Topic 6.11: Integrating Using Integration by Parts

Fifty-two factorial

Shuffle a deck of cards properly, and the order you end up holding is almost certainly one that nobody in history has ever held before. There are fifty-two choices for the top card, fifty-one for the next, fifty after that, all the way down to one, and multiplying all of those together gives you fifty-two factorial.

That’s a sixty-eight-digit number, bigger than the number of atoms in the entire Earth, which is why a good shuffle is almost certainly brand new. Your calculator can grind that number out, but I’d rather show you how to estimate it without doing all fifty-one multiplications. The whole trick comes down to the area under this one curve, y equals ln x, and not one rule you know can find it.

By the end of this lesson, you’ll be able to estimate fifty-two factorial with one integral and get its first digit right, and on the way you’ll pick up two techniques the BC exam keeps putting in its free-response questions. The key to that integral is a picture you’ve already seen, the growing rectangle from the product rule, so let’s run it again and keep everything it sweeps out.

The rectangle, run backwards

In the product rule lesson, this rectangle had a width u and a height v that were both changing, and each instant its area grew by a thin strip along the side and another along the top. This time let’s keep everything those borders paint. The side border is v tall, so each tiny d u it slides paints v d u, and the top border is u wide, so each tiny d v it climbs paints u d v.

Look where the paint ended up. Everything the side painted sits under the corner’s path, everything the top painted sits to the left of it, and together they fill exactly the new area, twelve minus the two we started with. In chunky steps there’s also a tiny white corner that belongs to neither strip, but as the steps shrink, those corners vanish, exactly like the corner in the product rule.

On paper, the product rule says the change in u times v is u d v plus v d u, which is just the two strips. Add those changes up over the whole run, from its start a to its end b, and the total change in u times v equals everything the top painted plus everything the side painted.

Now move the integral of v d u across, and you’ve got integration by parts. The integral of u d v from a to b equals u v from a to b, minus the integral of v d u. For antiderivatives we drop the limits, so the integral of u d v equals u v minus the integral of v d u. If the integral of u d v is one you can’t do, you trade it for the integral of v d u, and the whole art is making that a good trade.

Before we use it, test it yourself. Draw any path you like from the start to the end, as wiggly as you want, as long as it keeps moving up and to the right, and watch whether the two regions still add up to ten.

Now let’s run the corner along y equals e to the x, from zero to one. The region left of the path is now the integral of x times e to the x, which none of your rules can touch, so I’ve hidden its number.

Exactly one

The region under the path is just the area under e to the x, which you can find, it’s e minus one. How big is the region to the left of the path? This curvy, impossible-looking integral comes out to exactly one, and the rectangle found it with a subtraction. Let’s get the same answer on paper, the way the exam wants it.

I’ll pick u equals x, because x gets simpler when you differentiate it, and d v equals e to the x d x, because I know how to integrate that. That makes d u equal to d x, and v equal to e to the x. Plug into u v minus the integral of v d u, and you get x e to the x minus the integral of e to the x. That last integral is easy, so the answer is x e to the x minus e to the x, plus C.

You can always check by differentiating, and here the product rule gives back exactly x e to the x. What if you’d chosen u equals e to the x and d v equals x d x instead? Then v is x squared over two, and the leftover integral has an x squared in it, which is worse than where we started, so let’s cross that choice out.

For products like these, I want the leftover integral to get easier, so I check what happens when I differentiate each factor. Make u the part that gets simpler when you differentiate it, make d v something you can integrate, and if the leftover looks worse, switch. For the definite integral, the u v part gets the limits too. At one it’s e minus e, which is zero, at zero it’s negative one, and zero minus negative one is exactly one.

Pick the part that gets simpler

Let’s do one together, the integral of x cosine two x. x gets simpler when we differentiate it and we can integrate cosine two x, so u is x and d v is cosine two x d x. This is where most people slip, because the antiderivative of cosine two x is one half sine two x, and that half is there to cancel the two the chain rule brings out.

Then u v minus the integral of v d u is x over two times sine two x, minus the integral of one half sine two x. That integral is negative one quarter cosine two x, and the minus in front turns it into plus one quarter cosine two x, plus C. Now try one on your own, the integral from zero to pi of x sine x d x, as a decimal.

Sometimes one round of parts isn’t enough. With x squared e to the x, one round leaves the integral of two x e to the x, which still has an x in it, so we’d need parts again.

There’s a tidy way to run all the rounds at once. Differentiate u down one column until it hits zero, integrate d v down the other, then multiply along the diagonals with alternating signs, because each minus is another round of parts. That gives x squared e to the x minus two x e to the x plus two e to the x, plus C, and the table stops on its own because the polynomial runs out. Use a table to find the integral of x squared sine x.

No formula? No problem

The exam loves a version of parts with no formula at all. You only know that f has a continuous derivative, that f of one is two and f of three is four, and that the integral of f from one to three is five. Then it asks for the integral of x times f prime of x, from one to three. Parts doesn’t need a formula. Let u be x and d v be f prime of x d x, so v is just f of x.

The integral becomes x f of x from one to three, minus the integral of f, which is twelve minus two, minus five, so five. On a free-response question, this is what full credit looks like. The 2024 BC parts question gave a point for naming u and d v, a point for the parts line, and a point for the answer with its work, so write all three lines instead of jumping straight to the five.

Now you try one. This time g has a continuous derivative, g of one is four, g of three is two, and the integral of g from one to three is six. Find the integral of x times g prime of x from one to three.