The interval  divided into two sub-intervals gives two rectangles with vertices of the bases at 

For the approximated area, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or  and 

Because each sub-interval has width , the approximated area using rectangles is

As such, the approximated area is

 units squared

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the left endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 100.

In order to approximate the area under a curve using rectangles, one must take the sum of the areas of discrete rectangles under the curve. Taking the height of each rectangle as the function evaluated at the right endpoint, we obtain the following rectangle areas:

The sum of the individual rectangles yields an overall area approximation of 225.

The area under the curve of the function  is the definite integral from  to .

Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.

From here, we find the difference between the function values of the boundaries.

In order to solve a system of linear equations, we must start by solving one of the equations for a single variable:

We can now substitute this value for y into the other equation and solve for x:

Our last step is to plug this value of x into either equation to find y:

We first move  to the left side of the equation:

Subtract the bottom equation from the top one:

Left Side:

Right Side:

So

So dividing by a -1 we get our result.

For any system of linear equations, we can start by solving one equation for one of the variables, and then plug its value into the other equation. In this system, however, we can see that both equations are equal to y, so we can set them equal to each other:

Now we can plug this value for x back into either equation to solve for y:

So the solutions to the system, where the lines intersect, is at the following point:

In order to solve a system of linear equations, we can either solve one equation for one of the variables, and then substitute its value into the other equation, or we can solve both equations for the same variable so that we can set them equal to each other. Let's solve both equations for y so that we can set them equal to each other:

Now we just plug our value for x back into either equation to find y:

So the solution to the system is the point:

Start from equation 3 because it has the least number of variables. We see directly that .

Back substitute into the equation with the next fewest variables, equation 2. Then,

. Solving for , we get 

 or .

Then back substitute our  and  into equation 1 to get

.

Solving for x,

.

So our solution to the system is

We can solve the system using elimination. We can eliminate our  by multiplying the top equation by :

and then adding it to the bottom equation:

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We can now plug in our y-value into the top equation and solve for our x-value:

Our solution is then