anonymous
  • anonymous
Evaluate the integral by using substitution
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{?}^{?} x^3(x^4-10)^{45}dx \] by making the substitution that u=x^4-10
AravindG
  • AravindG
ok u substituted u now find what is dx in terms of du
klimenkov
  • klimenkov
\(u=x^4-10,\quad du=4x^3dx\) \(\int\limits x^3(x^4-10)^{45}dx=\frac14\int u^{45}du\)

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anonymous
  • 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
  • blurbendy
Don't forget to substitute (x^4 - 10) back in for you!
blurbendy
  • blurbendy
lol u*
anonymous
  • 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
  • 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
  • zepdrix
-10, sorry bout that. Shouldn't make much difference.
zepdrix
  • zepdrix
|dw:1358443764659:dw|\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]
anonymous
  • 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
  • anonymous
can i always divide how you did to get the 1/4?
zepdrix
  • 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
  • zepdrix
So I had to get the 4 OFF of it.
anonymous
  • anonymous
1/4x^3 dx?
anonymous
  • anonymous
or is it replaced as a C?
zepdrix
  • 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
  • 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
  • anonymous
So we need to balance it..?
zepdrix
  • 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
  • anonymous
Okay that makes sense.
anonymous
  • anonymous
what comes after balancing it though..?
zepdrix
  • 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
  • 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
  • anonymous
1/4(x^4-10)^(45)*x^3 ?
zepdrix
  • zepdrix
\[\huge \int\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right)\]
zepdrix
  • 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
  • anonymous
oh okay I see how you substituted it.
anonymous
  • anonymous
now i need to antidifferentiate?
zepdrix
  • zepdrix
Yes good good c:
anonymous
  • anonymous
umm..what part of that though? (x^4-10)^{45} ? and x^3dx
zepdrix
  • zepdrix
Did you make the substitution? We don't integrate until we've changed it over to U.
zepdrix
  • 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
  • zepdrix
don't worry about all that X garbage. We got rid of that with our substitution.
anonymous
  • anonymous
yeah i got that part.
zepdrix
  • 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
  • anonymous
yeah im having trouble integrating with respect to u... can you explain that more clearly
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
yeah that makes sense :) i forgot that formula
anonymous
  • anonymous
got it :)
zepdrix
  • zepdrix
yay team c:
anonymous
  • anonymous
wait but where does the x^3dx go then?...
zepdrix
  • zepdrix
You don't have to back substitute the DU. The DU disappears as a part of the integration process.
anonymous
  • anonymous
and thats why it becomes +C?
zepdrix
  • 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.

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