Integration - Fundamentals Thereom Of Calculus

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  • #1
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Homework Statement


[tex]
\int_0^3\ [/tex] (t-2)^1/3

Homework Equations



Second of Fundamental Thereom of Calculus

The Attempt at a Solution



I don't know what to do first because I'm not used to questions with square roots. Once someone help me with the beginning, I can probably do it because after that it's all the same process anyways.



Help, please?
 

Answers and Replies

  • #2
lurflurf
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use the power rule
(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'
or since v'=0
(u^v)'=v*u^(v-1)*u' (when v'=0)
in particular
[(t-2)^(4/3)]'=(4/3)(t-2)^(1/3)
 
  • #3
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"(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'"

seems like a crazy expression :smile:

OP,
Square roots work exactly the same way.
Try simple example first:

integrate (t-2)^2
 
  • #4
Dick
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seems like a crazy expression :smile:

OP,
Square roots work exactly the same way.
Try simple example first:

integrate (t-2)^2

May be crazy but it is true. Still I think rootX is just suggesting you try the u substitution u=(t-2) and then use the power law formula for integrals.
 
  • #5
HallsofIvy
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By the way- there is NO square root in this problem!

May be just me, but please do not use "square root" for all roots! It leads to things like people saying "3 squareroot of x" when they mean "cuberoot of x" and then it's time for the old two by four to come out!

In any case, a "root" is just a power- use the power rule:

Yes, make the substitution x- a= u and then
[tex]\int u^n du= \frac{1}{n+1} u^{n+1}+ C[/tex]
 

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