If F4 15 F is Continuous and 6 F X Dx 4 16 What is the Value of F6 F6

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Integration by parts conceptual problem

• Thread starter evsong
• Start date Sep 8, 2009

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]

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

Answers and Replies

What did you choose for u(x) and what did you choose for dv ?

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]

Homework Equations

IBP formula
[tex]

\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

I am not sure what you did there.
what is g(x)?

You should get
original integral = xf'(x) - int(f'(x)).dx and limits are defined all values are given.

oh would it be

u=x
du=dx

v= f'(x)
dv = f"(x)

so:

[tex]

xf'(x)- \int f'(x)dx
[/tex]

I don't know how to incorporate the given f'(1) =5 and f'(4) = 3

sorry for the double post. I accidently pressed submit before I was ready then clicked preview right after.

oh would it be

u=x
du=dx

v= f'(x)
dv = f"(x)

so:

[tex]

xf'(x)- \int f'(x)dx
[/tex]

I don't know how to incorporate the given f'(1) =5 and f'(4) = 3

oh would it be

u=x
du=dx

v= f'(x)
dv = f"(x)

This is a good start, because we don't know much about f''(x), but we do know some values for f'(x), so we want to get f'(x) in our result and get rid of f''(x).

so:

[tex]

xf'(x)- \int f'(x)dx
[/tex]

This doesn't work out for a simple reason; you have written here an indefinite integral, and the original expression is a definite integral. This expression still needs to be evaluated at limits in order to be a definite integral.

ok I need to evaluate it at the limits

[tex]
\left. xf'(x) \right|_1^{4} - \int_{1}^{4} f'(x) dx [/tex]

[tex]
\left. 4f'(x)-f'(x)- [f(x)] \right|_1^{4}
[/tex]

[tex]

4f'(4)-f'(1)-[f'(4)-f'(1)] [/tex]

[tex]
4*7 -(2)- [7-5] = 22 -2 = 20 [/tex]

how is this so far? I don't know how to incoorporate both the f'(1) and f'(4)

[tex]
\left. 4f'(x)-f'(x)- [f(x)] \right|_1^{4}
[/tex]

[tex]

4f(4)-f(1)-[f'(4)-f'(1)] [/tex]

These two lines are not equivalent. How did evaluating f at 4 and 1 turn into evaluating f' at 4 and 1 and vice versa? If you are confused about where to place the evaluation brace, note that by the fundamental theorem of calculus, the entire expression you got should be evaluated first at x=4 and then at x=1, with the latter being subtracted from the former.

I meant

[tex]
4f'(4) - f'(1) - [f'(4) - f'(1)]

[/tex]

i just forgot the '

I used integration by parts and evaluated the uv section at x=4 and x=1 then evaluated f(x) from 1 to 4

[tex]
4f'(4) - f'(1) - [f (4) - f (1)]
[/tex]

[tex]

(4*3) - 5 - [ 7-2 ] = 2 [/tex]

is this right??? ! :D !

IBP formula

[tex]
\int u(x)dv = u(x)v(x) - \int v(x) du

[/tex]

This is the formula for indefinite integration (i.e. without limits). What does the formula look like for definite integration (with limits)?

[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 shows a fundamental misunderstanding. [itex]f(4)[/itex] means [itex]f(x)[/itex] evaluated at [itex]x=4[/itex]. What you have written is meaningless I'm afraid!

Anyway first things first: what is the IBP formula for definite integration?

[tex]
4f'(4) - f'(1) - [f (4) - f (1)]
[/tex]

[tex]

(4*3) - 5 - [ 7-2 ] = 2 [/tex]

is this right??? ! :D !

That's it.

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Source: https://www.physicsforums.com/threads/integration-by-parts-conceptual-problem.335810/

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Source: https://gagebrokinights.blogspot.com/2022/10/if-f-4-12-f-is-continuous-and-7-f-x-dx.html

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