The area under a curve f(x) between two x-values  corresponds to the integral 

.

In this case, , and we can find by the fundamental theorem of calculus that...

 evaluated from  to 

The first step to find the area between two curves is to set them equal to each other and find their intersection points.

The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval . We find that  is on the top and  is on the bottom.

Now, to find the area between the two curves, we take the integral from  to  of the top curve minus the bottom curve.

evaluated from  to 

The definite integral of f(x) over the interval [0, 5] is the area under the curve.

A substitution makes this integral easier to evaluate. Let . Then . We can also change the limits of integration so that they are in terms of u. When x=5, . When x=0, . Making these substitutions, the integral becomes

The area under a curve  over the interval  is . 

In this example, this leads to the definite integral 

.

 A substitution makes the antiderivative of this function more obvious.

Let 

.

We can also convert the limits of integration to be in terms of  to simplify evaluation. When , and when .  

Making these substitutions results in

.

 Recall that in general , so evaluating the integral leads to

In order to find the anti-derivative, lets first factor the denominator.

Now we use partial-fraction decomposition to get this into a form we know how to take the anti-derivative.

Now we need to find the values of A and B. First lets write out the equation.

Now we can get our system of equations in order to find A and B.

From equation 1, we get

Now we can substitute this into equation 2.

Now we substitue B back into equation 1.

Now we can put the values of A, and B into our problem.

Now we can simply take the anti-derivative.

To evaluate the integral, we use inverse power law.

Remember that the inverse power law is

.

Lets apply this to our problem.

From here we plug in the bounds and subtract the lower bound function value from the upper bound function value.

Asking for the area under the curve means you are going to be doing an integral of the function provided. The bounds are given for you. The integral thus looks like this: .

When integrated, you get 

 evaluated at  and .

At  the expression evaluates to  and at  it evalutes to .

Subtracting the two gives you your final answer of .

For this particular function we will need to preform a "u-substitution".

In our case let

 which will make .

Now we will substitute these into our integral to get the following.

 if   

Then we integrate using the power rule which states,

Now plug back in the original variable and then subtract the function values at the bounds.

To find the area under the curve, we must integrate over the given bounds:

To integrate, we must do the following substitution:

The derivative was found using the following rule:

Now, rewrite the integral, and change the bounds in terms of u:

Note that during the rewriting, the bounds changed to the upper bound being -1 and the lower bound being 1, but the negative sign that came from the derivative of u allowed us to make the bounds as they are seen above. 

Now perform the definite integration:

The integral was found using the following rule:

and the definite integration was finished by plugging in the upper bound into the resulting function, plugging the lower bound into the resulting function, and subtracting the two, as seen above. 

To find the area underneath the curve of the function  on the interval , we find the definite integral

Because the function in this problem is always positive on the interval, or

 on the interval  the area underneath the curve can be found using the definite integral

and rewriting the function the definite integral is

Using the inverse power rule which states

we find the definite integral to be

And by the corollary of the Fundamental Theorem of Calculus, the definite integral equals

As such, the area is  square units