3 years ago

If f(x)=x^3-x what is the average rate of change of f(x) over the interval [1, 5]?

1 answer:

6 0

Average rate of change= [f (x2)-f (x1)]/(x2-x1)
(the same formula of the slope)
x2=5
x1=1
f (x2)=5^3-1 =124
f (x1)= 1^3-1 =0
(124-0)/(5-1) = 31

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Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the $i$ -th subinterval

$\ell_i=\dfrac{i-1}5$

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$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}$

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$2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5$

$\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)$

$\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}$

(compare to the actual value of the integral of about 14.45)

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