# I don't understand integration by parts

1. Feb 4, 2009

### fishingspree2

Hello everyone

Let's say I want to do this integral:

$$\int_{}^{}f(x)g(x)\,dx$$

We use this formula

$$\int f(x)g'(x)\, dx=f(x)g(x)-\int f'(x)g(x)\, dx$$

I don't understand the utility of this equation, since we want to find $$\int_{}^{}f(x)g(x)\,dx$$ and not $$\int f(x)g'(x)\, dx$$

Last edited: Feb 4, 2009
2. Feb 4, 2009

### rootX

d/dx [g . f ] = g'.f + g.f'
rearrange:
g.f' = d/dx [g . f ] - g'.f
integrate both sides
int {g.f'} = int { d/dx [g . f ] - g'.f}

Original question here was to integrate g.f' and you know how to integrate g'.f

That's how I see it. There isn't any formula or understanding for it. If above stuff is there, then you may use integration by parts

3. Feb 4, 2009

### fishingspree2

aren't we trying to integrate f.g (and not f'.g)???

sorry, I feel dumb:shy:

4. Feb 4, 2009

### rootX

not f.g but either f'.g or f.g'.

5. Feb 4, 2009

### AEM

Integration by parts is based on the formula for the derivative of a product. Here's how it goes. Suppose you take the derivative of the product of f(x) times g(x):

$$\frac{d}{dx} (f(x)g(x)) = f(x) \frac{d}{dx}g(x) + g(x) \frac{d}{dx} f(x)$$

or in the notation of differentials and suppressing indicating the x-dependence,

$$d(fg) = f dg + g df$$.

Now integrate both sides of this last equation:

$$\int d(fg) = \int f dg + \int g df$$

on the left hand side you have a perfect differential which the integration effectively cancels and you have

$$fg = \int f dg + \int g df$$

Let's rearrange this a little into

$$\int f dg = fg - \int g df$$

Why integration by parts is taught is that it often allows you to reduce an unmanageable integral into something manageable. What you have to do when you are using it is to match the pieces of your unknown integral with the parts of my last formula which I will first write out in more detail.

$$\int f(x) \frac{d}{dx}g(x) dx = f(x)g(x) - \int g(x) \frac{df(x)}{dx} dx$$

In general, the f(x) should be something that gets simpler when you take its derivative and the piece that you identify as the derivative of g(x) should be something that is easy to integrate so you can put it in where it goes on the right hand side.

Suppose you have as a problem to evaluate:

$$\int x e^x dx$$

The integral of $$e^x$$ is easy. It's just $$e^x$$. It's the darned x in the integral that gets in the way. But notice that if we identify x with f(x) then when we form the integral on the right hand side we'll be taking the derivative $$\frac{dx}{dx}$$ which is just 1 and it goes away. This is the simplification that I mentioned.

So, identify $$e^x dx$$ with $$\frac{d}{dx}g(x) dx$$ and identify x with f(x). Then fill in the pieces on the right hand side:

$$\int x e^x dx = x e^x - \int e^x (1) dx$$.

I put the "1" to remind you that the x disappeared when we took its derivative. That last integral is easy as I mentioned, so the final result is

$$\int x e^x dx = x e^x - e^x$$.

Make sure you see the idea that you want to match pieces so that when you take the derivative on the right hand side you get something simpler. Sometimes you have to integrate by parts twice. For example,

$$\int x^2 e^x dx$$.

The first time you reduce the exponent on the x from 2 to 1 and then the next time you reduce it from 1 to zero.

Make sure that you take the time to master integration by parts. It is an extremely useful and powerful technique that is used quite often in higher level courses.

Personally when I learned the technique (a long long time ago), I liked the differential notation that I showed you earlier. It's quite compact and easy to remember, but my impression is that the use of differentials in mathematics has fallen out of favor and is regarded as old fashioned. One other thing. It is almost always true that you want the derivative on the right hand side to be simpler than what you started with. I can think of one case where this is not true, but that's a story for another time...