How do I find the mean value of the function 2X^2 - 5X +3 on the interval [1,2]?
And how do i find the area bounded by f(x)= -x^2 +4x -21 and g(x)= sqrt2x? 3.5 > x > 0 (3.5 is greater than or equal to x which is greater than or equal to zero)
2007-07-29 12:06:15 · 3 answers · asked by LaLa 1 in Science & Mathematics ➔ Mathematics
Integrate over the interval.
Average Value = Integral from a to b / (b - a)
Average Value = ∫(2x² - 5x + 3)dx / (2 - 1)
= ∫(2x² - 5x + 3)dx = (2/3)x³ - (5/2)x² + 3x | [Eval from 1 to 2]
= (16/3 - 10 + 6) - (2/3 - 5/2 + 3) = 4/3 - 7/6 = 1/6
2007-07-29 12:18:14 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Mean Value = [f(2)-f(1)]/(2-1) = (1-0)/1= 1
It will be equal to the area below the curve y= (2x)^.5 plus the area above the curve y = -x^2+4x-21 over the range
0=
Answer should be about 71.63
2007-07-29 12:49:24 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
For the second one use a double integral SS(dy dx)
your inner limits will be the two given equations in terms of x
your outer limits will be the given boundaries (0, 3.5)
it helps to graph them first
2007-07-29 12:41:48 · answer #3 · answered by matt d 1 · 0⤊ 0⤋