anonymous
  • anonymous
Can some please help me with this: Suppose f(3)=4, f(8)=8, f'(3)=2, f'(8)=9, and f'' is continuous. Find the value of the definite integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{3}^{8}\] xf''(x)
anonymous
  • anonymous
I need this broken down in layman terms
amistre64
  • amistre64
well, if we try to use this according to the definition of integration .. what can we get?

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amistre64
  • amistre64
\[f(x)=\int f'(x)dx\]is the basic understanding of an integral
anonymous
  • anonymous
I don't understand what to do with the function values
amistre64
  • amistre64
it would appear to me that the 3 to 8 is the limits of integration
amistre64
  • amistre64
\[\int_{3}^{8}x\ f''(x) dx\] is what we have to work with and all we have to determine is what to do with the "x" part
amistre64
  • amistre64
i might have to logout and back in to do this ...
anonymous
  • anonymous
ok
amistre64
  • amistre64
the math still aint processing from the latex on my end; you see normal equations on your end?
anonymous
  • anonymous
yes I can see them.
amistre64
  • amistre64
good :) well then, lets see if we can move ahead with this then i think we can apply integration by parts to this .. just a gut felling
amistre64
  • amistre64
\[\int u dv =uv-\int v du\]
anonymous
  • anonymous
ok so would i just use xf''(x) as my function to integrate?
amistre64
  • amistre64
to wit: u = x v = f'(x) du = dx dv = f''(x) dx yes
anonymous
  • anonymous
so when I integrate that is where the values that were given will be used?
amistre64
  • amistre64
\[\int xf''(x)=xf'(x)-\int f'(x)dx\] \[\int xf''(x)=xf'(x)-f(x)\]
amistre64
  • amistre64
yes
amistre64
  • amistre64
im sure that looks alot more inspiring on your end lol
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anonymous
  • anonymous
I shows up normal on my end
amistre64
  • amistre64
(8 f'(8)) - f(8) ) - (3 f'(3)-f(3)) should be our results
anonymous
  • anonymous
ok I didn't get that far but how did you get rid of the x?
amistre64
  • amistre64
I didnt; i just used it in the integration by parts formula. there is no reason to get rid of it since the interval speaks in x to begin with
anonymous
  • anonymous
ok i see know you used the limit that you are evaluating the integral?
amistre64
  • amistre64
yep
anonymous
  • anonymous
I got 53, is that correct?
anonymous
  • anonymous
I see my error, I am about to recalculate it
amistre64
  • amistre64
8*9 - 8 - (3*2-4) 72-8 - (6-4) 64 - 2 = 62
amistre64
  • amistre64
i hope that makes more sense, I have to be heading out now. good luck :)
anonymous
  • anonymous
I accidently used 8*8-9-)-(3*2-4). Thank you that was really helpful.

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