All Calculus 3 Resources
Example Questions
Example Question #44 : Integration
Evaluate the following integral:
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 #151 : Calculus Review
Integrate:
Example Question #152 : Calculus Review
Evaluate:
Rewrite the function in terms of ln:
Plug in the high level and lower level limits:
Subtract higher from lower:
Answer=
Example Question #153 : Calculus Review
Solve:
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:
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
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
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
To integrate, we simply expand the function and integrate:
Example Question #154 : Calculus Review
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
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:
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