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swin2013

  • 3 years ago

Calculus integration with substitution problem

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  1. swin2013
    • 3 years ago
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    General integration t^2 (t^3 - 3)^10 dt is the u sub = t^3-3? du/dt = 3t^2 du = 3t^2dx

  2. zepdrix
    • 3 years ago
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    du=3t^2dt, yah looks good so far. From here, you want to solve for `t^2dt`, so you can replace that part in your integral. So get the 3 off of it! :O

  3. swin2013
    • 3 years ago
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    do you mean when you divide by 1/3?

  4. zepdrix
    • 3 years ago
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    mutiply by 1/3, or divide by 3, yes.

  5. swin2013
    • 3 years ago
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    i did 1/3 integral 3t^2 * t^2 (t^3-3)^10 dt and 3t^2 dt is du so 1/3 integral t^2 u^10 du?

  6. zepdrix
    • 3 years ago
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    \[\large du=3t^2 dt \qquad \rightarrow \qquad \color{orangered}{\frac{1}{3}du=t^2dt}\]I would have changed the 1/3 in this step, but the way you did it works fine also! :) It looks like you replaced dt with du, you didn't replace the t^2 hmm

  7. zepdrix
    • 3 years ago
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    Oh you have t^2 twice in there for some reason..? :O

  8. zepdrix
    • 3 years ago
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    \[\large \int \frac{1}{3}(t^3-3)^{10}\color{orangered}{3t^2\; dt}\]\[\large \color{orangered}{du=3t^2dt}\]

  9. swin2013
    • 3 years ago
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    because the original equation has t^2 * (t^3-3)^10 dt

  10. swin2013
    • 3 years ago
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    and u sub = 3t^2

  11. zepdrix
    • 3 years ago
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    So how did that change to, 1/3 integral 3t^2 * `t^2` (t^3-3)^10 dt? See how you have an extra t^2?

  12. zepdrix
    • 3 years ago
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    \[\large \int\limits t^2(t^3-3)^{10}dt \qquad \rightarrow \qquad \frac{1}{3}\int\limits 3t^2(t^3-3)^{10}dt\]I understand what you were doing here, you put a 1/3 and 3 in to fix things up. But you also threw in a t^2! woops! :)

  13. swin2013
    • 3 years ago
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    OH yea! i only put the 3! instead i put du there -_-

  14. zepdrix
    • 3 years ago
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    oh i see XD heh

  15. swin2013
    • 3 years ago
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    oh ok so it's 1/3 integral 3t^2 (t^3-3)^10 dt 1/3 integral u^10 du?

  16. zepdrix
    • 3 years ago
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    yes looks good!

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