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anonymous
 4 years ago
Evaluate the integral by using substitution
anonymous
 4 years ago
Evaluate the integral by using substitution

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?} x^3(x^410)^{45}dx \] by making the substitution that u=x^410

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0ok u substituted u now find what is dx in terms of du

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0\(u=x^410,\quad du=4x^3dx\) \(\int\limits x^3(x^410)^{45}dx=\frac14\int u^{45}du\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u=x^410 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
 4 years ago
Best ResponseYou've already chosen the best response.0Don't forget to substitute (x^4  10) back in for you!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay okay.. I completely understand until you guys got to the parts after finding du.. can someone explain everything after that step by step?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2\[\large \color{orangered}{u=x^4+10}\]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
 4 years ago
Best ResponseYou've already chosen the best response.210, sorry bout that. Shouldn't make much difference.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1358443764659:dw\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now do i substitute u and du in for the integral? and I need it to be in respect to x not u

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can i always divide how you did to get the 1/4?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2U 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
 4 years ago
Best ResponseYou've already chosen the best response.2So I had to get the 4 OFF of it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or is it replaced as a C?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2So 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 \(\large u=x^410\). Now we need to find a u'. Taking the derivative of u gave us, \(\large du=4x^3dx\).

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2See 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
 4 years ago
Best ResponseYou've already chosen the best response.0So we need to balance it..?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2Yes, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Okay that makes sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what comes after balancing it though..?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2So 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[\large \color{cornflowerblue}{u=x^410} \qquad \color{orangered}{\frac{1}{4}du=x^3dx}\]Using these pieces to make our substitution. I'll color code it so it's a little easier to understand.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/4(x^410)^(45)*x^3 ?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2\[\huge \int\limits (\color{cornflowerblue}{x^410})^{45}\cdot \left(\color{orangered}{x^3dx}\right)\]

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2Are 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay I see how you substituted it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i need to antidifferentiate?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm..what part of that though? (x^410)^{45} ? and x^3dx

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2Did you make the substitution? We don't integrate until we've changed it over to U.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2\[\large \int\limits\limits (\color{cornflowerblue}{x^410})^{45}\cdot \left(\color{orangered}{x^3dx}\right) \qquad \rightarrow \qquad \int\limits (\color{cornflowerblue}{u})^{45}\cdot \left(\color{orangered}{\frac{1}{4}du}\right)\]Integrate with respect to u.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2don't worry about all that X garbage. We got rid of that with our substitution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i got that part.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2So 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah im having trouble integrating with respect to u... can you explain that more clearly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay I got the back substitute.. im completely confused on the integrating with respect to u. i got 1/184(x^410)^(46)+C

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2We 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
 4 years ago
Best ResponseYou've already chosen the best response.2We 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
 4 years ago
Best ResponseYou've already chosen the best response.2Try 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that makes sense :) i forgot that formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait but where does the x^3dx go then?...

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2You don't have to back substitute the DU. The DU disappears as a part of the integration process.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and thats why it becomes +C?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.2The +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,\[\large g(x)=x^245\]\[\large g'(x)=2x\] 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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