All Calculus 2 Resources
Example Questions
Example Question #21 :Definite Integrals
Evaluate the indefinite integral.
None of the other answers
None of the other answers
The correct answer is.
The integral itself is not too difficult to take, simply use the Power Rule on theandterms. The trick is to be careful when integrating.is a constant value (about) not a variable, so it must be integrated accordingly.
Example Question #22 :Definite Integrals
Evaluate.
None of the other answers
This integral requires integration by parts followed by u-substitution. Here are the details
.Start
.Factor out the
Set up integration by parts with,.We then haveand.随后,我们利用分部积分formula.
.
Now at this point we use u-substitution to evaluate the 2nd integral. Let, thenand therefore.Substituting into the integral we have
.(Don't forget to change the bounds of integration by plugging them intofor our equation for.)
Example Question #23 :Definite Integrals
Evaluate.
Not possible without a calculator
This integral isn't possible to integrate directly using antiderivatives, but we can still find its value by noticing thatis an odd function, and that our limits of integration are negatives of each other.
Hence
.(Sinceis an odd function)
Example Question #24 :Definite Integrals
Evaluate
None of the other answers.
We can use u-substitution for this integral
.Start
Let, then, and our integral becomes
.(Don't forget to change the bounds of integration by plugging them into ourequation for)
Example Question #25 :Definite Integrals
Evaluate.
None of the other answers
We proceed by using integration by parts.
.Start
Let, then we get
.Then using the integration by parts formula, we get
Example Question #26 :Definite Integrals
Evaluate the following integral:
To evaluate this integral, we must integrate by parts, according to the following formula:
So, we must assign our u and dv, and differentiate and integrate to find du and v, respectively:
导数和积分被发现使用following rules:
,
Note that we ignore the constant of integration.
Now, use the above formula:
Note that both the product of u and vandthe integral are being evaluated from zero to.
The integral was performed using the following rule:
Simplifying the above results, we get.
Example Question #27 :Definite Integrals
Find the area betweenandbetween
We can write this problem as:
Integrating:
By the fundamental theorem of calculus:
Example Question #28 :Definite Integrals
Compute the Indefinite Integral
Evaluate the integral
Example Question #29 :Definite Integrals
Suppose, whereis a constant
Findsuch that
By the fundamental theorem of calculus:
Example Question #30 :Definite Integrals
Evaluate this integral.
Answer not listed
In order to evaluate this integral, first find the antiderivative of
In this case,.
The antiderivative is.
Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:
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