High School Math : High School Math

Example Questions

Example Question #21 : Finding Indefinite Integrals

What is the indefinite integral of  with respect to ?

Possible Answers:

Correct answer:

Explanation:

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

Be sure to include  to compensate for any constant!

Example Question #21 : Finding Indefinite Integrals

Possible Answers:

Correct answer:

Explanation:

To find the indefinite integral, or anti-derivative, we can use the reverse power rule. We raise the exponent of each variable by one and divide by that new exponent.

Don't forget to include a  to cover any constant!

Simplify.

Example Question #2241 : High School Math

Possible Answers:

Correct answer:

Explanation:

To find the indefinite integral of , we can use the reverse power rule. To do this, we raise our exponent by one and then divide the variable by that new exponent.

Don't forget to include a  to cover any constant!

Example Question #61 : Asymptotic And Unbounded Behavior

Determine the indefinite integral:

Possible Answers:

Correct answer:

Explanation:

, so this can be rewritten as

Set . Then

and

Substitute:

The outer factor can be absorbed into the constant, and we can substitute back:

Example Question #62 : Asymptotic And Unbounded Behavior

Evaluate:

Possible Answers:

Correct answer:

Explanation:

Set . Then

and

Also, since , the limits of integration change to  and .

Substitute:

Example Question #1 : Finding Integrals By Substitution

Determine the indefinite integral:

Possible Answers:

Correct answer:

Explanation:

Set . Then

.

and

The integral becomes:

Substitute back:

Example Question #82 : Calculus Ii — Integrals

Possible Answers:

Correct answer:

Explanation:

This integral will require a u-substitution.

Let .

Then, differentiating both sides, .

We need to solve for dx in order to replace all x terms with u terms.

.

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from  to . Because , the bounds will change to  and .

Essentially, we have made the following transformation:

.

The latter integral is easier to evaluate.

At this point, we can separate the integral into two smaller integrals.

.

The integral evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite  as the sum of two fractions.

We need to solve for the values of A and B.

This means that  and . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that .

Let's now go back to the integral .

Distribute the 2 to both integrals and separate it into two integrals.

.

Remember we need to add this value back to the value of , which we already determined to be -2.

The final answer is .

Example Question #81 : Calculus Ii — Integrals

Evaluate:

Possible Answers:

Correct answer:

Explanation:

Set .

Then and .

Also, since , the limits of integration change to  and .

Substitute:

Example Question #1 : Understanding Taylor Series

Give the   term of the Maclaurin series of the function

Possible Answers:

Correct answer:

Explanation:

The   term of the Maclaurin series of a function  has coefficient

The second derivative of  can be found as follows:

The coeficient of  in the Maclaurin series is therefore

Example Question #2243 : High School Math

Give the   term of the Taylor series expansion of the function  about .

Possible Answers:

Correct answer:

Explanation:

The  term of a Taylor series expansion about  is

.

We can find  by differentiating twice in succession:

so the  term is