In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

We plug in the equation of the curve in for , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

Next, we integrate our expression, ignoring our  and  values for now. We get:

Now, to find the area under the part of the curve we're looking for, we plug our  and  values into our integrated expression and find the difference, using the following skeleton:

In this problem, this looks like:

Which plugged into our integrated expression is:

With our final answer being:

First, integrate the expression. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator as well: . Then, evaluate that expression by plugging in 2 and then subtract from that the result from plugging in 0: .

The first step I'd recommend is to chop up the expression into 3 terms since we have one denominator: . Now, integrate each term, remembering to add one to the exponent and then putting that result in the denominator as well: Tack on a "+C" at the end because it is an indefinite integral.

The first step I'd recommend is to chop up the expression into 3 terms since we have one denominator: . Now, integrate each term, remembering to add one to the exponent and then putting that result in the denominator as well: Tack on a "+C" at the end because it is an indefinite integral.

The integral is equal to 

and was found using the following rules:

Note that all of the constants of integration added up to a single C.

To evaluate the integral, we can split the integral into the sum of two integrals:

The first integral can be solved using a substitution:

and the integral rewritten in terms of u is

The integral was solved using the identical rule.

Now, we replace u with our original term:

.

Solving the second integral, we get

using the following rule:

Summing the results, we get

To evaluate the integral, we must make the following substitution:

The derivative was found using the following rule:

Next, we rewrite the integral and integrate:

The integral was found using the following rule:

Finally, replace u with our original x term:

To integrate, we can split the integral into three integrals, for ease of making substitutions later:

To integrate the first integral, we must make the following substitution:

Next, rewrite the integral and integrate:

The integral was performed using the following rule:

Finally, replace u with the original term:

The result of all three integrals, added together, is

and the other two integrals were integrated using the following rules:

,

To evaluate the integral, we must make the following substitution:

The derivative was found using the following rule:

Next, rewrite the integral in terms of u and integrate:

The integral was found using the following rule:

Finally, replace u with our original x term:

To evaluate the integral, we first must perform the following substitution:

The derivative was found using the following rule:

Now, rewrite the integral and integrate:

The integral was performed using the following rule:

Finally, replace u with our original term: