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An example of integration by parts.
We want to find the antiderivative of ∫ x^5 ex
2 dx.
There is more than one way to do this. The point of this handout is to try two different ways, and to practice combining substitution with integration by parts.
The first method is to use substitution to make the integral easier, and then use inte- gration by parts.
The second is to use integration by parts directly. Here it might be a little harder to see how to choose the parts.
1. Substitution, then integration by parts.
Starting with u = x^2 , we compute du = 2x dx. Solving for dx gives dx =
du 2 x
. If we
substitute the formula for dx into the original integral, we get
∫ x^5 ex
2 du 2 x
x^4 ex
2 du.
We now want to write everything leftover in terms of u. Since u = x^2 , we can write ex 2 as eu, and x^4 as u^2. So, after substitution, the integral becomes ∫ 1 2
u^2 eu^ du.
This looks a lot like the integrals we’ve been considering, and seems a good candidate for integration by parts.
Depending on how you like to remember integration by parts, you might run into a little notational problem trying to solve the integral above. If you like the u dv method, then it’s going to be a bit awkward to apply – there’s already a variable called “u” in the equation, so it’s going to be confusing trying to keep track of which is the “old” u, and which is the “new” u.
A way out is to change the u to any other name (other than v, of course). For example, we could call it w, so that we’re looking for the integral
∫ 1 2
w^2 ew^ dw;
after we solve the integral we’ll go back and replace the w’s by x^2 , just as we would have done with the u’s.
We’ve seen integrals like the one above before. We know that the way to apply inte- gration by parts is to make sure that the w^2 gets differentiated, and the ew^ integrated. The resulting integral will be simpler than the one we started with.
In the u dv notation, that means that we set
u = 12 w^2 , diff du = w dv = ew, int v = ew.
and that gives us
( 1 )
w^2 ew^ dw =
w^2 ew^ −
wew^ dw.
To deal with the second integral,
wew^ dw, we use integration by parts again. We use the same kind of pattern – differentiate w to reduce the power of w in the integral.
Picking
u = w, diff du = 1 dv = ew, int^ v = ew.
we get
∫ wew^ dw = wew^ −
ew^ dw
= wew^ − ew.
Now we can substitute the integral of wew^ back into equation (1), to get
∫ 1 2
w^2 ew^ dw =
w^2 ew^ − wew^ + ew.
Finally, now that we know how to solve the integral in w, we substitute back in w = x^2 to get the solution
∫ x^5 ex
2 dx =
x^4 ex
2 − x^2 ex
2
2 .
2. Integration by parts directly.
The trouble here is to figure out how to pick the parts to start off.
It’s tempting to try the same kind of trick – differentiate the x^5 part and integrate ex
2
to try and simplify the integral. One problem is that there is no way to integrate ex 2
by itself – it just can’t be done (and it’s a famous example of a function which can’t).
Another common guess is to note that (^) dxd ex
2 = 2xex
2 and try and “fix it up” by adding 2 x to the denominator: (^) ∫
ex
2 dx =
ex 2
2 x
That isn’t right either – when differentiating something like e
x^2 2 x the quotient rule has to be used, and its derivative is
d dx
ex 2
2 x
2 xex 2 · 2 x − ex 2 · 2 (2x)^2
= ex
2 −
ex 2
2 x^2
which isn’t ex 2 .
In fact, ex 2 has no antiderivative that we can write down using any functions we know. It’s a fact that’s worth remembering – if you see ex 2 in an integral, you know you’re going to have to work around it somehow.
The only thing close to ex 2 which we can integrate is xex 2
. It doesn’t seem so different from ex 2 , but actually it is quite different, it has antiderivative we can find: ∫ xex
2 dx =
ex
2 .
That fact is why we had to pick the parts the way we did in section 2.
This handout can (soon) be found at
http://www.mast.queensu.ca/ ∼ mikeroth/calculus/calculus.html
E-mail address: mikeroth@mast.queensu.ca
