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

##### 1 Answer

The area is

#### Explanation:

The area between a curve and the

Here, we have

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

We can integrate term by term and then evaluate from

#=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

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

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