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

There are 2 things to note.

1. This function is positive only from  to . From  to , this function is negative.

2. This function is not differentiable at  because of the absolute value.

Therefore, this function needs to be broken up into 2 pieces.

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 values.

1. Using the power rule which states,

 to each term we find,

.

2. Plug in the respective upper and lower limits for x and then take the difference.

Now take the sum of the integrals of the two pieces.

Find the area under  where .

Take the integral of :

Next, evaluate it on the interval:

To find this area, you can either take the integral of  from 1 to , or you can redefine the integral in terms of  and solve that way, as the integral of the natural log is more difficult to find.

To do this, rewrite  as  and then redefine the bounds of the integral, so instead of integrating on  you can integrate on .

The final integral is 

.

This however, is the area between the curve and the y-axis. To find the remaining area, subtract this from the area from  to the -axis on , which has an area of .

Thus, the final expression resulting in the answer is,

  .

To find the area under the curve, we need to perform a definite integral. Essentially, this integral will be summing up all the infinitesimally small rectangles that make up the region. The entire region is in the first quadrant, so we don't have to worry about splitting our region up. 

When we take the integral we will need to use,

 then plug in the upper and lower bounds into the function and take the difference.

Therefore,

To find the area under the curve, evaluate 

.

Remember the integral of  is .

Integrating yields:

Now, evaluate:

 If  on the interval , then the area bounded by the two functions is given by 

.

To find the area bounded by  and , we must first find when they intersect.  

Setting the functions equal , we get .

Therefore the two functions intersect at  and  and we wish to find the area of the region bounded by  and  between these two points. By plotting the two functions, we see that on this interval, .

Therefore, the area of the region bounded by these functions is given by

.

The area for the region bounded by the graph of  and the x-axis on the interval  is given as

.

Because  on the interval  the area is

 .

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

 .

As such,

And by the corollary of the First Fundamental Theorem of Calculus,

.

Hence the area is

 square units.

The area for the region bounded by the graph of  and the x-axis on the interval  is given as

.

As such,

.

And because the two regions are symmetric about the y-axis,

.

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

 .

As such,

.

And by the corollary of the First Fundamental Theorem of Calculus,

.

Hence the area is  units squared.

The area under the curve of any function is given by the integral of the function. Note that the function evaluated from x=0 to x=1 is always positive, so the integration over the entire interval can be done using one integral. The integral of the function is given by

and is equal to 

.

The integration comes from the rules 

 and .

Evauating the integral between the two points is done by plugging in the upper limit of integration (x=1) into the function and then subtracting by the function when the lower limit of integration (x=0) is plugged in. When evaluated from the point x=0 to x=1, the integral is equal to 

.

The dot product of  and  is the sum of the products of its individual corresponding components, or

.