Question

Can someone help explain something about the integral in my textbook?

Answer 1

The lesson is on change of variable, so they gave me an example: “Change the variable to evaluate \[\int\limits_{1}^{0} x \sin (\pi x^2) dx\]

The next step it shows like this: \[\frac{ 1 }{ 2\pi } \int\limits_{1}^{0} (\sin(\pi x^2) (2\pi x dx)\]

My question is where did the \[\frac{ 1 }{ 2\pi }\] come from?

Answer 2

Oops, flipped that 0 and 1 for the integral.

Answer 3

Oh, they are using the "magic one" basically.

Answer 4

wait, is it from 1 to 0, or from 0 to 1?

Answer 5

I flipped it haha. Sorry. 0 to 1.

Answer 6

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

Answer 7

I can assume you know u-substitution, ok?

Answer 8

Yes.

Answer 9

(Keep in mind, that dx is also part of the product, not just the notation of the integral. It is the change in x that you multiply times the "rectangles" from x=0 to x=1.... blah blah blah Reinmann sums that you hae learned earlier).

Your problem is:

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

If I were to set,
\(u=\pi x^2\)

Then what should my \(du\) be?

Answer 10

If;
\(\color{#000000}{\displaystyle u=\pi x^2}\)

then
\(\color{#000000}{\displaystyle du=?}\) (you tell me)

Answer 11

1

Answer 12

du=1 ??

Answer 13

Otherwise, 2pix. x_x

Answer 14

\[2 \pi x dx\]

Answer 15

yes, was just about to explain that, lol

Answer 16

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

What they want, is that they want to prepare the expression for du

Answer 17

my du would be = 2πx dx

by I don't have the 2π inside the integral, so they multiply and divide by 2π
("magic one" technique)

Answer 18

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

\(\color{#000000}{\displaystyle \frac{2 \pi}{2 \pi} \int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

\(\color{#000000}{\displaystyle \frac{1}{2 \pi} \int\limits_{0}^{1}2\pi x\sin(\pi x^2)~dx}\)

\(\color{#000000}{\displaystyle \frac{1}{2 \pi} \int\limits_{0}^{1}\sin(\pi x^2)\cdot \left[2\pi x ~dx\right]}\)

Answer 19

\(\color{#000000}{\displaystyle \frac{1}{2 \pi} \int\limits_{0}^{1}\sin(\color{#0000ff}{\displaystyle \pi x^2})\cdot \left[\color{#ff0000}{\displaystyle 2\pi x ~dx}\right]}\)

Now, if we set, \(\color{#0000ff}{\displaystyle u=\pi x^2}\)
then we have the du, ready,
\(\color{#ff0000}{\displaystyle du=2\pi x ~dx}\)

\(\color{#000000}{\displaystyle \frac{1}{2 \pi} \int\limits_{0}^{1}\sin(\color{#0000ff}{\displaystyle u})\cdot \left[\color{#ff0000}{\displaystyle du}\right]}\)

Answer 20

Before we talk about the limits of integration, tell me whether or not you understand this little algebraic manipulation?

Answer 21

Yes.

Answer 22

Ok, now the limit of integration can be delt with in two ways ...

Answer 23

Way 1) You don't change the limits of integration,
And you substitute the original variable back after integrating.

Way 2) You [do] change the limits of integration,
and you DON'T substitute the original variable
back after integrating.

Answer 24

So...\[\frac{ 1 }{ \pi }\]

Answer 25

what do you mean?

Answer 26

Is that the correct answer? The integral?

Answer 27

Yes, very good!

Answer 28

Thanks for the help and the time! :) I think I understand now.

Answer 29

Anytime :)

Answer 30

You didn't need that manipulation btw

Answer 31

You don't really need to "prepare" the expression for du, as we have done (multiply times 2π/2π and all that)

Answer 32

I will show you briefly..

Answer 33

Ok.

Answer 34

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}x\sin(\pi x^2)~dx}\)

You can set, \(\color{#000000}{\displaystyle u=\pi x^2}\)
And you get that,
\(\color{#000000}{\displaystyle du=2\pi x~dx}\)

You only have the \(\color{#000000}{\displaystyle x~dx}\) inside the integral, so you can divide by 2π on both sides,

\(\color{#000000}{\displaystyle \color{red}{\frac{\color{#000000}{du}}{2\pi}}=\color{red}{\frac{\color{#000000}{2\pi x~dx}}{2\pi}}}\)

\(\color{#000000}{\displaystyle \frac{1}{\pi}du=x~dx}\)

I will highlight the substituted part accordingly for better vision...

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}\color{red}{x}\sin(\color{blue}{\pi x^2})~\color{red}{dx}}\)

\(\color{#000000}{\displaystyle\int\limits_{0}^{1}\color{red}{\frac{1}{2\pi}}\sin(\color{blue}{u})~\color{red}{du}}\)

Answer 35

Oh, there is a mistake, the line that says

\(\color{#000000}{\displaystyle \frac{1}{\pi}du=x~dx}\)

should say

\(\color{#000000}{\displaystyle \frac{1}{2\pi}du=x~dx}\)

Answer 36

Lemme attempt to do a question like this, is it okay if you check it for me?

Answer 37

Sure

Answer 38

Can I do it with this, or is this different?\[\int\limits_{0}^{1}e^{\cos(5x)}[\sin(5x)] dx\]

Answer 39

by "this" you are referring to the u-substitution?

