Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #21 : Improper Integrals

Determine 

Possible Answers:

Correct answer:

Explanation:

By the fundamental theorem of calculus:

Example Question #22 : Improper Integrals

Evaluate 

Possible Answers:

1

Does Not Exist

0

Correct answer:

Explanation:

First compute the indefinite integral. Remember that

Evaluate the integral from the limits given

 is always  and any number to the power of 0 is always 1.

therefore: 

Example Question #23 : Improper Integrals

Let  be a constant real number. Evaluate .

Possible Answers:

The integral does not exist

Correct answer:

Explanation:

This intergral is improper because of the upper limit at infinity. To handle this, we use the theorem that if an integral 

 

exists, then it is equal to the value of 

.

Applying this to our given integral, we get:

Using the power rule for integration, which is valid since ,we obtain:

Here the second term in the limit didn't depend on , so it evaluates to 

.

To evaluate the limit for the first term, we note that from , we have that .

Therefore the power on  is always less than  for any . We have the theorem that 

.

Therefore the first term in the limit as  vanishes, and we are left with the result 

.

Example Question #23 : Improper Integrals

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

The integral we are given is improper due to the upper bound being infinity. To integrate, we must rewrite the integral with a limit for the upper bound:

Note that we moved the constant outside of the integral and the limit. 

Now, we can integrate, remembering to keep the limit in front of every step in the process:

The integral was performed using the following rule:

Now, we can definitely integrate by plugging in the upper and lower bounds and subtracting the results:

Note that the absolute value goes away because the bounds of the integral are both positive.

The function goes to infinity as the natural log of infinity is infinity.

Example Question #24 : Improper Integrals

Evaluate the following improper integral:

Possible Answers:

Correct answer:

Explanation:

Substitute a in for infinity and solve the integral:

Now, take the limit of this expression as a approaches infinity:

Example Question #25 : Improper Integrals

Integrate:

Possible Answers:

Correct answer:

Explanation:

To evaluate the improper integral (infinity is its upper limit of integration which makes it improper), we must use a limit for the upper bound.

But before we do this, we can use a u-substitution to rewrite the integral as something we can easily integrate:

The derivative is found using the identical rule to the derivative.

Now, rewrite the integral, with a limit for the upper bound, and integrate

The integral was found using the following rule:

Next, we can replace u with our x terms, and perform the definite integration:

As t goes to infinity, the denominator of the first term goes to infinity, which means the entire term goes to zero. We are left with

Example Question #2661 : Calculus Ii

Integrate:

Possible Answers:

Correct answer:

Explanation:

To evaluate the improper integral (infinity is its upper bound), we must use a limit on the upper bound:

Note that we are approaching infinity from the right because it is the upper bound of integration. 

We can now integrate (don't forget the limit!):

The following rule was used for integrating:

Now, the definite part of the integration:

Note that because the bounds of the integral were both positive, we removed the absolute value sign. When t goes to infinity, the natural logarithm function goes to infinity. The natural logarithm of 1 is zero, so we are left with infinity.

Example Question #26 : Improper Integrals

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

This integral is improper because the lower bound creates a zero in the denominator. To integrate, we must use a limit:

The following rule was used:

Now, the definite part of the integral:

The natural logarithm of 1 equals zero, and as the natural logarithm approaches zero, the function goes to negative infinity. The negative sign in front of the function makes it go to infinity.

Example Question #21 : Improper Integrals

Determine whether the series converges absolutely or conditionally

Possible Answers:

Only conditional convergence

No it neither converges absolutely nor conditionally

Converges absolutely and conditionally

Only absolute convergence

Correct answer:

No it neither converges absolutely nor conditionally

Explanation:

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Example Question #28 : Improper Integrals

Evaluate the integral 

Possible Answers:

Correct answer:

Explanation:

We can evaluate this integral using integration by parts

. (Start)

 (Use integration by parts, with )

.

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