Intergrals?

Question

Intergrals?

johnweldon1993 · Answer

What about em? :)

askme12345 · Answer

$$\int\limits_{}^{}x^2\sqrt{X^3+1dx}$$

johnweldon1993 · Answer

Alright...so here is a hint...u-substitution

$$\large u = x^3 + 1$$

askme12345 · Answer

(I just started learning about intergrals so bare with me i will be very slow)
So do I just substitute that in..?
$$\int\limits_{}^{}x^2\sqrt{Udx}$$

johnweldon1993 · Answer

Oh okay....alright step by step :)

Alright so...

u-substitution is a method you use when you see that inside an integral...is a function...AND it's derivative

johnweldon1993 · Answer

For example...I see that in

$$\large \int x^2\sqrt{x^3 + 1}dx$$

we have an x^3   ...and we know that 3x^2 is it's derivative...well we see that we have an 'x^2' in here as well...so that is what we want

askme12345 · Answer

okay makes sense!

johnweldon1993 · Answer

So what we do...is make what is called a "u-substitution"

You make something = U...and then you take the derivative of U

So where I had

$$\large U = x^3 + 1$$
$$\large dU = 3x^2dx$$

dU meaning the derivative of U

Now...remember when I said that the derivative is 3x^2   ....we only want the x^2 part.....so we divide both sides by that 3 to get

$$\large \frac{dU}{3} = x^2dx$$

johnweldon1993 · Answer

Alright...that last thing we got

$$\large \color \red{\frac{dU}{3} = x^2dx}$$    look familiar?

$$\large \int \color \red{x^2} \sqrt{x^3 + 1}\color \red{dx} $$

Alright from here....we know thatthat U we have = x^3 + 1   so lets replace that in that integral

$$\large \int x^2 \sqrt{U}dx$$

And also we see that $\large \frac{dU}{3} = x^2dx$   so lets replace that as well

$$\large x^3dx \sqrt{U} \rightarrow \int \sqrt{U}du$$

johnweldon1993 · Answer

Oops forgot something there....we knew that

$$\large \frac{dU}{\color \red{3}} = x^2 dx$$ so that means that last integral SHOULD be

$$\large \int \sqrt{U}\frac{dU}{3}$$

We are almost done..just 1 more thing...in an integral...constants can be taken out...because they only affect the final result...not anything inbetween...so lets factor out a $\large \frac{1}{3}$

$$\large \frac{1}{3} \int \sqrt{U}du$$

johnweldon1993 · Answer

Now we are all ready to integrate :)

Just like in derivatives...integration has sort of a power rule too...except instead of

$$\large nx^{n - 1}$$
We instead have
$$\large \frac{x^{n + 1}}{n + 1}$$

So lets look at what we have

$$\large \frac{1}{3}\int \sqrt{u}du$$

How else can we write $\large \sqrt{u}$?

$$\large u^\frac{1}{2}$$ right?

So we have

$$\large \frac{1}{3} \int u^{1/2}du$$

johnweldon1993 · Answer

Now we apply that sort of "power rule"

$$\large \frac{x^{n + 1}}{n + 1}$$ means that here...n = (1/2)

$$\large \frac{x^{1/2 + 1}}{1/2 + 1} \rightarrow \frac{x^{3/2}}{3/2} \rightarrow \frac{2x^{3/2}}{3}$$

johnweldon1993 · Answer

Whew...that was alot to type...aright so now...we integrated...altogether we have left

$$\large \frac{1}{3} \frac{2u^{3/2}}{3}$$

Let combine that to make it look nice...

$$\large \frac{2u^{3/2}}{9}$$

******yes I know I crossed from U to X..sorry didnt mean to...here we are still at U :)

johnweldon1993 · Answer

Alright...so now...the original substitution we made

$$\large U=x^3+1$$


We want to plug that back in here...because we dont want U in our answer...we want 'x'

so from that 

$$\large \frac{2u^{3/2}}{9}$$

we will then have

$$\large \frac{2(x^3+1)^{3/2}}{9}$$

johnweldon1993 · Answer

And finally...the 1 thing you ALWAYS want to remember in integration *because I guarantee your professors will knock points* is to add a + C at the end of problems..

$$\large \frac{2(x^3 + 1)^{3/2}}{9} + C$$ would be your final answer! :)

askme12345 · Answer

Wow!!!! Thank you!! That simplified it for me haha, thank you soooo much!! I do have a question though, I mainly understand the steps that you took im just confused at this part (attached)

