Main content

## Calculus 2

### Unit 1: Lesson 10

Reverse power rule- Reverse power rule
- Reverse power rule
- Reverse power rule: negative and fractional powers
- Indefinite integrals: sums & multiples
- Reverse power rule: sums & multiples
- Rewriting before integrating
- Reverse power rule: rewriting before integrating
- Rewriting before integrating: challenge problem
- Reverse power rule review

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Rewriting before integrating: challenge problem

In this example, we find the antiderivative an expression which is not so simple. Created by Sal Khan.

## Want to join the conversation?

- Correct me if I'm wrong, but when you apply the sum rule to break the antiderivative into smaller pieces, you would technically have an added constant for each piece. At4:40when you add the constant, that constant would technically represent the sum of all smaller constants found when taking the general antiderivative of each of the smaller pieces, correct? It would be almost impossible to find a particular solution given an initial condition, since it we would be solving for multiple constant C's.(89 votes)
- c can be anything, as opposed to x, which refers to a single, unknown value. We add c to generalise the (INDEFINITE) integral, so c can be anything, and everything, it doesn't matter!!

c = 0, 1, 2, 3, 4...(to infinity) and all the values in between, and all the negative numbers and and and...(you get the idea...basically all the numbers in the universe - can someone please let me know if this includes complex numbers...?)

To conclude,, any number, plus any number is still, any number! (c + c = c)

I emphasised INDEFINITE, because definite integrals refer to the area under the curve/line. (gross oversimplification!!) Clearly, changing c (or the height) would change the area. So c for definite integrals can only be one, specific, value, if any..(5 votes)

- why is it necessary to put that dx after every integral sign. Whats the purpose?(7 votes)
- I would disagree that the dx is mostly for show.

You can think of integration simply as the sum of a lot of really skinny (infinitely skinny, but not zero) rectangles. Now, how do you get the area of a rectangle? Well, you multiply its length times its width. Integration just takes the length times the width of a bunch of skinny rectangles and sums them. So how does that connect to the integration expression?

Let's say you have an integral that looks like this: ∫x^2dx. There are three parts to the integral: 1) the function part, x^2, represents the length of the rectangles--you can also think of it as the height of the function, or the y-value; 2) the dx in the integral represents the width of each rectangle; 3) and the elegant elongated S-symbol, ∫, tells you to sum the areas of the rectangles.

Watch Sal's videos on Riemann sums and it will become obvious what dx means and why you need to have it there. Without the dx, you can't have any width. And if you don't have any width, then you can't have any area. And no area means nothing to sum. And nothing to sum means no integration. And no integration means no differentiation (you get a glimpse of the link between the two in this video, but you will see it in a profoundly beautiful way in the Fundamental Theorem of Calculus) . And no differentiation means no calculus. And think of how horrible the world would be without calculus. :)(30 votes)

- Why are these called indefinite integrals?(4 votes)
- Because the Integral doesn't have any Boundaries it needs to relate to.(22 votes)

- Is there a way to solve for C in the equation, or for any antiderivative?(4 votes)
- If you've given any initial conditions, then yes. If not, then not.

For example, suppose you're given that f´(x) = 2x and f(0) = 1. Taking the antiderivative of f´(x) gives you x^2 + C, and with the initial condition that f(0) = 1 you can solve for f(x) = x^2 + 1.(7 votes)

- Holy cow, I was sitting here wondering what a "hairier expression" is. I was thinking like a Fourier or something. I get it now.(2 votes)
- Hi Sal,

Upto my understanding Integration is the inverse of Differentiation. So if we integrate 2x, we will get x^2. It means that d/dx (x^2) is 2x. My question is why should we neglect constants in integration? why not we can integrate some constant, say integration of 8 dx and it should be 8x accordingly . Because differentiation of 8x is 8 (a constant) right? Hope you will understand my doubt. Please reply me sal

thanks for your nobel service sal

Srini(3 votes)- You are correct that if we are told to find the indefinite integral of 8, it would be 8x (because the derivative of 8x is 8). But if you are told to find the derivative of x^2 + 8, your answer is simply 2x. What about finding the derivative of x^2 + 9? Again, your answer is 2x. So, some information about the original function is lost when looking only at its derivative. This is the reason why we write "+ C" when finding the anti-derivative, because the derivative of x^2 + any constant is 2x. If there is no constant in the original function, you could still think of it as x^2 + 0.(3 votes)

- I would really appreciate to look at the graph of such kind of an indefinite integral as to where do you actually put that constant C in that graph, i want to see it located in that particular graph cuz in definite integrals we have area under that specific part of that curve according to the limits we put up Its really confusing about the representation of C in the graph of indefinite integrals :-0(3 votes)
- Graphically the integrals gives you the a "family of "parallel' curves" not a unique curve... that's what the "C" is doing here(2 votes)

- Hi Sir! I truly admire the way you teach, but I think their is something in this video of hairy integration that had been overlooked by Sir Khan. Please correct me if I'm wrong cause I'm just starting to learn this subject. The thing you had overlooked is in the 3rd term. the exponent you had written is -3/2 instead of -5/2. thank you :D(3 votes)
- Lets take the anti-derivative of the third term alone, but first, some algebraic manipulation:

(18x^(1/2))/(x^3)=18x^(1/2-3)=18x^(1/2-6/2)=18x^(-5/2)

Now we take the anti derivative of 18x^(-5/2):

Here we raise the exponent with 1 power:

-5/2+1=-5/2+2/2=-3/2

And divide by the exponent:

[18x^(-3/2)]/[-3/2]=-12x^(-3/2) (+c if you will)

For negative constants of exponents it might seem confusing to raise a power by one, but this is how we would do it.

