### All Calculus 1 Resources

## Example Questions

### Example Question #123 : How To Find Area Of A Region

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

**Possible Answers:**

**Correct answer:**

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 #124 : How To Find Area Of A Region

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

**Possible Answers:**

**Correct answer:**

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 #4121 : Calculus

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

**Possible Answers:**

**Correct answer:**

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 #212 : Regions

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

**Possible Answers:**

**Correct answer:**

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 #4122 : Calculus

Find the area between the curves of and from .

**Possible Answers:**

**Correct answer:**

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 #4123 : Calculus

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

**Possible Answers:**

**Correct answer:**

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 #4124 : Calculus

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 .

**Possible Answers:**

It is not possible

**Correct answer:**

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 #4125 : Calculus

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

**Possible Answers:**

The answer is not shown.

The question is invalid.

**Correct answer:**

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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