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

Question

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

anonymous · Answer

$$\int\limits_{3}^{8}$$   xf''(x)

anonymous · Answer

I need this broken down in layman terms

amistre64 · Answer

well, if we try to use this according to the definition of integration .. what can we get?

amistre64 · Answer

$$f(x)=\int f'(x)dx$$is the basic understanding of an integral

anonymous · Answer

I don't understand what to do with the function values

amistre64 · Answer

it would appear to me that the 3 to 8 is the limits of integration

amistre64 · Answer

$$\int_{3}^{8}x\ f''(x) dx$$ is what we have to work with and all we have to determine is what to do with the "x" part

amistre64 · Answer

i might have to logout and back in to do this ...

anonymous · Answer

ok

amistre64 · Answer

the math still aint processing from the latex on my end; you see normal equations on your end?

anonymous · Answer

yes I can see them.

amistre64 · Answer

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 · Answer

$$\int u dv =uv-\int v du$$

anonymous · Answer

ok so would i just use xf''(x) as my function to integrate?

amistre64 · Answer

to wit:

u = x          v = f'(x)

du = dx     dv = f''(x) dx


yes

anonymous · Answer

so when I integrate that is where the values that were given will be used?

amistre64 · Answer

$$\int xf''(x)=xf'(x)-\int f'(x)dx$$
$$\int xf''(x)=xf'(x)-f(x)$$

amistre64 · Answer

yes

amistre64 · Answer

im sure that looks alot more inspiring on your end lol

anonymous · Answer

I shows up normal on my end

amistre64 · Answer

(8 f'(8)) - f(8) ) - (3 f'(3)-f(3))  should be our results

anonymous · Answer

ok I didn't get that far but how did you get rid of the x?

amistre64 · Answer

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 · Answer

ok i see know you used the limit that you are evaluating the integral?

amistre64 · Answer

yep

anonymous · Answer

I got 53, is that correct?

anonymous · Answer

I see my error, I am about to recalculate it

amistre64 · Answer

8*9 - 8 - (3*2-4)

72-8 - (6-4)

64 - 2 = 62

amistre64 · Answer

i hope that makes more sense, I have to be heading out now.  good luck :)

anonymous · Answer

I accidently used 8*8-9-)-(3*2-4).  Thank you that was really helpful.