Calculus 3 : Integration

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #41 : Integration

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

The only way to evaluate this integral is through integration by parts. To do this, you must follow the equation:

You must assign values for u and dv from the original integrand and then find the values of du and v. Specifically:

From here, plug in the values into the equation for integrating by parts:

To evaluate this integral, a u-substitution is needed:

Now, the integrand from the equation can be rewritten as:

When you replace u with x and add the other half of the equation, the final answer is:

 

Example Question #42 : Integration

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

The only way to evaluate this integral is through integration by parts where the formula states:

Values for u and dv must be picked from the integrand and then the remaining values are found from those. Specifically:

Now, these values must be plugged into the equation:

The integral that now must be evaluated can be done so using a u-substitution:

Next, every x must be replaced with u and integrated:

After replacing u with x and adding the remainder of the equation, the final answer is:

Example Question #43 : Integration

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

The only way to evaluate this integral is by splitting it in to two separate integrals:

The first integral can be evaluated by using a u-substitution:

Then, x must be replaced with u and evaluated:

For the second integral, the integrand resembles the form:

In this case,  and the integral of it is 

After adding all of the evaluated integrals, the final answer is:

 

Example Question #44 : Integration

Evaluate the following integral:

Possible Answers:

Correct answer:

Explanation:

The only way to evaluate this integral is by recognizing that the integral resembles the following:

To evaluate the integral, the x-term must be replaced with u:

Now that the denominator is in the proper form, the integral can be evaluated according to the first equation:

U must be replaced with x to give the final answer:

Example Question #152 : Calculus Review

Integrate: 

Possible Answers:

Correct answer:

Explanation:

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Example Question #45 : Integration

Evaluate: 

Possible Answers:

Correct answer:

Explanation:

Rewrite the function in terms of ln:


Plug in the high level and lower level limits:


Subtract higher from lower:

Answer=

Example Question #154 : Calculus Review

Solve:

Possible Answers:

Correct answer:

Explanation:

The feature of this integral that may initially be confusing is the bounds of integration, because they contain a variable. However, this is something that will be used frequently in later integration problems. Treat the bounds as normal, and integrate:

After the definite integration, we get

 

Example Question #46 : Integration

Integrate:

Possible Answers:

Correct answer:

Explanation:

To integrate, we must use integration by parts, which states that

We must designate a u and a dv from the integral given, and then differentiate and integrate, respectively:

Now, use the above formula and integrate again:

which simplified becomes

Example Question #156 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

This question requires the use of a complex conjugate. as the other methods of integration do not seem to work. In the case, we take the complex conjuage of the denominator, which is sec(x)+ 1: 

Simplyifying the above equation results in: 

We now notice that the denominator is a trig identity: , we can replace it with 

Now we can separate the above equation into two different integrals: 

Simplifying each integral, now you get: 

For the first integral, , we can use u-substitution to integrate it: 

Crossing out cos(x), we now get: 

Now integrate as normal: 

Substitute u = sin(x) 

Simplifying:

 

For the second integral,  , we would use the trig identity  . Rearranging the equation gets us 

Now we can integrate separately:

 

Now combining the result from the first integral, , we now get 

Example Question #157 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

At first, we notice that the power in the numerator is smaller than the one in the denominator, but we can't quite use u-substitution yet, as the derivative of the denominator is 2t - not 2t -1. To do this we must separate the integrals: 

Now, we can use u-substitution for the first integral by setting 

Which results in: 

The 2t's cross out and now we get: 

Now integrate as normal: 

Substitution back for , we now get: 

For the second integral, , we have to use a special formula: 

, or in this case translates to: 

Combining the two results get us:

 

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