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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 5

Lesson 1: Integration by parts

# Integration by parts intro

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.E (LO)
,
FUN‑6.E.1 (EK)
By looking at the product rule for derivatives in reverse, we get a powerful integration tool. Created by Sal Khan.

## Want to join the conversation?

• So why are you solving for the integral of f(x)g'(x) dx and not of the integral of f'(x)g(x) dx? Can you solve it for the other integral? This part confuses me.
(46 votes)
• f(x) and g(x) are arbitrary functions. You can solve for the other integral and the result will not change.
You are solving for the integral of (function 1 * derivative of function 2) dx. If you call them f(x) and g(x) or g(x) and f(x) does not matter.
(69 votes)
• is anti derivative the same as integration ?
(11 votes)
• The antiderivative is ONE type of integral, but there are others. Thus, not all integrals are antiderivatives, but antiderivatives are a type of integral.

The antiderivative is also called the "primitive integral" or the "indefinite integral".
(28 votes)
• what is the point of intergration?
(7 votes)
• At this level, integration translates into area under a curve, volume under a surface and volume and surface area of an arbitrary shaped solid. In multivariable calculus, it can be used for calculating flow and flux in and out of areas, and so much more it is impossible to list. Take a look at the multivarible calculus program: https://www.khanacademy.org/math/multivariable-calculus.
But where they REALLY come in handy is in solving differential equations (DEs) which is the math we use to describe our world.

DEs are everywhere in our lives. Light can be described by a wave equation, and similarly quantum particles (in your computer, for instance) are also described by a [slightly different] wave equation. Anywhere where there is water flowing can be described by a DE. Aerodynamics, vibrations, propulsion, electronics, sprinklers, traffic jams, population growth and decay, image processing, machine vision, neural networks, weather, heat transfer, engine efficiency, climate change, structural integrity, nuclear weapons, artillery trajectory, solar cells, financial derivatives pricing, and even the coffee cooling in your cup are all described by differential equations.

The EQ on your iPod boosts the sub bass boom on your favorite Hip Hop tune by breaking the sound into small little waves, amplifying just the sub bass waves and combining them back into music again every few microseconds.

That act of breaking and combining those waves so very fast is one daily life application of differentiation and integration.
(25 votes)
• I don't get this. At all. If I want to find the antiderivative of x*cos(x), why/how can I put it in the formula where Sal solved for the antiderivative of f(x) * g'(x)?

Why can't I simply take the antiderivative of x, and multiply that with the antiderivative of cos(x)?

Why is 0.5x^2 * sin(x) wrong, while x*sin(x) + cos(x) is right?

The derivative of 0.5x^2 * sin(x) is x*cos(x), while the derivative of x*sin(x)+cos(x) is cos(x)-sin(x), right?
(7 votes)
• The derivative of 0.5 x² sin(x) is 0.5 x² cos(x) + xsin(x)
The derivative of xsin(x) + cos(x) is xcos(x)
So, it is NOT true that the antiderivative of f(x)*g(x) is the product of their antiderivatives.

Let us look at the derivative of xsin(x) + cos(x) and maybe you'll see the error you made. Since the two portions are added (not multiplied) the derivative of their sum is the sum of their derivatives.
d/dx [cos(x)] = -sin(x)
d/dx [xsin(x)] = sin(x) +xcos(x)
Adding these together: - sin(x) + sin(x) +xcos(x) = xcos(x)
If you take these steps in reverse order, hopefully you'll see why the calculus doesn't work the way you suggest.
(17 votes)
• I think the "integration by parts rule" is missing a C.
Is it not necessary because the right side of the equation also has an indefinite integral?
(8 votes)
• It seems it was left out in order to ease the explanation of the formula. It is mentioned at .
(11 votes)
• at why does sal choose to solve for the anti derivative of f(x)g'(x) dx?
(9 votes)
• Sal could have solved for the anti derivative of f'(x)g(x)dx but the answer would be the same. You can call those functions whatever you want, they could be d(x)r'(x) or book(x)tree'(x). Those names are just symbolic.
(7 votes)
• Why is there no video introducing integration by parts? This video explains a formula which hasn't yet been introduced. I am doing this course in the suggested order and integration by parts has not been addressed yet.
(2 votes)
• This is the introduction, it introduces the concept by way of the product rule in differential calculus, and how you can derive the IBP formula from the PR. The next videos will show how to use it.
It is very common to be introduced to a new subject via theorems and definitions (and this will be the case more often has you get into higher math), then, once you understand the "whys" of how something works, you can apply it to the "wheres", that is, to situations where it comes in handy.

