- #1
evsong
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1. Suppose : f(1) = 2, f(4) =7 , f'(1)=5, f'(4) = 3 and f"(x) is continuous. Find the value of:
[tex]
\int_{1}^{4} xf''(x)dx
[/tex]
[tex]
IBP formula
\int u(x)dv = u(x)v(x) - \int v(x) du
[/tex]
I re-wrote the IBP formula from
[tex]
= f(x) \int g(x) - \int\int g(x) f(x)'
[/tex]
I turned it into that so I can see so I can see that the derivative of [tex] u [/tex] is [tex] du [/tex] And derivatives [tex]f'(1)=5[/tex] [tex] f'(4)=3 [/tex] are given.
so does that mean that two du's are given?
well I set
[tex]
f(1) = \int_{1}^{4} xf''(x)dx = 2
\newline
[/tex]
and
[tex]
\newline
f(4) = \int_{1}^{4} xf''(x)dx = 7
[/tex]
This is where I am stuck. I don't know what to do next. Thanks for the help in advanced :D
[tex]
\int_{1}^{4} xf''(x)dx
[/tex]
Homework Equations
[tex]
IBP formula
\int u(x)dv = u(x)v(x) - \int v(x) du
[/tex]
The Attempt at a Solution
I re-wrote the IBP formula from
[tex]
= f(x) \int g(x) - \int\int g(x) f(x)'
[/tex]
I turned it into that so I can see so I can see that the derivative of [tex] u [/tex] is [tex] du [/tex] And derivatives [tex]f'(1)=5[/tex] [tex] f'(4)=3 [/tex] are given.
so does that mean that two du's are given?
well I set
[tex]
f(1) = \int_{1}^{4} xf''(x)dx = 2
\newline
[/tex]
and
[tex]
\newline
f(4) = \int_{1}^{4} xf''(x)dx = 7
[/tex]
This is where I am stuck. I don't know what to do next. Thanks for the help in advanced :D
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