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### Course: Class 12 math (India) > Unit 10

Lesson 9: Challenging definite integrals# Challenging definite integration

A rather difficult integration by parts problem. Created by Sal Khan.

## Want to join the conversation?

- But Zero isn't an even number. From Wikipedia "An even number is an integer that is evenly divisible by 2." Shouldn't the function have had a third definition for when the function equals zero?(0 votes)
- 0/2 = 0. Therefore, 0 is evenly divisible by 2 and is even.(97 votes)

- I don't understand how the antiderivative of cos πx is 1/π sin πx. I know the antiderivative of cos x is sin x, but where does the 1/π come in? I would be grateful if anybody could point me to a resource that would help with this.(22 votes)
- This is best to understand by u substitution.

let u = πx

thus du = π dx

thus dx = du/π

So, ∫ cos (πx) dx = ∫ cos (u) du/π = 1/π ∫ cos (u) du

= (1/π)(sin u) + C

back substituting

= (1/π)(sin πx) + C(39 votes)

- I'm thoroughly confused. At10:06, Sal gives a formula for integration by parts which he says he's gone through many times, but I just watched the integration by parts videos and didn't find it written it in this form? And I really thought I was getting the hang of these du, dv notations :(

Can someone please explain how ∫f(x)g'(x)dx=∫udv? I mean, I get it's some sort of substitution, but can't understand how f, g, and dx can be reduced to u and dv.

Any help is greatly appreciated! :)(9 votes)- Nvm, I was able to derive it:

(uv)'=uv'+u'v

d(uv)/dx=u*dv/dx+du/dx*v

u*dv/dx=d(uv)/dx-du/dx*v

∫(u*dv/dx)dx=∫d(uv)/dx*dx-∫(du/dx*v)dx

∫u*dv=∫d(uv)-∫du*v

∫udv=uv-∫vdu(13 votes)

- why there is no "c"in the progress?even in the end, thanks in advance(4 votes)
- You do not use the + C with definite integrals, just with indefinite integrals. If you use the Fundamental Theorem of Calculus to compute the definite integral, there is a + C but it cancels out and we can ignore it.

For example, let us use a simple integral:`F(x)= x² + 3x + C ←`

**the solution to an indefinite integral**

Now let compute this as a definite integral with

bounds [2, 3]:

F(3) - F(2) = [3²+(3)(3)+C] - [2²+3(2)+C]

= 18 + C - 10 - C ←**Notice I am adding and subtracting C**.

= 8

So, though you technically do have a +C when computing a definite integral via the Fundamental Theorem of Calculus, the + C always is both added and subtracted, thus dropping out. Therefore, we usually just leave out mentioning it altogether.(16 votes)

- The pop-up comment said the second term towards the end should be positive, not negative, but I don't see why it'd be positive... Can someone explain? I think I'm missing a step.(5 votes)
- Since it is minus that integral, the signs switch. The integral had a negative, so minus the integral is supposed to have a positive.(5 votes)

- At9:00isn't it (x-1) not (1-x)? That would completely change the answer wouldn't it?(6 votes)
- It is 1-x. The slope of that line is negative.(2 votes)

- it will take atleast 6 to 7 min to solve it in exam but we are given only 3 min per question .is there any easier way to do it.Or should I leave these kind of question(4 votes)
- Practice a lot, so you know many different techniques and can tell in advance whether they will work.

Second, memorize as many integral forms as you can. After all, if you know the solution to a challenging integral based on an integral form, you don't have to figure out how to get the answer.

Here is a table of many integral forms. Just study these, maybe make flashcards, and hopefully you will be in good shape:

http://integral-table.com/downloads/single-page-integral-table.pdf(5 votes)

- At12:25why does Sal put a negative in front of x/pi*sin(pi*x)?(3 votes)
- This is a mistake.

