int_1^2e^(3x)dx-2int_1^2xdx+int_1^2dx`

The first integral is the most difficult to find. We will want to get it into the form

`inte^udu=e^u+C`

Thus, we set `u=3x`, so `du=3dx`.

Manipulating just the first integral, we see:

`int_1^2e^(3x)dx=1/3int_1^2e^(3x)*3dx=1/3int_3^6e^udu`

It is very important to note that switching from `x` to `u` requires us to switch the bounds of the definite integral. The `1` and `2` have both been plugged into the `u=3x` expression, becoming `3` and `6`, respectively.

Now that we've found this integral, we can simply integrate the other two as well. All together, we have the expression:

`=1/3[e^u]_3^6-2[x^2/2]_1^2+[x]_1^2`

Which gives:

`=1/3(e^6-e^3)-2(2^2/2-1^2/2)+(2-1)`

`=(e^6-e^3)/3-3+1=(e^6-e^3)/3-2=(e^6-e^3-6)/3`

Approximately, this is `~~125.7811`.