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How do you find the average value of #f(x)=-x^5+4x^3-5x-3# as x varies between #[-2,0]#?

Answer 1

Charles Arocha

The answer is #=-2/3#

The average value is #1/(b-a)int_a^bf(x)dx#

We need #intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

Therefore,

The average value is #1/(0--2)int_-2^0(-x^5+4x^3-5x-3)dx#

#=1/2[-x^6/6+4x^4/4-5x^2/2-3x]_-2^0#

#=1/2(0-(-2^6/6+16-10+6))#

#=1/2(32/3-12)#

#=-4/6=-2/3#

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Answer 2

Isaiah Allen

To find the average value of ( f(x) = -x^5 + 4x^3 - 5x - 3 ) over the interval ([-2,0]), you need to compute the definite integral of ( f(x) ) over that interval and then divide the result by the width of the interval.

Calculate the definite integral of ( f(x) ) over ([-2,0]) using the Fundamental Theorem of Calculus.
Compute the difference of ( f(x) ) evaluated at the upper and lower bounds of the interval.
Divide the result from step 2 by the width of the interval, which is ( 0 - (-2) = 2 ).

[ \text{Average value} = \frac{1}{2} \int_{-2}^{0} (-x^5 + 4x^3 - 5x - 3) , dx ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.