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monroe17

  • one year ago

Evaluate the integral by using substitution

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  1. monroe17
    • one year ago
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    \[\int\limits_{?}^{?} x^3(x^4-10)^{45}dx \] by making the substitution that u=x^4-10

  2. AravindG
    • one year ago
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    ok u substituted u now find what is dx in terms of du

  3. klimenkov
    • one year ago
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    \(u=x^4-10,\quad du=4x^3dx\) \(\int\limits x^3(x^4-10)^{45}dx=\frac14\int u^{45}du\)

  4. ASAAD123
    • one year ago
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    u=x^4-10 du/dx=4x^3 \[\int\limits \frac{ u ^{45} }{4 }dx=\frac{ 1 }{4 }\int\limits u ^{45}=\frac{ 1 }{4 }\times \frac{ u ^{46} }{ 46 }+c=\frac{ u ^{46} }{ 184 }+c\]

  5. blurbendy
    • one year ago
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    Don't forget to substitute (x^4 - 10) back in for you!

  6. blurbendy
    • one year ago
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    lol u*

  7. monroe17
    • one year ago
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    okay okay.. I completely understand until you guys got to the parts after finding du.. can someone explain everything after that step by step?

  8. zepdrix
    • one year ago
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    \[\large \color{orangered}{u=x^4+10}\]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.

  9. zepdrix
    • one year ago
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    -10, sorry bout that. Shouldn't make much difference.

  10. zepdrix
    • one year ago
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    |dw:1358443764659:dw|\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]

  11. monroe17
    • one year ago
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    now do i substitute u and du in for the integral? and I need it to be in respect to x not u

  12. monroe17
    • one year ago
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    can i always divide how you did to get the 1/4?

  13. zepdrix
    • one year ago
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    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 \(\large x^3 dx\).

  14. zepdrix
    • one year ago
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    So I had to get the 4 OFF of it.

  15. monroe17
    • one year ago
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    1/4x^3 dx?

  16. monroe17
    • one year ago
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    or is it replaced as a C?

  17. zepdrix
    • one year ago
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    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, \(\large du=4x^3dx\).

  18. zepdrix
    • one year ago
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    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.

  19. monroe17
    • one year ago
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    So we need to balance it..?

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

  21. monroe17
    • one year ago
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    Okay that makes sense.

  22. monroe17
    • one year ago
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    what comes after balancing it though..?

  23. zepdrix
    • one year ago
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    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 ).

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

  25. monroe17
    • one year ago
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    1/4(x^4-10)^(45)*x^3 ?

  26. zepdrix
    • one year ago
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    \[\huge \int\limits (\color{cornflowerblue}{x^4-10})^{45}\cdot \left(\color{orangered}{x^3dx}\right)\]

  27. zepdrix
    • one year ago
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    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.

  28. monroe17
    • one year ago
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    oh okay I see how you substituted it.

  29. monroe17
    • one year ago
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    now i need to antidifferentiate?

  30. zepdrix
    • one year ago
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    Yes good good c:

  31. monroe17
    • one year ago
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    umm..what part of that though? (x^4-10)^{45} ? and x^3dx

  32. zepdrix
    • one year ago
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    Did you make the substitution? We don't integrate until we've changed it over to U.

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

  34. zepdrix
    • one year ago
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    don't worry about all that X garbage. We got rid of that with our substitution.

  35. monroe17
    • one year ago
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    yeah i got that part.

  36. zepdrix
    • one year ago
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    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.

  37. monroe17
    • one year ago
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    yeah im having trouble integrating with respect to u... can you explain that more clearly

  38. monroe17
    • one year ago
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    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

  39. zepdrix
    • one year ago
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    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,\[\large \frac{1}{n+1}x^{n+1}+C\]

  40. zepdrix
    • one year ago
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    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?

  41. zepdrix
    • one year ago
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    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.

  42. monroe17
    • one year ago
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    yeah that makes sense :) i forgot that formula

  43. monroe17
    • one year ago
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    got it :)

  44. zepdrix
    • one year ago
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    yay team c:

  45. monroe17
    • one year ago
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    wait but where does the x^3dx go then?...

  46. zepdrix
    • one year ago
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    You don't have to back substitute the DU. The DU disappears as a part of the integration process.

  47. monroe17
    • one year ago
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    and thats why it becomes +C?

  48. zepdrix
    • one year ago
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    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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