integral = 2 x - x^3/3 at sqrt 2 - at 0
= 2 * 2^.5 - 2^(1.5)/3 - 0
length of rectangle = 2^.5
so
y * [ 2^.5 ] = 2 * 2^.5 - 2^(1.5)/3
y * [ 2^.5 ] = 2 * 2^.5 - 2^.5 *2 /3
y * [ 2^.5 ] = 2 * 2^.5 [ 1-1/3)
y * [ 2^.5 ] =(4/3) (2^.5)
y = 4/3
where is that?
2-x^2 = 4/3
x^2 = 2/3
d
Determine if the Mean Value Theorem for Integrals applies to the function f(x)=2-x^2 on the interval [0,√2). If so, find the x-coordinates of the point(s) guaranteed by the theorem
a) No, the Mean Value Theorem for Integrals does not apply
b) Yes, x=4/3
c) Yes, x= √5/3
d) Yes, x=√(2/3)
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