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Evaluate the integral by using substitution

Mathematics
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\[\int\limits_{?}^{?} x^3(x^4-10)^{45}dx \] by making the substitution that u=x^4-10
ok u substituted u now find what is dx in terms of du
\(u=x^4-10,\quad du=4x^3dx\) \(\int\limits x^3(x^4-10)^{45}dx=\frac14\int u^{45}du\)

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Other answers:

u=x^4-10 du/dx=4x^3 \[\int\limits \frac{ u ^{45} }{4 }dx=\frac{ 1 }{4 }\int\limits u ^{45}=\frac{ 1 }{4 }\times \frac{ u ^{46} }{ 46 }+c=\frac{ u ^{46} }{ 184 }+c\]
Don't forget to substitute (x^4 - 10) back in for you!
lol u*
okay okay.. I completely understand until you guys got to the parts after finding du.. can someone explain everything after that step by step?
\[\large \color{orangered}{u=x^4+10}\]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.
-10, sorry bout that. Shouldn't make much difference.
|dw:1358443764659:dw|\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]
now do i substitute u and du in for the integral? and I need it to be in respect to x not u
can i always divide how you did to get the 1/4?
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 \(\large x^3 dx\).
So I had to get the 4 OFF of it.
1/4x^3 dx?
or is it replaced as a C?
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, \(\large du=4x^3dx\).
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.
So we need to balance it..?
Yes, if I'm understanding you correctly :) To solve for x^3 dx we can divide both sides by 4.\[\large du=4x^3dx \qquad \rightarrow \qquad \frac{1}{4}du=\cancel{\frac{4}{4}}x^3dx\]
Okay that makes sense.
what comes after balancing it though..?
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 ).
\[\large \color{cornflowerblue}{u=x^4-10} \qquad \color{orangered}{\frac{1}{4}du=x^3dx}\]Using these pieces to make our substitution. I'll color code it so it's a little easier to understand.
1/4(x^4-10)^(45)*x^3 ?
\[\huge \int\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right)\]
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.
oh okay I see how you substituted it.
now i need to antidifferentiate?
Yes good good c:
umm..what part of that though? (x^4-10)^{45} ? and x^3dx
Did you make the substitution? We don't integrate until we've changed it over to U.
\[\large \int\limits\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right) \qquad \rightarrow \qquad \int\limits (\color{cornflowerblue}{u})^{45}\cdot \left(\color{orangered}{\frac{1}{4}du}\right)\]Integrate with respect to u.
don't worry about all that X garbage. We got rid of that with our substitution.
yeah i got that part.
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.
yeah im having trouble integrating with respect to u... can you explain that more clearly
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
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,\[\large \frac{1}{n+1}x^{n+1}+C\]
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?
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.
yeah that makes sense :) i forgot that formula
got it :)
yay team c:
wait but where does the x^3dx go then?...
You don't have to back substitute the DU. The DU disappears as a part of the integration process.
and thats why it becomes +C?
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.

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