Find the area bound by the x and y axes, the function g(x), and the line 

Because we are asked to find area, we want to set up an integral. Be sure to include the correct limits of integration:

So our final answer is:

Find the intersection points created by the bounded lines.  The intersection of , ,  and  is .    The intersection of lines  and  is .

Since the area does not include the region below , we must subtract the areas of the top curve with the area of the bottom curve.  The top curve is  and the bottom curve is .

Integrate the function  with the respect to  from  to .

Integrate the function  with the respect to  from  to .

Subtract both areas to get the area of the bounded region.

To solve, simply integrate the function from to . Thus,

Since  is only defined when  Also, the graph of the function  is the upper half of the circle with center as  and radius . By definition of integral, we can find the value of  is the same as the area of the semi circle. Therefore, it is equal to 

To solve, simply integrate from  to . Thus,

To find the area between two curves, we first need to determine on what interval we will be taking their integrals. This interval has endpoints where the two functions intersect, so we need to find the intersection points.

To find intersection points of two curves, set the two functions equal (since the -values are the same at the intersection points) and solve for the -value(s).

Here:

So or , and we integrate over the interval .

We also need to know which curve is above the other. Find this by graphing or by testing -values in the interval of interest. In this case, the linear function is the uppermost.

Then take the integral of the upper function (the linear here), and subtract the integral of the lower function (the quadratic here), so that we subtract out the area we do not want to count.

To find the area under the curve, we need to take the definite integral. But in order to do that, we first need to determine the limits of integration.

In this case, since the graph of the function is a parabola opening downward, we need to find the places where it intersects the -axis -- the zeros of the function.

To find zeros, set the function equal to 0 and solve for :

  //  divide by -1 to make factoring easier

  //  factor

  //  solve

 or

So we want to take the integral from to  of .

The antiderivative is . (We omit  when finding the definite integral.)

We want , or

To find the area of a curve, you are not finding the slope (thus not differentiating) but rather integrating. Specifically, you only differentiate once to find 2-dimensional area. Thus, the answer is "integrate the function once."

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:

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: