#$&*
course MTH 174
1/28 5:44 PM
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
Optional preliminary questions:
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Question: `q001. From the product rule we know that (u v) ' = u ' v + u v '.
Applying the product rule to the expression t sin(t) we obtain
(t sin(t)) ' = sin(t) + t cos(t) so t cos(t) = sin(t) - (t sin(t)) '.
It follows that the indefinite integral of t cos(t) is
integral ( t cos(t) dt ) = integral ( sin(t) dt) - integral ( (t sin(t)) ' dt).
The integral of sin(t) is easily seen to be -cos(t) + c.
Since the an antiderivative of (t sin(t)) ' with respect to t is t sin(t), the right-hand side of the above is
integral ( sin(t) dt) - integral ( (t sin(t)) ' dt) = -cos(t) + t sin(t) + c
so
integral ( t cos(t) dt ) = -cos(t) + t sin(t) + c .
This is a good result, though since we started with the function t sin(t), we weren't actually looking looking for the integral of t cos(t). We could develop a table of integrals by starting with various product functions and proceeding as above. However it will be more useful to develop this idea into a method for integrating the function we want. We will do so in the next few problems.
Apply the reasoning above to the product function t e^t.
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Your solution:
confidence rating #$&*:
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Self-critique Rating:
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Question: `q002. The method illustrated in the preceding uses the product rule (u v) ' = u ' v + u v ', and the fact that an antiderivative of (u v) ' is immediately seen to be u v. We will consider u and v to be functions of t, and will use ' to indicate the derivative with respect to t
So whatever the functions u and v, (u v) ' is very easily integrated.
This means that if we can integrate either of the two expression u ' v and u v ', we can very easily find the integral of the other.
Specifically, suppose we can integrate u ' v. Then since we can rearrange (u v) ' = u ' v + u v ' to the form
u v ' = (u v) ' - u ' v
we can integrate u v ', obtaining
.integral( u v ' dt) = integral ( (u v) ' dt) - integral ( u ' v dt)
To apply the method we need to first be able to express a given function as a product of two functions. As we will see this doesn't always guarantee that the method will work, but we need to be able to take the first step.
There are often multiple ways to express a given function as a product of two functions. For example, t^2 cos(t) sin(t) can be expressed as the product of t^2 and cos(t) sin(t), or a the product of t cos(t) * t sin(t), or as the product of cos(t) with t^2 sin(t), and in a number of other ways. It could even be expressed as the product 1 * t^2 cos(t) sin(t).
Express each of the following functions as a product of two functions, in all possible ways:
t^2 * e^t
sin^2(x)
t^3 / (t^2 - 1)
x^2 ln(x)
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Your solution:
confidence rating #$&*:
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Self-critique Rating:
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Question: `q003. Now let's consider the function t^3 * e^(t^2).
We want to apply the fact that
u v ' = (u v) ' - u ' v
so we will want to express our function as u * v '. This means that we will express our function in various ways as a product of two functions, letting one of the two be u and the other v '. Right now we really don't know where this is heading and we don't yet have the experience needed to make good choice of u and v ', so we're going to use a scattershot approach, being prepared to try out every possibility. We will soon develop a 'feel' for the sorts of choices do and do not tend to work.
Among the several ways to express the function t^3 * e^(t^2) as a product u * v of two functions we will consider only the following candidates:
u = t^3, v ' = e^(t^2)
u = e^(t^2), v ' = t^3 (noting that it's going to matter which function is u and which is v)
u = t^2, v ' = t e^(t^2)
u = t e^(t^2), v ' = t^2.
Start with the second candidate.
If u = e^(t^2) and v ' = t^3, then what are the functions u' and v? Note that to find v we need to integrate our expression for v '.
In order to apply the formula, we need only be able to integrate the expression u ' v. What is the expression u ' v, and can we integrate it?
Among the remaining three choices there is one function v ' that cannot be integrated, another that is easily integrated, and another that isn't difficult to integrate by substitution.
For each of the remaining choices:
Find the functions u ' and v, if possible. If this isn't possible, state why, and since the procedure won't work if you can't find u ' and v, you can move on to the next function.
Determine if you can integrate u ' * v. If this isn't possible, state why, and since the procedure won't work if you can't integrate u ' * v, you can move on to the next function.
If you can integrate u ' * v, then since ( u v ) ' is easy to integrate you can apply the fact that
u v ' = (u v) ' - u ' v
to integrate u v ', obtaining
integral ( u v ' ) = integral ( (u v) ' ) - integral ( u ' v ).
What is your result?
Take the derivative of your result. Does it match the original function?
Note: In case you weren't able to integrate t e^(t^2), the result is 1/2 e^(t^2). This result is easily obtained using straightfoward substitution, as covered in the preceding assignment.
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Your solution:
confidence rating #$&*:
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Self-critique Rating:
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Question: `q004. Each of the following can be integrated by the method outlined in the preceding problems.
t^2 * e^(t^3)
sin(x) cos^2(x)
t / (t^2 - 1)
x ln(x)
The recommended process is:
Express the function as the product u * v ' in every possible way.
