# Calculus 1 : Functions

## Example Questions

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### Example Question #128 : Area

Find the area of the region bound by the lines , the x-axis, and the function v(x)

Explanation:

Find the area of the region bound by the lines , the x-axis, and the function v(x)

First, we need to set up our integral.

It should look like this:

Next, recall the rule for integrating polynomials: Increase each exponent by 1 and divide by that number:

Simplify this to get

Next, evaluate our integral by finding the difference between V(10) and V(5)

### Example Question #129 : Area

Find the area below the function  and above the function  from x=0 to x=5.

Explanation:

To find the area between the curves, we must integrate the difference between the upper and lower function between the x values specified:

The integral was found using the following rule:

Evaluated between the x values, we get

### Example Question #130 : Area

Find the area from x=0 to x=4, under the function of  and above the function of .

Explanation:

To determine the area between the two functions, we take the integral of the difference between the upper and lower functions evaluated between the bounds stated:

The integration was performed using the following rule:

When we evaluate it definitely, we get

The result of the integration is -24, but the area itself is the magnitude of this:

.

### Example Question #131 : Area

Calculate the area of the region between the function and the x-axis on the interval .

Explanation:

The area of the region between a function and the x-axis on the interval   is given as the definite integral

Because    on the interval  the area formula becomes

In order to solve the integral we use the inverse power rule which says

And because integrating is a linear operation we apply the power rule term-by-term to get

And by the corollary of the Fundamental Theorem of Calculus

### Example Question #132 : Area

Find the area of the region bound by the y-axis, h(t), and the lines

Explanation:

Find the area of the region bound by the y-axis, h(t), and the lines

First, we want to set up our integral:

Next, recall that  integrates to , and that to integrate a polynomial, we simply take each term, increase its exponent by 1, and divide by the new number. Doing so yields the following:

Now, to find the area, we need to find the H(6)-H(5)

Simplify to get:

So, we can round our answer to get:

### Example Question #133 : Area

Find the area between the curves of  and  from

Explanation:

Set this up as a definite integral to find volume. In order to do this, we need to know which function is above the other. In this case,   for all

To solve for area

Notice that

Therefore:

By the fundamental theorem of calculus:

### Example Question #134 : Area

Calculate the area between  and  round to the second decimal place.

Explanation:

We can find the interval that we are supposed to integrate over by setting the two equations equal to each other.

Manipulating the equation you get

.

Factoring this gives you,

.

Therefore we integrate between 1 and 9.

Next we see that if we were to graph both equations,  is above  and we know that we need to have the upper function minus the lower so our integral will be the following:

### Example Question #135 : Area

Many times people forget that area under a curve can be found, not only by integration, but also sometimes by using simple geometry.

Find the area under the following function

on the interval .

It is not possible

Explanation:

There are two different ways to solve this problem.  You can integrate both pieces of the piecewise function separately and add them together to get the area or you can look at the problem in a more geometric way.

First we graph the equation:

.

We can actually make this into two right triangles

.

The area of a triangle is known as

.

The red triangle has a height of  and a base of length two.

The blue triangle has a height of  and a base of one.

Therefore the area of this function on the interval  is

.

### Example Question #136 : Area

Which of the following is an expression that describes the area enclosed by  and .

The question is invalid.

Explanation:

The first step is to write the first equation as a function of x.  You will note however that this is the equation for a circle and is thus not a function of x.  This ends up not being a problem however since  is only located in quadrant 1 and 2 and thus we only need the piece of the circle in quadrant 1 and 2 which happens to be a function, namely .

The next step is to find the points in which these two functions intersect by setting them equal to each other.

This yields the points . All that is left now is to plug them into the correct formula:

.

Yet since it is symmetric, we can rewrite it as

.

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