I understand replacing the first part, putting in the U in place of the original equation, but what happens to the x^2 ? i just dont get the  part before the arrow

johnweldon1993 · Answer

Oh okay :)

So right...you got that U = x^3 + 1

That left us with the

$$\large \int x^2 \sqrt{u}dx$$

Right?

askme12345 · Answer

yes!

johnweldon1993 · Answer

Alright...

so remember where we had the fact that

$$\large U = x^3 + 1$$

and when we take the derivative of that...we have

$$\large dU = 3x^2dx$$ 

that make sense right?

askme12345 · Answer

yup, we ignore the 3 bc its a constant?

johnweldon1993 · Answer

Hmm not here...we don't ignore anything

$$\large \frac{d}{dx} x^3 + 1 = 3x^2dx$$

remeber the power rule of derivatives?

$$\large nx^{n - 1}$$

so that means if 'n' = 3  ...then 

$$\large 3x^{3 - 1} = 3x^2$$


and the + 1   ... $\large x^3 + 1$  there turns to 0..because the derivative of a constant is 0

johnweldon1993 · Answer

That is how we went from

$$\large u = x^3 + 1$$ to $$\large du = 3x^2dx$$

Just took the derivative of 'u' :)

johnweldon1993 · Answer

Don't forget..right now I'm just focusing on 1 thing after another...not jumping ahead to when we "ignore" the $\large \frac{1}{3}$ because it is a constant....just doing the derivative portion here :)

askme12345 · Answer

oh okay! my question was more about why you plugged it it where you did but i actually just reread it and can say that i misread it the first time -- thank you!!!!
I have another one to solve , ill try to solve it and show u my reasoning if you wouldnt mind looking it over when im done thanks!

johnweldon1993 · Answer

Mmhmm I was getting to where I plugged it in :)

But alright...and yes of course let me know :)

askme12345 · Answer

$$\int\limits_{}^{}$$ 4x^3 e^x^4 dx

askme12345 · Answer

Is the problem .. I'm not sure if i can do it on my own haha

so do i make U=4x^3e^x^4

johnweldon1993 · Answer

$$\large \int 4x^3 e^{x^4}dx$$

Like that?

askme12345 · Answer

Yup! Sorry I didn't know how to write the e^x^4

johnweldon1993 · Answer

No problem hun :)

alright...first thing is first...factor out the constants...4 is being multiplied to everything...so we just take that out until the very end...

$$\large 4\int x^3e^{x^4}dx$$

Okay? :)

johnweldon1993 · Answer

From there,

We want to make a u-substitution again (because I see a x^4 and I also see an x^3 so I know something close to the derivative is there...)

johnweldon1993 · Answer

Lets make

$$\large u = x^4$$
so the derivative of that is
$$\large du = 4x^3dx$$ right?

askme12345 · Answer

yes! (Sorry my internet was lagging)

johnweldon1993 · Answer

Mine as well :P

Alright so lets write everything we have

$$\large 4\int x^3e^{x^4}dx$$
$$\large u = x^4$$
$$\large du = 4x^3dx$$

Alright looking at the du equation...we have $4x^3dx$

Now in our integral we have $x^3$ so that means we need to find someway to get rid of that extra 4 in our du equation... so we divide everything by that 4 to get

$$\large \frac{du}{4} = x^3dx$$

Alright..looking at our integral

$$\large 4 \int x^3e^{x^4}$$

We have the 'u' we made to be x^4...so lets put U there
$$\large 4 \int x^3e^{u}dx$$

And also now look at this
$$\large 4 \int \color \red{x^3}e^{u}\color\red{dx}$$

Looks alot like what we just solved for du...

$$\large \frac{du}{4} = x^3dx$$

So lets replace that as well

$$\large 4 \int e^u \frac{du}{4}$$

Now we dont nee that 4 there under the du...so lets factor that out..remember...constants can always come out of the integral

$$\large \frac{4}{4} \int e^u du$$
We are left with
$$\large \int e^u du$$

Now what is the integral of e^u? well it is still just e^u...so we have

$$\large e^u + C$$   *** remember that + C*** :P

And just like last time....we want to relace the 'u back with what we made u = in the beginning...which was x^4...so our final answer would be

$$\large e^{x^4} + C$$