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monroe17

  • 2 years ago

Evaluate the integral by using substitution

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  1. monroe17
    • 2 years 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
    • 2 years ago
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    ok u substituted u now find what is dx in terms of du

  3. klimenkov
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Don't forget to substitute (x^4 - 10) back in for you!

  6. blurbendy
    • 2 years ago
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    lol u*

  7. monroe17
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    -10, sorry bout that. Shouldn't make much difference.

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

  11. monroe17
    • 2 years 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
    • 2 years ago
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    can i always divide how you did to get the 1/4?

  13. zepdrix
    • 2 years 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
    • 2 years ago
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    So I had to get the 4 OFF of it.

  15. monroe17
    • 2 years ago
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    1/4x^3 dx?

  16. monroe17
    • 2 years ago
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    or is it replaced as a C?

  17. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    So we need to balance it..?

  20. zepdrix
    • 2 years 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
    • 2 years ago
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    Okay that makes sense.

  22. monroe17
    • 2 years ago
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    what comes after balancing it though..?

  23. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    1/4(x^4-10)^(45)*x^3 ?

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

  27. zepdrix
    • 2 years 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
    • 2 years ago
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    oh okay I see how you substituted it.

  29. monroe17
    • 2 years ago
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    now i need to antidifferentiate?

  30. zepdrix
    • 2 years ago
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    Yes good good c:

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

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

  33. zepdrix
    • 2 years 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
    • 2 years ago
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    don't worry about all that X garbage. We got rid of that with our substitution.

  35. monroe17
    • 2 years ago
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    yeah i got that part.

  36. zepdrix
    • 2 years 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
    • 2 years ago
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    yeah im having trouble integrating with respect to u... can you explain that more clearly

  38. monroe17
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yeah that makes sense :) i forgot that formula

  43. monroe17
    • 2 years ago
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    got it :)

  44. zepdrix
    • 2 years ago
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    yay team c:

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

  46. zepdrix
    • 2 years 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
    • 2 years ago
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    and thats why it becomes +C?

  48. zepdrix
    • 2 years 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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