Evaluate the integral by using substitution

- anonymous

Evaluate the integral by using substitution

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[\int\limits_{?}^{?} x^3(x^4-10)^{45}dx \]
by making the substitution that u=x^4-10

- AravindG

ok u substituted u
now find what is dx in terms of du

- klimenkov

\(u=x^4-10,\quad du=4x^3dx\)
\(\int\limits x^3(x^4-10)^{45}dx=\frac14\int u^{45}du\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

u=x^4-10
du/dx=4x^3
\[\int\limits \frac{ u ^{45} }{4 }dx=\frac{ 1 }{4 }\int\limits u ^{45}=\frac{ 1 }{4 }\times \frac{ u ^{46} }{ 46 }+c=\frac{ u ^{46} }{ 184 }+c\]

- blurbendy

Don't forget to substitute (x^4 - 10) back in for you!

- blurbendy

lol u*

- anonymous

okay okay.. I completely understand until you guys got to the parts after finding du.. can someone explain everything after that step by step?

- zepdrix

\[\large \color{orangered}{u=x^4+10}\]Taking the derivative with respect to x,\[\large \frac{du}{dx}=4x^3\]Moving the dx to the other side gives us,\[\large du=4x^3dx\]Dividing both sides by 4 gives,\[\large \color{orangered}{\frac{1}{4}du=x^3dx}\]
Use the two orange pieces to make your substitution.

- zepdrix

-10, sorry bout that. Shouldn't make much difference.

- zepdrix

|dw:1358443764659:dw|\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]

- anonymous

now do i substitute u and du in for the integral? and I need it to be in respect to x not u

- anonymous

can i always divide how you did to get the 1/4?

- zepdrix

U substitutions can be tricky.
We're not replacing one little piece at a time.
We're replacing a big chunk of stuff with another big chunk.
So the first part was ok, we have a U=something.
It's pretty clear how to plug that in.
But then we also wanted to replace \(\large x^3 dx\) with something involving DU.
To do so, we had to essentially SOLVE FOR \(\large x^3 dx\).

- zepdrix

So I had to get the 4 OFF of it.

- anonymous

1/4x^3 dx?

- anonymous

or is it replaced as a C?

- zepdrix

So we start with U.
From there, we took the derivative of U.
Our main goal here is to find a suitable U and then also a U' somewhere in the integral. We're hoping that by making the substitution, it'll get easier.
So our \(\large u=x^4-10\). Now we need to find a u'.
Taking the derivative of u gave us, \(\large du=4x^3dx\).

- zepdrix

See that second part?
We need to SOLVE for x^3dx. See how the 4 is in the way? It doesn't quite match what's in our integral because of that annoying 4.

- anonymous

So we need to balance it..?

- zepdrix

Yes, if I'm understanding you correctly :)
To solve for x^3 dx we can divide both sides by 4.\[\large du=4x^3dx \qquad \rightarrow \qquad \frac{1}{4}du=\cancel{\frac{4}{4}}x^3dx\]

- anonymous

Okay that makes sense.

- anonymous

what comes after balancing it though..?

- zepdrix

So we've got it to a point where we'll now be able to replace EVERY x and dx with something involving u and du.
And we're hoping it will be simpler than it was in X (and it will be :D ).

- zepdrix

\[\large \color{cornflowerblue}{u=x^4-10} \qquad \color{orangered}{\frac{1}{4}du=x^3dx}\]Using these pieces to make our substitution.
I'll color code it so it's a little easier to understand.

- anonymous

1/4(x^4-10)^(45)*x^3
?

- zepdrix

\[\huge \int\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right)\]

- zepdrix

Are you able to see how we're going to replace the pieces?
I just rearranged some things so it's a little easier to match up the pieces.

- anonymous

oh okay I see how you substituted it.

- anonymous

now i need to antidifferentiate?

- zepdrix

Yes good good c:

- anonymous

umm..what part of that though? (x^4-10)^{45} ? and x^3dx

- zepdrix

Did you make the substitution?
We don't integrate until we've changed it over to U.

- zepdrix

\[\large \int\limits\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right) \qquad \rightarrow \qquad \int\limits (\color{cornflowerblue}{u})^{45}\cdot \left(\color{orangered}{\frac{1}{4}du}\right)\]Integrate with respect to u.

- zepdrix

don't worry about all that X garbage. We got rid of that with our substitution.

- anonymous

yeah i got that part.

- zepdrix

So integrating with respect to u gives us,\[\large \frac{1}{4}\cdot \frac{1}{46}u^{46}+C\]Having trouble with that part?
We'll have one more minor step after you integrate.
We were given the integral in terms of X. So we want our final answer to also be in terms of X. From here, we need to BACK substitute.

- anonymous

yeah im having trouble integrating with respect to u... can you explain that more clearly

- anonymous

okay I got the back substitute.. im completely confused on the integrating with respect to u.
i got 1/184(x^4-10)^(46)+C

- zepdrix

We can simply apply the Power Rule for Integration.\[\large \int\limits x^n dx \quad = \quad \frac{x^{n+1}}{n+1}+C\]Which I prefer to write this way,\[\large \frac{1}{n+1}x^{n+1}+C\]

- zepdrix

We raise the power by 1, then divide by the NEW power.
Example:\[\large \int\limits x^3 dx \quad = \quad \frac{1}{4}x^4+C\]See how the power increased to 4, then we divide by 4 also?

- zepdrix

Try to completely ignore the 1/4 in our U integral. It's a constant, it can be pulled outside of the integral. It won't affect the process at all.
We'll just multiply by 1/4 after integrating.

- anonymous

yeah that makes sense :) i forgot that formula

- anonymous

got it :)

- zepdrix

yay team c:

- anonymous

wait but where does the x^3dx go then?...

- zepdrix

You don't have to back substitute the DU.
The DU disappears as a part of the integration process.

- anonymous

and thats why it becomes +C?

- zepdrix

The +C comes from the fact that it's an INDEFINITE integral.
See how there are no upper or lower limits of integration?
Here is a quick example:
\[\large f(x)=x^2+7\]\[\large f'(x)=2x\]
Let's look at another function,\[\large g(x)=x^2-45\]\[\large g'(x)=2x\]
See how the constants become 0 when you take a derivative? Alright so let's look at this integral a moment.
\[\large \int\limits 2x \;dx\quad = \quad x^2+ ?\]
Is it +7? Is it -45???
We don't know. This integral solution represents a FAMILY of functions. ALL of the possible x^2's.
So we have to throw a +C on the end to compensate for any constant that may have been there originally.

Looking for something else?

Not the answer you are looking for? Search for more explanations.