To solve this integral we must first realize that an integral is the opposite operation of a derivative and that the derivative of  is .

This means that  is our answer where A is a constant. We must take the chain rule into account to figure out the constant associated with our answer.

The derivative of  is  so that means we must multiply our integral by  to get our answer.

Also, we can't forget about the original 2 in the numerator.

Our integral then becomes 

.

To solve a definite integral we must take the difference of the function at the upper and lower limits. The answer is then

.

By the logarithmic divison rules, this answer is

.

Because we know that , this problem becomes easier than it looks.  To solve this, we actually use something known as the u-substitution method.  By setting , we are one step closer to solving this.  After deriving both sides with respect to , we have .

By multiplying both sides by , we have . 

 This allows us to be able to rewrite the problem as 

, giving us our final answer.

To find the average value of this function, we must find the total value of the area under the curve, and divide it by the number that represents the range of our interval , which is .  We can then use our understanding of integral calculus to find the anti-derivative and use the end points to find the total area under the curve.

If we divide this number by , we will come out with our average value, .

The only difference between this problem and a normal integral is the absolute value, which makes a small change in the computation and analytical understanding of the problem.  Graphically, the part of the function that would normally be under the x-axis is now flipped and positive.  This only means that we need to break this integral up into at least two parts.

First, we need to find the roots of our function. Because our function can be factored like this , we can see that our roots will be .  If we quickly analyze the graph, we can see that any value inside the interval  would normally be negative while values outside of this interval would normally be positive.  In order to solve the problem, we need to change all values of our function that are inside the given interval to positive values.

Once we have done this, we can add up the area of these three sections  (by using anti-derivatives) and come up with the answer of .

Which of the following integral expressions could be used to find the area bound by g(t), the y-axis, and the line .

The question asks us to set up an integral to find the area of a given region. Because we are given limits of integration, we must set up a definite integral (one with numbers at the top and bottom of the integral sign). 

Furthermore, we want our lower limit to be 0 and our upper limit to be 5. We know this because we need the area from the y-axis to 5.

So our integral should start like this:

Next, we want to put the function we are given within the integral as well, making the anwser:

Which of the following is the correct integral expression for this statement: Find the area of the region bound by f(z), the y and z axes, and the line .

Let's begin here by not getting too worried. Note that although this function may look intimidating, we don't actually have to calculate anything. All we are going to do is set this puppy up.

The first step is to set up the correct limits of integration. We want the smallest of the two to go on bottom. 

Now, you may be used to always having zero on the bottom, but -6 is smaller, so we will put -6 down below.

Next, it's just a matter of putting in f(z) and we have our integral:

Area of a circle is given by 

 and its circumference 

. 

Notice how circumference is the derivative of area with respect to radius . 

Using the power rule we know that:

, where and are constants.

Writing this as an integral,

The integral is equal to 

and was found using the following rules:

, 

To evaluate the integral we must perform the following substitution:

Now, rewrite the integral and integrate:

The integral was performed using the following rule:

Finally, replace u with our original term:

Compute the indefinite integral,

Compute the boundaries,

Simplify,