To integrate this function, we need to know a few things.

1) Any constant term can be pulled out of the integral and multiplied through at the end. So our original expression becomes this:

2) Integrating one over x doesn't follow the classic integrating rules. In this case, the integral of 1 over x is equal to the ln of x. So we do the following

We need to remember our plus c, otherwise we will not quite be correct. 

We could then evaluate the function on any interval to find the area underneath the curve.

To find the area of a curve, we can use an integral.

Set up the following:

Because we want the area between the y axis and 4, we need to evaluate the integral from 0 to 4. (Hence the limits on our integral sign)

Recall that to integrate a polynomial, we need to increase each exponent by 1 and divide by that number, so our integral changes as follows.

So, we need to evaluate our expression from 0 to 4, which means we will plug in 4 and 0 into the equation and take the difference. However, this gets even easier when we consider that plugging in 0 will result in 0. It goes as follows:

So if we simplify, we get the following:

So our answer is 29696 units squared

Looking at the graph, we see that the area lies between the points (0,0) and (1,1). Then, you can either differentiate with respect to y or x. With respect to y with  to the right. We can just use the positive square root because of the domain of x being only positive. With respect to x with  on top. 

To find the area of a region, set up an integral and evaluate it. In this case, we need the area under the curve of g(t) on the interval [5,17]. Therfore, our integral should look like the following:

Recall to evaluate integrals with polynomials, we need to increase each exponent by 1, and then divide the term by that number. Then, once we have integrated, we need to evaluate on the given interval by plugging in each number and finding the difference between them.

This looks a little messy, but once we simplify it, it will look much better.

Okay, still a little messy, but not too bad!

Making the area under the curve 97968 units squared

Notice that the second equation is not a function.  It would be easier to integrate with respect to y instead of x.

The following formulas are for area under the curve for two scenarios:

It will be easier to use the second scenario.  The right curve is .

Find the intersection of the two curves by setting both equations equal to each other.

 and 

These will be the lower and upper bounds of this integral.  Set up the integral and evaluate.

The area of the region to the nearest integer is .

Given the information that the region is bounded by ,  and , it is necessary to evaluate the bounds of the integral.

Substitute  into  to determine the values of .

These values will be the lower and upper bounds of the integral.  

Set up the integral and evaluate.

First, we find the points of intersection by setting the two equations equal to each other and solving. So, we solve  to get  and .

Then, we use the points of intersection as our limits of integration and find the definite integral of the top function - the bottom function; if you are unsure which is which, you can either graph the two together for a visual answer, or evaluate both functions at a point between the points of intersection. The larger answer will correspond to the top function .

So, proceeding with the integral, we get

 at

 , and at  

.

So, we have that our answer is 

.

To find the area underneath a curve within a given integral, integrate the curve between the upper and lower limits:

The integral of 3x² is x³ and the integral of - sin(x) is cos(x), giving:

 from  to .

or

 which simplifies to  ³ 

To find the area between the curve and the x-axis, we will need to integrate the curve with respect to x:

 from x = 0 to 1

Introducing are upper and lower bounds, we can find the area to be:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following: 

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two.

1. Using the power rule which states,

 to each term we find,

.

2.  , which is  after rounding.