Eliminate any breakdown for which it is not possible to find v or u '. (In almost every case it will be possible to find u ', since the rules of differentiation work for any properly formed expression, but in many cases most choices of v ' cannot be integrated.)
For those breakdowns where u ' and v can be found, see if u ' v can be integrated.
If u ' v can be integrated, apply the formula to integrate the original function u v '.
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Your solution:
confidence rating #$&*:
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Self-critique Rating:
Note: A number of additional practice problems involving derivatives and integrals are included at the end of this document, listed under 'Supplemental Questions'.
Questions from text assignment:
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Question: Section 7.2 Problem 3
7.2.3 (previously 7.2.12. (3d edition 7.2.11, 2d edition 7.3.12)) Give an antiderivative of sin^2 x
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Your solution:
Sin^2x = ½(1 - cos(2x))
U = 2x
Du = 2
Du/2 = dx
Int (1) = x
Int cos(u) = sin(u)
Replace u and distribute the 1/2; ½(x - (sin(2x))/2))
(x/2)-(sin(2x)/4)
confidence rating #$&*: 3
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Given Solution: Good student solution:
The answer is -1/2 (sinx * cosx) + x/2 + C
I arrived at this using integration by parts:
u= sinx u' = cosx
v'= sinx v = -cosx
int(sin^2x)= sinx(-cosx) - int(cos x (-cos x))
int(sin^2x)= -sinx(cosx) +int(cos^2(x))
cos^2(x) = 1-sin^2(x) therefore
int(sin^2x)= -sinx(cosx) + int(1-sin^2(x))
int(sin^2x)= -sinx(cosx) + int(1) - int(sin^2(x))
2int(sin^2x)= -sinx(cosx) + int(1dx)
2int(sin^2x)= -sinx(cosx) + x
int(sin^2x)= -1/2 sinx(-cosx) + x/2
INSTRUCTOR COMMENT: This is the appropriate method to use in this section.
You could alternatively use trigonometric identities such as
sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x.
Solution by trigonometric identities:
sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is
1/2 ( x - sin(2x) / 2 ) + c =
1/2 ( x - sin x cos x) + c.
note that sin(2x) = 2 sin x cos x.
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Self-critique (if necessary):
Sin(2x) = 2 sinxcosx
So (x/2)-(sin(2x)/4) simplifies to:
(x/2) - ((sinxcosx)/2)
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Self-critique Rating: OK
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Question: Section 7.2 Problem 4
problem 7.2.4 (previously 7.2.16 was 7.3.18) antiderivative of (t+2) `sqrt(2+3t)
**** what is the requested antiderivative?
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Your solution:
(t+2)(2+3t)^(1/2)
U = t + 2 v = (2/9)(2+3t)^(3/2)
Du = 1 dv = (2 + 3t)^(1/2)
(t+2) (2/9)(2+3t)^(3/2) - INT[(2/9)(2+3t)^(3/2)]
(2/9)INT[(2+3t)^(3/2)]
U = 2+3t
Du = 3
Du/3 = dx
(U ^(3/2 + 1))/(5/2)
(2/9)(1/3)(2(2+3t)^(5/2)/5)
(4/135)(2+3t)^(5/2)
(t+2) (2/9)(2+3t)^(3/2) - (4/135)(2+3t)^(5/2)
confidence rating #$&*: 3
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Given Solution: If you use
u=t+2
u'=1
v'=(2+3t)^(1/2)
v=2/9 (3t+2)^(3/2)
then you get
2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or
2/9 (t+2) (3t+2)^(3/2) - 2 / (3 * 5/2 * 9) (3t+2)^(5/2) or
2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get
(3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to
2( 9t + 26) ( 3t+2)^(3/2) / 135.
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Self-critique (if necessary):
Needed to simplify further
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Self-critique Rating:3
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Question: Section 7.2 Problem 8
**** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3)
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Your solution:
Deriving cos(x^3) or integrating it seems to be the wrong way, so I’m lost.
confidence rating #$&*: 1
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Given Solution:
It usually takes some trial and error to get this one:
We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v.
We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with.
We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc..
The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following:
Let u = x^3, v' = x^2 cos(x^3).
Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have
1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx).
Now let u = x^3 so du/dx = 3x^2. You get
1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ).
It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one.
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Self-critique (if necessary):
I understand now.
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Self-critique Rating: 3
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Question: **** What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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Your solution:
U = x^5 v =
Du = 5x^4 dv = cosx^3
U = cosx^3 v= (x^6)/6
Du = dv = x^5
confidence rating #$&*: 3
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Given Solution:
TYPICAL STUDENT COMMENT:
I tried several things:
v'=cos(x^3)
v=int of v'
u=x^5
u'=5x^4
I tried to figure out the int of cos(x^3), but I keep getting confused:
It becomes the int of 1/3cosudu/u^(1/3)
I feel like I`m going in circles with some of these.
INSTRUCTOR RESPONSE:
As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.
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Self-critique (if necessary): OK
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Self-critique Rating: 3
&#This looks good. Let me know if you have any questions. &#