Answer 40

Well I know U sub works here. But I mean, can I do the same thing you did? The shortened thingy?

Answer 41

when I solved for x dx?

When I set, u=πx²
and, du=2πx dx -->(solving for x dx) --> (1/2π) du = dx

Answer 42

You mean this 'manipulation' ?

Answer 43

Yes.

Answer 44

yes, definitely valid by any u-substitution integral.

Although there are ((definitielly .. later on)) some problems where u-sub wouldn't work at all, but whenever your initial way works (going straight u=.... and du=...), then doing manipulation such as the one I showed is also going to work.

Answer 45

I am pretty sure you know what the u-substitution here, is, but I am going to ask just in case. What do you substitute as u?

Answer 46

cos(5x)

Answer 47

Yes, u=cos(5x)

Answer 48

Then,

du = ?

Answer 49

5 sin (5x)

Answer 50

5 sin(5x) \( \color{red}{\sf dx}\)

Answer 51

agh.

Answer 52

and it'snegative

Answer 53

d/dx cos(x) = - sin(x)
(right ?)

Answer 54

negative...

Answer 55

So if you set
u = cos(5x)

then,
du = -5 sin(5x) dx

Answer 56

\(\color{red}{\sf + Attachment}\)
Example of a u-substition integral where I change the limits of integration.

Answer 57

Ohhhh.....................Ai ai. Why would they teach me this now and not waaaay back. xD hahaha.

Answer 58

you are referring to the shortcut for du?

Answer 59

Haha yeah.

Answer 60

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1}e^{\cos(5x)}\sin(5x) dx}\)

\(\color{#000000 }{ \displaystyle u=\cos(5x)}\)
\(\color{#000000 }{ \displaystyle du=-5\sin(5x)~dx\quad \Longrightarrow\quad -(1/5)du=\sin(5x)~dx}\)

\(\color{#000000 }{ \displaystyle x=0\quad \quad \Longrightarrow\quad\quad u=\cos(5\cdot 0)=\cos(0)=1 }\)
\(\color{#000000 }{ \displaystyle x=1\quad \quad \Longrightarrow\quad\quad u=\cos(5\cdot 1)=\cos(5) }\) (transcendental number)

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1}e^{\color{blue}{\cos(5x)}}\color{red}{\sin(5x) dx}\quad \quad \Longleftrightarrow\quad\quad \color{red}{-\frac{1}{5}}\int\limits_{1}^{\cos(5)}e^{\color{blue}{u}}\color{red}{du} }\)

Answer 61

Notice that I changed the limits of integration and you don't need to substitute the x back after integrating.

Answer 62

Here are some interesting u-substitution problems, if you want ...

\(\color{#000000 }{ \displaystyle \int\limits_{\large 1}^{\large 2} x^{-1/2}e^{x^{1/2}} dx}\)

\(\color{#000000 }{ \displaystyle \int\limits_{\large e^2}^{\large e^e}\frac{1}{x\ln x} dx}\)

Answer 63

haha the bottom one hurts my heart.

Answer 64

what is the derivative of ln(x) ?

Answer 65

\[\frac{ 1 }{ x }\]

Answer 66

what is the derivative of x^(1/2) ?

Answer 67

And that is all ... nothing hard really :)

Answer 68

you just have to notice it, that's all ...

Answer 69

1/2

Answer 70

oops.

Answer 71

1-2x^-1/2

Answer 72

i think there is an oops again there

Answer 73

-.- 1/2x^-1/2

Answer 74

yes, and that is also the key for u-sub for example problem 1....

Answer 75

anyway, did you do your problem?

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1}e^{\cos(5x)}\sin(5x) dx}\)

Answer 76

oh, I have to go, it's almost 12....

I will put up more material to the post when I come online....

Answer 77

I am sorry for leaving you like this.

Answer 78

\[-\frac{ 1 }{ 5 }[e^u]_{1}^{\cos(5)}\]

Answer 79

Yes, very good, you [do] end up getting,

\(\color{#000000 }{ \displaystyle -\frac{1}{5}\left[e^u\color{white}{\LARGE |} \right] ^{\color{blue}{u=\cos(5)}}_{\color{red}{u=1}}\quad \Longrightarrow \quad -\frac{1}{5}\left[e^{\color{blue}{\cos(5)}}-e^{\color{red}{1}} \right] }\)

and this simplifies to,
\(\color{#000000 }{ \displaystyle -\frac{ 1 }{5}e^{\cos(5)}+\frac{ 1 }{5}e }\)

Answer 80

\(\bf\color{#13b3ff}{\href{http:///tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinitePtII.aspx}{\sf U-substitution~\color{#1e9d1f}{Examples-Problems}~\color{#000000}{\left(Lamar~Tutorial/Notes\right)}}}\)

\(\bf\color{#13b3ff}{\href{https:///www.khanacademy.org/math/integral-calculus/integration-techniques/u_substitution/v/u-substitution}{\color{#13b3ff}{\sf U-substitution ~~ \color{#1e9d1f}{Video~\text{&}~Practice}~\color{#000000}{\left(Khan~Academy\right)}}}}\)

If you want more material, then I can also make up some examples, and if you like explanations/proofs/whatever \(\sf ....\)