Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #341 : Calculus 3

Integrate: 

Possible Answers:

Correct answer:

Explanation:

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Example Question #152 : Calculus Review

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 #342 : Calculus 3

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 #154 : Calculus Review

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 #49 : Integration

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 #155 : 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:

 

Example Question #156 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

To integrate, we simply expand the function and integrate:

 

 

Example Question #154 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

To solve this problem, we would have to apply u-substitution: 

Substituting for u and dx, get us: 

The  terms cross out and simplifying we get: 

We can then 'isolate' the constant: 

To integrate , we have to convert it to 

Since the derivative of cos(x) is -sin(x), we would have to apply u-substitution again, but using a different variable to prevent confusion: 

Replacing for v and du gets us: 

Simplifying get us: 

Integrating gets us: 

Now substitute v = cos(u) from earlier: 

Now substitute u = 2/x from earlier as well and now we get: 

Example Question #155 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

We notice that this problem can be solved with a simple u -substitution: 

Substituting for u and dx get us: 

The  terms cross out, and simplifying we get: 

Isolating the -1 outside the integral we get: 

This is now simply: 

Substituting , from earlier gets us our final answer: 

Example Question #161 : Calculus Review

Possible Answers:

Correct answer:

Explanation:

This problem can be solved using u-substitution:

Substituting for u and dx in the original integral gets us: 

However the factors x+1 and 2x+2 don't cross out, but we can factor 2x+2 to cross out x+1: 

Simplifying gets us: 

Isolating the 1/2 from the integral gets us: 

Substituting  from earlier, we now get: 

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