1. ## Integration by parts

Question is....

Use Integration by parts to show that the

integral f(x)dx = xf(x) - integral xf'(x)dx

What do I need to make my U? DV? I have tried all kinds of combinations of the substitution method and just can't figure it out. HELP!!!! TIA John

2. We will assume that f(x) is a differenciable function on some open interval 'I'.
Therefore, it is countinous.
Therefore, there exists a function F(x) definied on open interval 'I' such that, F'(x)=f(x).
Now we can proceede.
---
You are given,
Integral f(x)dx,
You can express it as,
Intergral (1)f(x)dx,
Let u'=x and v=f(x)
Then,
u=1 and v'=f'(x)
By integration by parts,
f(x)-Integral xf'(x)dx

3. Originally Posted by ThePerfectHacker
We will assume that f(x) is a differenciable function on some open interval 'I'.
Therefore, it is countinous.
Therefore, there exists a function F(x) definied on open interval 'I' such that, F'(x)=f(x).
Now we can proceede.
---
You are given,
Integral f(x)dx,
You can express it as,
Intergral (1)f(x)dx,
Let u'=x and v=f(x)
Then,
u=1 and v'=f'(x)

if u=1, u'=0 or if u'=x u=(x^2)/2

RonL

4. Originally Posted by CaptainBlack
if u=1, u'=0 or if u'=x u=(x^2)/2

RonL
My mistake,
u'=1 and v=f(x)
Then,
u=x and v'=f'(x)

You have to forgive me, I do intergrations in a formal way.

5. Originally Posted by TreeMoney
Question is....

Use Integration by parts to show that the

integral f(x)dx = xf(x) - integral xf'(x)dx

What do I need to make my U? DV? I have tried all kinds of combinations of the substitution method and just can't figure it out. HELP!!!! TIA John
Consider:

int x f'(x) dx,

let u=x and v'=f', then:

int x f'(x) dx = uv - int u' v dx=x f(x) - int f(x) dx.

Now rearrange:

int f(x) dx = x f(x) - int x f'(x) dx

RonL

6. ## Part B :o(

There is another part to this question... It says.

If f and g are inverse functions and f' is continuous, prove that

the integral(from a to b) f(x)dx = bf(b) - af(a) - integral(from f(a) to f(b)) g(y)dy

Then there is a hint, it says to use part (a) and make the substitution y = f(x)

I tried to figure this out, but kept getting stuck, To be honest I probably wasn't even close to having the right answer. PLEASE HELP!!!

7. Originally Posted by TreeMoney
There is another part to this question... It says.

If f and g are inverse functions and f' is continuous, prove that

the integral(from a to b) f(x)dx = bf(b) - af(a) - integral(from f(a) to f(b)) g(y)dy
The definite integral that we get from what we have already found is:

int(a,b) f(x) dx = b f(b) - a f(a) -int(a,b) x f'(x) dx

Now let y=f(x), then dy = f'(x) dx, x=g(y), and the limits of integration
become y1=f(a) and y2=f(b), so putting all of this into the integral on the
right hand side we have:

int(a,b) x f'(x) dx = int(f(a),f(b)) g(y) dy,

so:

int(a,b) f(x) dx = b f(b) - a f(a) - int(f(a),f(b)) g(y) dy

which is what was to be prooved.

RonL