Why Is My Calculation of the Antiderivative for sqrt(20-x) Incorrect?

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Homework Help Overview

The discussion revolves around the calculation of the antiderivative of the function sqrt(20-x) over the interval from 0 to 20. Participants are examining the correctness of the original poster's approach and the evaluation of the antiderivative.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the antiderivative but expresses confusion regarding the evaluation at the limits. Some participants question the correctness of the original expression and suggest that the presence of the square root may be misinterpreted in the context of the exponent.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering corrections and clarifications. There is a recognition of errors in the initial evaluation, and some guidance has been provided regarding the proper form of the antiderivative.

Contextual Notes

There is an acknowledgment of potential misunderstandings related to the evaluation of the antiderivative at the specified limits, as well as the impact of fatigue on the original poster's calculations.

hekoshi
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I can't seem to get this antiderivitive correct: sqrt(20-x) from 0 to 20.

I end up with -(2/3)(sqrt(20-x)^(3/2) which is zero evaluated at either of the two limits.

This is not correct since the area below the curve of the original equation is definitely not 0.
 
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You shouldn't have sqrt in your answer since that's already accounted for in the 3/2 exponent. Also, the antiderivative functionis not zero for x = 0. Check it again.
 
Yes and also it is not zero at both the limits; it should work out now.
 
oops, i meant to write -(2/3)(20-x)^(3/2), but yeah, i didn't evaluate it correctly the first time. It should be:

F(b)-F(a)=-(2/3)(20-20)^(3/2)-(-(2/3)(20-0)^(3/2))

*Facepalm* I blame it on lack of sleep, haha.

thanks.
 

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