The integral of:`integral(x*cos(pi*x)dx) = xsin(pi*x)/pi + cos(pi*x)/pi^2`

Source: http://www.wolframalpha.com/input/?i=integral+xcos%28pi*x%29+dx(4 votes)

- What is the greatest integer function? Is it the same as the absolute value?(3 votes)
- No, because when it is evaluated with a negative number, its output is still a negative number. It is sometimes referred to as the "floor" function, since the number being evaluated is always rounded down unless it is a whole integer. Here is an example with a negative number:

If f(x)=[x] (greatest integer function), evaluate f(-2.8).

f(-2.8)=[-2.8]. Remember, round down if it is not a whole number, even when negative; so f(-2.8)=-3.(2 votes)

- In this video Sal uses a geometric interpretation of the two-part function. Is there any other way to integrate two-part function ?? A formula maybe???(4 votes)
- If by two part function you mean a piecewise function, yes. You would just break the integral up into two separate integrals. If I have f(x) as a piecewise function where f(x)=2x-5 if x<2 and f(x)=x² if x is greater than or equal to 2. and I'm asked to integrate f(x) from 0 to 4, I can simply integrate 2x-5 from 0 to 2 and add that with the integral of x² from 2 to 4.(1 vote)

## Video transcript

For any real number
x, let brackets around x denote the
largest integer less than or equal to x, often known as
the greatest integer function. Let f be a real valued
function defined on the interval
negative 10 to 10, including the
boundaries by f of x is equal to x minus the greatest
integer of x, if the greatest integer of x is odd, and 1
plus the greatest integer of x minus x, if the greatest
integer of x is even. Then the value of pi
squared over 10 times a definite integral
from negative 10 to 10 of f of x cosine of pi of
x dx is-- so before even try to attempt to
evaluate this integral, let's see if we can at least
visualize this function, f of x, right over here. So let's do our best
to visualize it. So let me draw my x-axis. And let me draw my y-axis. So let me draw my y-axis. And then let's think about what
this function will look like. So this is x is equal to
0, this is x is equal to 0, this is x is equal to 1, x is
equal to 2, x is equal to 3. We could go down to
negative 1, negative 2. We could just keep
going, if we like. Hopefully we'll see
some type of pattern, because it seems to
change from odd to even. So between 0 and 1, what is
the greatest integer of x? So let me just
write it over here. So between 0 and 1,
until you get to 1-- so maybe I should do this--
from including 0 until 1, the greatest integer
of x is equal to 0. If I'm at 0.5, the greatest
integer below 0.5 is 0. As I go from 1 to 2,
this brackets around x is equal to 1. It's the greatest integer. If I'm at 1.5, the
greatest integer is 1. If I'm at 1.9, the
greatest integer is 1. And then if I go to above
from between 2 and 3, then the greatest
integer is going to be 2. If I'm at 2.5, greatest
integer is going to be 2. So with that, let's
try to at least draw this function over
these intervals. So between 0 and 1, the
greatest integer is 0. 0 we can consider to be even. 0 is even, especially
if we're alternating. 1 is odd, 2 is even, 3 is odd. So 0 is even. So we would look at this
circumstance right over here, if x is even. And then over this time frame
or over this part of the x-axis, the greatest integer
of x is just 0. So the equation or the
line or the function is just going to be 1
minus x over this interval, because the greatest
integer is 0. So 1 minus x will
look like this. If this is 1, 1 minus x
just goes down like that. That's what it looks
like from 0 to 1. Now let's think about what
happens as we go from 1 to 2. As we go from 1 to 2, not
including 2 but including 1, all the way up to
2, not including it, the greatest integer is 1. The greatest integer is odd. So we use this case. And over here, we're going
to have x minus the greatest integer of x over this interval. The greatest integer of x is 1. So it's going to be
the graph of x minus 1. So x minus 1 at 1
is going to be 0. And at 2, it's
going to be 1 again. So it's going to be this. It's going to look
just like that. So this right here is x minus 1. And then this over here,
essentially was 1 minus x. And we could keep doing it. As we go from 2 to 3, the
greatest integer of x is 2. We would look at this case. So we're going to have 1 plus 2. So we're going to
have 3 here minus x. So when we start
over here when x is 2 or a little bit above that,
