How do you find the average value of #f(x)=-2x^3+10x^2-7x+5# as x varies between #[1/2,3]#?

Question

Christopher Allers · Accepted Answer

#595/48#

f(x) is continuous on the closed interval #[1/2,3]#

The average value of f(x) is found using.

#color(red)(bar(ul(|color(white)(2/2)color(black)(1/(b-a)int_a^bf(x)dx)color(white)(2/2)|)))# #"where " [a,b]=[1/2,3]#

#rArr1/(5/2)int_(1/2)^3(-2x^3+10x^2-7x+5)dx#

#=2/5[-1/2x^4+10/3x^3-7/2x^2+5x]_(1/2)^3#

#=2/5[(-81/2+90-63/2+15)-(-1/32+5/12-7/8+5/2)]#

#=2/5[33-193/96]=595/48#

Ezekiel Alexander · Answer

undefined To find the average value of $ f(x) = -2x^3 + 10x^2 - 7x + 5 $ over the interval $ [1/2, 3] $, we use the formula:

$$ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx $$

where $ a $ and $ b $ are the endpoints of the interval.

In this case, $ a = 1/2 $ and $ b = 3 $, and the function is $ f(x) = -2x^3 + 10x^2 - 7x + 5 $.

So, the average value is:

$$ \text{Average value} = \frac{1}{3 - 1/2} \int_{1/2}^{3} (-2x^3 + 10x^2 - 7x + 5) \, dx $$

$$ = \frac{1}{5/2} \int_{1/2}^{3} (-2x^3 + 10x^2 - 7x + 5) \, dx $$

$$ = \frac{2}{5} \left[ -\frac{1}{2}x^4 + \frac{10}{3}x^3 - \frac{7}{2}x^2 + 5x \right]_{1/2}^{3} $$

Now, we substitute the endpoints:

$$ = \frac{2}{5} \left[ \left( -\frac{1}{2}(3)^4 + \frac{10}{3}(3)^3 - \frac{7}{2}(3)^2 + 5(3) \right) - \left( -\frac{1}{2}(1/2)^4 + \frac{10}{3}(1/2)^3 - \frac{7}{2}(1/2)^2 + 5(1/2) \right) \right] $$

$$ = \frac{2}{5} \left[ \left( -\frac{1}{2}(81) + \frac{10}{3}(27) - \frac{7}{2}(9) + 15 \right) - \left( -\frac{1}{2}(1/16) + \frac{10}{3}(1/8) - \frac{7}{2}(1/4) + \frac{5}{2} \right) \right] $$

$$ = \frac{2}{5} \left[ \left( -\frac{81}{2} + \frac{270}{3} - \frac{63}{2} + 15 \right) - \left( -\frac{1}{32} + \frac{10}{24} - \frac{7}{8} + \frac{5}{2} \right) \right] $$

$$ = \frac{2}{5} \left[ \left( -\frac{81}{2} + 90 - \frac{63}{2} + 15 \right) - \left( -\frac{1}{32} + \frac{5}{6} - \frac{7}{8} + \frac{5}{2} \right) \right] $$

$$ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{3}{96} + \frac{40}{96} - \frac{42}{96} + \frac{120}{96} \right] $$

$$ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{-3 + 40 - 42 + 120}{96} \right] $$

$$ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{115}{96} \right] $$

$$ = \frac{2}{5} \left[ \frac{-162 + 180 - 63 + 30 + 115}{96} \right] $$

$$ = \frac{2}{5} \left[ \frac{100}{96} \right] $$

$$ = \frac{5}{12} $$

So, the average value of $ f(x) = -2x^3 + 10x^2 - 7x + 5 $ over the interval $ [1/2, 3] $ is $ \frac{5}{12} $.