In general, in lower math you are shown how to use a tool without getting into why it works and where it came from. In higher math it becomes more difficult to use a tool if you don't know how and why it works first.
(15 votes)
• In many places it is written as a general rule of the thumb to select the first function in the order LIATE where
L - Logarithmic functions
I - Inverse trigonometric functions
A - Algebraic functions
T - Trigonometric functions
E - Exponential functions
While in some other places it is written as ILATE where the inverse trig functions goes first in preference while log goes down.
Which one is supposed to be followed ILATE or LIATE and how reliable are they ?
(6 votes)
• I use ALPoET never failed been using it since calculus never had a flaw plus I like edgar allen poe and it reminds me of it.

A - ArcTrigFunctions ex: arcsine or sin^-1x
L - Logs/Natural Logs ex: Log ex: lnx
Po - Polynomials ex: x^2+1
e - Exponentials ex: e^x ex: 2^x
T - TrigFunctions ex: sinx ex: coshx
(4 votes)
• Do you have a video explaining basic integration?
(3 votes)
• how do you integrate (x+1/x)^2 with respect to x ?
(1 vote)
• It's always simpler to integrate expanded polynomials, so the first step is to expand your squared binomial:
``(x + 1/x)² = x² + 2 + 1/x²``

Now you can integrate each term individually:
``∫(x² + 2 + 1/x²)dx = ∫x²dx + ∫2dx + ∫(1/x²)dx``

Each of those terms are simple polynomials, so they can be integrated with the formula:
``∫axⁿdx = a/(n+1) xⁿ⁺¹ + C``

So the final result is:
``∫(x + 1/x)²dx = 1/3 x³ + 2x - 1/x + C``
(7 votes)

## Video transcript

What we're going to do in this video is review the product rule that you probably learned a while ago. And from that, we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by parts. So let's say that I start with some function that can be expressed as the product f of x, can be expressed as a product of two other functions, f of x times g of x. Now let's take the derivative of this function, let's apply the derivative operator right over here. And this, once again, just a review of the product rule. It's going to be the derivative of the first function times the second function. So it's going to be f-- no, I'm going to do that blue color-- it's going to be f-- that's not blue-- it's going to be f prime of x times g of x times-- that's not the same color-- times g of x plus the first function times the derivative of the second, plus the first function, f of x, times the derivative of the second. This is all a review right over here. The derivative of the first times the second function plus the first function times the derivative of the second function. Now, let's take the antiderivative of both sides of this equation. Well if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to-- well what's the antiderivative of this? This is going to be the antiderivative of f prime of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now, what I want to do is I'm going to solve for this part right over here. And to solve for that, I just have to subtract this business. I just have to subtract this business from both sides. And then if I subtract that from both sides, I'm left with f of x times g of x minus this, minus the antiderivative of f prime of x g of x-- let me do that in a pink color-- g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go. And then let me copy and paste the other side. So let me copy and paste it. So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book. So this is essentially the formula for integration by parts. I will square it off. You'll often see it squared off in a traditional textbook. So I will do the same. So this right over here tells us that if we have an integral or an antiderivative of the form f of x times the derivative of some other function, we can apply this right over here. And you might say, well this doesn't seem that useful. First I have to identify a function that's like this. And then still I have an integral in it. But what we'll see in the next video is that this can actually simplify a whole bunch of things that you're trying to take the antiderivative of.