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Mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Question: Section 7.2 Problem 13
problem7.2.13 (previously 7.2.50 was 7.3.48) f(0)=6, f(1) = 5, f '(1) = 2; find int( x f '', x, 0, 1).
**** What is the value of the requested integral?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I begin by letting u = x and u’=1
I let v’=f’(X) and v = f(x)
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The integrand is x f '', not x f '.
Unfortunately the font and the spacing of the document appear to obscure this. The first ' often looks like part of the letter f.
Having chosen u = x, you would therefore need to let v ' = f '' (x).
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So we let u = x and v ' = f '', leading to v = f ' and du = dx.
An antiderivative is therefore
u v - integral ( v du )
= x f ' - integral (f ' (x) dx )
= x f ' (x) - f(x).
The definite integral from 0 to 1 is thus
( 1 * f ' (1) - f(1) ) - (0 * f ' (0) - f(0))
= 1 * 2 - 5 - (0 - 6)
= 2 - 5 + 6
= 3.
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So now I use u*v-integral(u’v)
So I have xf’(x)-integral 1*f(x)
Xf(X)-integral f(x) the integral of f(x) is f(x) so we have
xf(x)-f(x)
we are told that f’(1)=2, F(1)=5 and f(0)=6
At this point I am confused on how to proceed, I
##??? I had the same steps as the given solution up until this point :
xf'(x)-integral of f'(x) ( I got this step)
Integral of f'(x) is f(x). So antiderivative is ( I got this step)
x f ' (x)-f(x) Why did the antiderivative become xf”(x) - f(x) instead of xf(x) - f(x)???
Besides this I am still having a hard time figuring the value of the interval. I think I should use the first fundamental theorem of calculus, and I know I evaluate at each value of the integral but I am confused. Could you walk me through this process and also if available provide me a formula like you did with the chain rule that would clarify for me???
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See my notes. You read f ' (x) where the problem had f '' (x).
This isn't entirely your fault, as the font and spacing of the original document made this difficult to distinguish.
I think that's your main point of confusion. Check my notes and let me know if you have additional questions on this.
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#$&*
mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Practice test 1 emailed
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I emailed you a practice test #1 as an email attachment. I wondered if you could give me and idea of what my grade would have been if this were the real test,and also ask that you give me suggestions for improvement, and let me know which ones I got wrong or need to correct. I hope through taking a few practice tests I will feel more confident and be better prepared for the actual tests.
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I got the email. However I can't insert comments into a PDF.
On #4 you have the right general behaviors and did a great job translating the positive/negative behavior and positive/negative slopes into the increasing/decreasing behaviors and concavity of the integral.
However I don't think the graph of the integral ever goes negative, and if it does is doesn't go very far into negative values. The positive area accumulated between x = 0 and x = about 3 appears to exceed the negative area between about x = 3 and about x = 5.5, so there is never a point to which the accumulated negative area exceeds the positive area previously accumulated.
Past x = 5.5 the positive area overwhelms any negative area that follows. So even if there is a short interval over which the graph of the integral is negative, the value will never get close to being negative again.
I can't comment further on this because the problem is split between two pages and it's just too inefficient for me to flip back and forth. If you have additional questions on this problem and/or want to send a revision you're welcome to include a copy of your graphs on a single page so I can compare them directly.
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For #5 you have a good table of values for F ' (t), but to estimate F(t) you would forst need to graph F ' (t) and approximate the areas beneath the curve for the interval from t = 0 to t = .3, and from t = .3 to t = 1.6. It would be worth your effort to make these estimates.
Those areas correspond to the definite integrals of F'(t), so it's while it would be instructive and I recommend it, it's not actually necessary to estimate the areas associated with your graph. You can simply integrate F ' (t) from 0 to .3, then from .3 to 1.6, and use your results to find F(.3) and F(1.6).
Reminder on notation: You used f ' (t) for the derivative of F(t). The derivative of F(t) would be F ' (t), not f ' (t).
There is a convention in the text that f(t) can stand for the derivative of F(t), and with this convention you could label your derivative function f(t). But you would not want to label it f ' (t), which would be likely to cause you confusion.
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You did a very good job on #6, and would have received full credit, but your notation leaves me a little nervous. You will be more secure on problems of this type of you use a little clearer notation.
If dy/dt = t/3 + 1/3, then
y = integral(t/3 + 1/3 with respect to t) = t^3 / 6 + t / 3 + c.
You have this result but you called it dt.
Note that if y = t^3 / 6 + t / 3 + c, then
dy / dt = t / 3 + 1/3,
so it is indeed appropriate to say that for this situation, y = t^3 / 6 + t / 3 + c, or if we want to emphasize that y is a function of t, we can write
y (t) = t^3 / 6 + t / 3 + c.
Now, y(1.3) = 1.5 gives you the equation
1.5 = 1.3^2 / 6 + 1.3/3 + c
which matches your equation, and has solution c = .79.
Thus
y(t) = t^3 / 6 + t / 3 + .79.
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I think the last problem looks very good. I can't check it in full detail because it's split between pages and I can't keep track of details in the question while flipping back and forth. You're welcome to send a copy of the problem and solution on one page so I can check all the details. I think they're all there, but I really need to see the questions and your solutions at the same time to make sure you're answering the questions exactly as they are asked.
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I've included notes on the problems that required them. Problems on which I didn't include notes (and even some on which I did) would have received full credit.
You would have made a solid B on this test, and you're easily within reach of A work.
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Do sent me a text copy of the test questions themselves so I can file them for future use, along with my comments.
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