I hope this was a little helpful!(2 votes)

- Correct me if I'm wrong, but is this the same as integration by parts?(2 votes)
- No, this is not integration by parts. It is just the use of the rule that the integral of sums is equal to the sum of the integrals, EG:
`∫ Ax1 + Bx2 + Cx3 = ∫Ax1 + ∫Bx2 + ∫Cx3`

Go here to learn Integration by Parts: https://www.khanacademy.org/math/integral-calculus/integration-techniques/integration_by_parts/v/deriving-integration-by-parts-formula(3 votes)

- At 3.06, why is it that we don't take the antiderivative of x^3?(2 votes)
- you can't integrate the numerator and the denominator seperately.

you can integrate if it is, let's say, int(x + 3), and get x^2 / 2 + 3x (+C)

but for this particular problem, you need to combine the x terms(2 votes)

## Video transcript

So our goal in this video is
to take the antiderivative of this fairly crazy
looking expression. Or another way of saying it is
to find the indefinite integral of this crazy
looking expression. And the key realization
right over here is that this expression is
made up of a bunch of terms. And the indefinite integral
of the entire expression is going to be equal to the
indefinite integral of each of the term. So this is going to be equal
to, we could look at this term right over here, and just
take the indefinite that, 7x to the third dx. And then from that,
we can subtract the indefinite
integral of this thing. So we could say
this is, and then minus the indefinite integral
of 5 times the square root of x dx. And then we can look at
this one right over here. So then we could say plus
the indefinite integral of 18 square root of x. Square roots of x,
over x to the third dx. And then finally, I'm
running out of colors here, finally I need more
colors in my thing. We can take the
antiderivative of this. So plus the antiderivative of x
to the negative 40th power dx. So I've just rewritten this
and color-coded things. So let's take the
antiderivative of each of these. And you'll see that we'll
be able to do it using our whatever we want to call it. The inverse of the power
rule, or the anti-power rule, whatever you might
want to call it. So let's look at the first one. So we have-- what
I'm going to do is, I'm just going to
find the antiderivative without the constant, and just
add the constant at the end. For the sake of this one. Just to make sure we get the
most general antiderivative. So here the exponent is a 3. So we can increase it by 1. So it's going to
be x to the 4th. Let me do that same purple
color, or pink color. It's going to be x
to the 4th, or we're going to divide by x to the 4th. So it's x to the 4th over 4
is the antiderivative of x to the 3rd. And you just had this scaling
quantity, the seven out front. So we can still just
have the seven out front. So we get 7x to the 4th over 4. Fair enough. From that, we're
going to subtract the antiderivative of this. Now at first this
might not be obvious, that you could use our
inverse power rule, or anti-power rule here. But then you just
need to realize that 5 times the
principal root of x is the same thing as 5
times x to the 1/2 power. And so once again, the
exponent here is 1/2. We can increment it by 1. So it's going to
be x to the 3/2. And then divide by the
incremented exponent. So divide by 3/2. And of course we had
this 5 out front, so we still want to
have the 5 out front. Now this next expression
looks even wackier. But once again, we can
simplify a little bit. This is the same thing-- let
me do it right over here-- this is the same thing as
18 times x to the 1/2 times x to the negative 3 power. x to the 3rd in the
denominator is the same thing as x to the negative 3. We have the same base, we
could just add the exponents. So this is going to be equal
to 18 times x to the 2 and 1/2 power. Or another way of
thinking about it, this is the same thing as
18 times x to the 5/2 power. Did I do that right? Yeah. Negative 3. Oh sorry, this is
negative 2 and 1/2. Let me make this very clear. And this is going to be
the negative 5/2 power. x to the negative 3 is the same
thing as x to the negative 6/2. Negative 6/2 plus
1/2 is negative 5/2. So once again, we just have
to increment this exponent. So negative 5/2 plus 1 is
going to be negative 3/2. So you're going to have
x to the negative 3/2. And then you divide
by what your exponent is when you increment it. So divide it by negative 3/2. And then you had
the 18 out front. And we obviously are going
to have to simplify this. And then finally, our
exponent in this term. Let me not use that
purple anymore. The exponent in this term
right over here is negative 40. If we increment it, we get
x to the negative 39 power, all of that over negative 39. And now we can add
our constant, c. And all we need to do is
simplify all of this craziness. So the first one is
fairly simplified. We can write it as
7/4 x to the 4th. Now this term right over
here is essentially negative 5 divided by 3/2. So 5 over 3/2 is
equal to 5 times 2/3, which is equal to 10 over 3. So this thing right over here
simplifies to negative 10/3 x to the 3/2. And then we have all
of this craziness. Now 18 divided by
negative 3/2 is equal to 18 times negative 2/3. Which is equal to, well we can
simplify this a little bit, this is the same thing
as 6 times negative 2. Which is equal to negative 12. So this expression
right over here is negative 12x to
the negative 3/2. And then finally, this
one right over here, we can just rewrite it
as, if we want, we could, well, we could just
write negative 1/39 x to the negative 39 plus c. And we're done. We've found the indefinite
integral of all this craziness. And I encourage you to take
the derivative of this. And you can do it using
really just the power rule, to take the derivative of this. And verify that it does
indeed equal this expression that we took the
antiderivative of.