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Using the mean value theorem, obtain the mean value of the function f(x) = x^2 in the...
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(Level 1) Associate Educator, Expert, Newton
The average value of the function f(x) on the interval [a,b] is
`1/(b-a) int_a^b f(x) dx`
Thus, the average value for your question is:
`1/(2+1) int_(-1)^2 x^2 dx`
`=1/3 [1/3 x^3 ]|_(-1)^2`
`=1/9 (8- (-1))`
now, if f(x) is continuous on [a,b], then there is a point c in [a.b] where the average of f(x) equals f(c)
here, f(x) is a polynomial, which is continuous. So, there is some c in [-1,2] at which f(c) =1
`f(c)=c^2 = 1`
`c= +- 1`
The gist is:
The area under the curve f(x) on [-1,2] is the same as the area of the rectangle that has width 2+1 and height f(c):
The area under the parabola is the same as the area under the straight line.
This is different from the mean value theorem.
The mean value theorem states:
if f(x) is continuous on [a,b], and differentiable on (a,b) , then there is some c in (a,b) where
`f'(c) = (f(b)-f(a))/(b-a)`
The gist is this:
If you drew a line through the points (a,f(a)) and (b,f(b)), then somewhere between (a,b) there is a place where the tangent line has the same slope as the line you drew
So, for us, f(a) = f(-1) = 1, f(b) = f(2) = 4
`(f(b)-f(a))/(b-a) = (3-1)/(2+1) = 3/3=1`
`f'(c) = 2c`
2c = 1
c = 1/2
That is, at c=1/2, f(c) has the same tangent slope as the slope of the line connecting (-1,1) to (2,4):
Posted by mlehuzzah on December 11, 2012 at 4:17 AM (Answer #1)
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