we're going to have 3 minus 2. We're going go to 1. It's going to be right at 1. And then as x is equal
to 3, 3 minus 3 is 0. It's going to oscillate
back down like this. I think we have an
appreciation for what this graph is
going to look like. It's going to keep going
up and down like this, with a slope of negative 1, then
positive 1, then negative 1, then positive 1. It's just going to keep doing
that over and over again. You can keep trying it
out with other intervals, but it's pretty clear
that this is the pattern. Now, what we want to do is
evaluate the integral from 10 to negative 10 of this function
times cosine of pi of x. So let's think about
cosine of pi of x and think about whether
that also is periodic. And of course, it is. And then if we can
simplify this integral so we don't have to evaluate
it over this entire period over here, maybe we can simplify
it into a simpler integral. So cosine of pi x,
cosine of 0 is 1. Don't want you to
get that wrong. Cosine pi 0 is cosine of 0. So that's 1. Cosine of pi is negative 1. So when x is equal to 1,
this becomes cosine of pi. So then the value of the
function is negative 1. It'll be over here. And then cosine of 2 pi, 2
times pi, is then 1 again. So it'll look like this. This is at 1/2. When you put it over here,
it'll become pi over 2. Cosine of pi over 2 is 0. So it'll look like this. Let me draw it as
neatly as possible. So it will look like this. Cosine, and then
it'll keep doing that, and then it'll go like this. So it is also periodic. So if we wanted to figure out
the integral of the product of these two periodic
functions from all the way from negative 10 to 10,
can we simplify that? And it looks like
it would just be, because we have this
interval, let's look at this interval over here. Let's look at just from 0 to 1. So just from 0 to 1, we're
going to take this function and take the product
of this cosine times essentially
1 minus x, and then find the area under that
curve, whatever it might be. Then when we go
from 1 to 2, when we take the product
of this and x minus 1, it's actually going to be the
same area, because these two, going from 0 to 1 and
going from 1 to 2, it's completely symmetric. You can flip it over
this line of symmetry, and both functions are
completely symmetric. So you're going to
have the same area when you take their product. So what we see is,
over every interval, when you go from 2 to 3,
first of all, the integral from 2 to 3 is
clearly the same thing as the integral from 0 to 1. Both functions look
identical over that interval. But it will also be the
same as going from 1 to 2, because it's
completely symmetric. When you take the
products of the function, that function will be completely
symmetric around this axis. So the integral
from here to here will be the same as the
integral from there to there. So with that said, we can
rewrite this thing over here. So what we want to evaluate,
pi squared over 10 times the integral from negative 10
to 10 of f of x cosine of pi x, using the logic we
just talked about. This is going to be
the same thing as being equal to pi squared over 10
times the integral from 0 to 1, but 20 times that,
because we have 20 integers between
negative 10 and 10. We have 20 intervals
of length 1. So times 20 times the
integral from 0 to 1 of f of x cosine of pi x dx. I forgot to write
the dx over there. I want to make sure
you understand, because this is really the
hard part of the problem, just realizing that the integral
over this interval is just 1/20 of the whole thing, because over
every interval, from 0 to 1, the integral is going to
evaluate to the same thing as going from 1 to 2, which
will be the same thing as going from 2 to 3, or going from
negative 2 to negative 1. So instead of doing
the whole interval from negative 10
to 10, we're just doing 20 times the
interval from 0 to 1. From negative 10 to 10, there's
a difference of 20 here. So we're multiplying by 20. And this simplifies
it a good bit. First of all, this part
over here simplifies to 20 divided by 10 is 2. So it's 2 pi squared. So it becomes 2 pi
squared-- that's just this part over here-- times
the integral from 0 to 1. Now, from 0 to 1,
what is f of x? Well, we just figured out,
from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x
from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate
this integral right over here. So let's do that. So 1 minus x times
cosine of pi x is the same thing as cosine of
pi x minus x cosine of pi x. Now, this right here,
