In order to find the area under  on the interval of  to , you must evaluate the definite integral

First, integrate the function.

Then, substitute values for .

Finally, evaluate while cancelling negative signs.

The area described is the integral 

.

This can be evaluated using the linear properties of integrals and the Power Law, which says

.

Thus the integral above is equal to

.

To the find the area of the region between  and , one needs to first figure out which function is above the other in the interval . Plug in  and  into each function to check which one has the higher value at these respective points.

 has a higher value than at both of these x-values, and since both functions are increasing during this interval, we can conclude that  is above  during this interval.

After that, set the definite integral 

.

Using the general rule for integrals,

and for exponential functions,

on our function we get

  from  to .

This equals to 

.

Area of a region between two graphs  and , and between two  values  and  can be given as

Since we want to find the area between  and  between  and ,

we can set this up as one definite integral. 

We can ignore the value until we have a numerical answer.

First, we know that by the sum rule

By the power rule, we know that

, where are constants and  is a variable.

Therefore,

We also know as an identity that

, when 

Therefore,

We also know that for definite integrals,

, where 

In our case, 

In order to find the area of the region between the curve of the function and the x-axis on the interval [,], we solve for the integral

.

For this problem the integral becomes

And since the fuction is always positive on the interval, the integral becomes

.

When taking the integral, we will use the inverse power rule which states,

 .

Applying this rule we get

 .

And by the corollary of the First Fundamental Theorem of Calculus

 .

As such, the area is 

 square units.

In order to find the area of the region between the curve of the function and the x-axis on the interval , we solve for the integral

.

For this problem the integral becomes

.

And since the function is always positive on the interval, the integral becomes

.

When taking the integral, we will use the inverse power rule which states,

 .

Applying this rule we get

 .

And by the corollary of the First Fundamental Theorem of Calculus

 .

As such, the area is 

 square units.

For the region to the right of the y-axis, the range of x is .  The next step is to find the integral of the function:

The method of integration by parts will serve here:

Now,  converges to zero; however, to determine what  conveges to, use L'Hopital's Rule:

Since  can be rewrittena s  and both the numerator and denominator approach  as x does:

The function that bounds a rectangle is a horizontal line.

In this case, , so the area of the shape is just the integral of this line over the bounds of the x-axis, which in this case is 3 and 7 since the rectangle starts at 3 and ends at 7.

Thus, is the solution. 

The area of the region bounded underneath by the x-axis can be found by integrating the function

over the specified interval of :

To integrate use the rule,

.

Applying this rule to our function we find the following.

To find the area of the region between the functions  , , the first step will be to determine the lower and upper x bounds. We're told the region is bounded by the y-axis on the left, so . The upper x-bound will be where the two functions intersect:

For the value of , each of the functions equals .

Over the interval of , , so the area of the region can be determined by the integral:

Use the following rules to integrate the function:

 and 

Applying the above rules we get,

.