### All Calculus 1 Resources

## Example Questions

### Example Question #1061 : Functions

Evaluate the indefinite integral.

**Possible Answers:**

**Correct answer:**

In this problem we are being asked to integrate the function.

In order to do this we need to remember,

.

Since

then the integral,

.

### Example Question #12 : Equations

Evaluate the indefinite integral.

**Possible Answers:**

**Correct answer:**

This problem is asking us to integrate the function.

In order to do this we need to remember,

.

Therefore,

### Example Question #13 : Equations

Which of the following integrals could be evaluated to find the area under the curve of the following function from 3 to 13?

**Possible Answers:**

**Correct answer:**

We are asked to set up an integral without solving it. We want to evaluate the function from 3 to 13, so 3 goes at the bottom of the integral and 13 goes at the top. No other options have this correct.

### Example Question #1062 : Functions

Write an integral expression which will solve the area under the curve of , bounded by , from .

**Possible Answers:**

**Correct answer:**

Recall that area under the curve requires integration from either top minus bottom curve, or right minus left curve.

The top curve is:

The bottom curve is:

The interval indicates that the lower bound is 2 and the upper bound is 5.

Write the integral.

### Example Question #15 : Equations

Write the correct expression to find the area of from .

**Possible Answers:**

**Correct answer:**

This is a tricky question.

If we chose , the region in the third quadrant of is negative area, since is no longer the top curve. Evaluating this integral will cancel out the negative area in the third quadrant with the positive area on the first quadrant. This integral will give zero area, which is incorrect.

The correct method is to split this interval into 2 separate integrations: One from and the other from .

From interval , the top curve is minus the bottom curve, , is .

From interval , the top curve is minus the bottom curve, , is .

Set up the integral.

### Example Question #11 : How To Find Integral Expressions

Which of the following expressions could be evaluated to find the area under on the interval ?

**Possible Answers:**

**Correct answer:**

To find the area under a curve, we usually want to use an integral. All of our answer choices are integrals though, so this doesn't narrow things down. Then next thing to look for are our limits of integration. In this case, we are on the interval [360,540], so look for an option that has the lower limit of integration as 360 and the higher limit as 540. Only one option has this:

For further clarification, to set up an integral, we want our function within the integral and our limits of integration on the integral sign itself. Once we have the correct integral set up, we can evaluate it to find the area under the curve.

### Example Question #17 : Equations

Suppose a student wants to find the area under the curve between functions , , and . Find the correct integral that will determine the area.

**Possible Answers:**

**Correct answer:**

The region bounded by the three functions represents a triangle. It will be easier to determine the bounded region by subtracting the left curve from the right curve, and using to integrate. In order to integrate in terms of , any equation in terms of must be rewritten in terms of .

Isolate .

Find the intersection of the lines and .

The correct bounds are from . Write the correct integral.

### Example Question #2091 : Calculus

Evaluate the following **definite** integral:

**Possible Answers:**

**Correct answer:**

There are two steps to evaluating a definite integral. The first step is finding the anti-derivative of the function that we are integrating. In this problem, we are integrating the function , and the anti-derivative of this function is . (To check that your anti-derivative is correct, derive it. If you do not arrive back at the original function, go back and find where you may have made a mistake).

The second step is evaluating the anti-derivative at the given endpoints, and subtracting the second from the first. To do this, we look at the two endpoints, or "limits of integration," and decide which is higher. In this case, the two limits are and , and so we first evaluate the anti-derivative at . So, we get . Next, we evaluate at , and get , and subtracting this from the first value, we get .

### Example Question #1063 : Functions

Suppose a function . Setup an integral that will evaluate the area under the curve if this parabola is bounded by .

**Possible Answers:**

**Correct answer:**

First, in order to find the zeros of the function, we need to use the quadratic formula.

In our case,

.

These will be the lower and upper bounds of the integral.

Set up the integral in terms of .

### Example Question #11 : Writing Equations

Set up the correct integral to solve for the volume of the bounded region rotated about the -axis: , , and

**Possible Answers:**

**Correct answer:**

To find the volume, it is necessary to find the area of the disk, which is:

The radius is , since the function is rotated about the x-axis.

The volume is the antiderivative of the area.

The lower and upper bounds are defined by and .

Complete the integral.