well, let's just focus on taking
the antiderivative. This is pretty easy. But let's try to do
this one, because it seems a little bit
more complicated. So let's take the antiderivative
of x cosine of pi x dx. And what should jump
in your mind is, well, this isn't that simple. But if I were able to
take the derivative of x, that would simplify. It's very easy to take the
antiderivative of cosine of pi x without making
it more complicated. So maybe integration by parts. And remember,
integration by parts tells us that the integral--
I'll write it up here-- the integral of udv is equal to
uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos
where I prove this and show examples of exactly
what that means. But let's apply it
right over here. And in general,
we're going to take the derivative of
whatever the u thing is. So we want u to be
something that's simpler when I take
the derivative. And then we're going to take
the antiderivative of dv. So we want something that does
not become more complicated when I take the antiderivative. So the thing that
becomes simpler when I take this
derivative is x. So if I set u is
equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going
to be the rest of this. This whole thing over
here is going to be dv. dv is equal to cosine pi x dx. And so v would just
be the antiderivative of this with respect to x.
v is going to be equal to 1 over pi sine of pi x. If I took the derivative here,
derivative of the inside, you get a pi, times 1
over pi, cancels out. Derivative of sine of pi
x becomes cosine of pi x. So that's our u,
that's our v. This is going to be equal to u times
v. So it's equal to x, this x times this. So x over pi sine of pi x
minus the integral of v, which is 1 over pi
sine of pi x du. du is just dx. And this is pretty
straightforward. The antiderivative of sine of
pi x is 1 over pi or negative 1 over pi cosine of pi x. And you could take
the derivative if you don't believe me. You could do u substitution. But hopefully you can start
to do these in your head, especially if you're going to
take the IIT joint entrance exam. So this whole
expression is going to be, this part
over here is going to be x over pi sine of pi x. And then this over here
is going to be-- well, you have the antiderivative
of sine of pi x is negative 1 over
pi cosine of pi x. The negatives cancel out. So you have plus, and then
the 1 over pi times 1 over pi, 1 over pi squared
cosine of pi x. That's the antiderivative
right there. And you can verify. Derivative of cosine
of pi x is going to be negative pi sine of x. One pi will cancel out here. You get a negative sign, and
then you have sine of pi x. So this is the
antiderivative of that. And so if we want
the antiderivative of this whole thing
right over here, this is what we
care about from 0 to 1 dx, the
antiderivative of cosine pi x-- pretty straightforward. We've actually already
done it right over here. It is 1 over pi sine of pi x. That's this first term. And the antiderivative
of x cosine pi x is this thing over here. But we're subtracting it. So we'll put a negative
sign out front. So minus x over pi sine of pi x
minus 1 over pi squared cosine of pi x. And of course, we took
the antiderivative, but it's a definite integral. We need to evaluate
it from 0 to 1. And we don't want to forget
that 2 pi squared out front. So let's evaluate this. So the first thing, we're going
to have 1 over pi sine of 1 pi. Sine of 1 pi is 0. So you have 0 minus 1 over
pi times sine of 1 pi again. That's again 0, minus 1 over
pi squared cosine of pi, or cosine of 1 pi. Cosine of pi is negative 1. Negative 1 times negative
1 over pi squared is plus 1 over pi squared. So we've evaluated it at 1. And from that, we want to
subtract it evaluated at 0. Sine of 0 is 0 minus--
this is clearly 0, because you have a 0
out front-- minus 0. And then you have
minus cosine of 0. Cosine of 0 is 1. So then you have a
minus 1 over pi squared. And so this term, we could
just say this is a negative. These don't matter--
negative, positive, positive. And we're just left with
a 1 over pi squared plus 1 over pi squared, which is
equal to 2 over pi squared. That's what this
part evaluates to. It's 2 over pi squared. We can't forget
that we were going to multiply this whole
thing times 2 pi squared. That's this thing
out front here. And so the pi squared
cancels out the pi squared. We're left with 2 times
2, which is equal to 4. And we're done. This thing that looked
pretty complicated just evaluates out to 4.