what is the average value of y for the part of the curve y=4x-x^3 which is in the first quadrant?
a)3/8
b)2/3
c)32/3
d)3/2
e)8/4

Question

what is the average value of y for the part of the curve y=4x-x^3 which is in the first quadrant?
a)3/8
b)2/3
c)32/3
d)3/2
e)8/4

luigi0210 · Answer

Do you know the formula for average value?

anonymous · Answer

Nope I forgot haha

luigi0210 · Answer

The average value of a function, say
$$f(x)$$
over the interval [a,b] is given by the the formula
$$\huge f_{ave}=\frac{1}{b−a} \int\limits_{x=a}^{b}f(x)dx$$
assuming
$$f(x)$$
is properly integrable over [a,b].

agent0smith · Answer

And your a and b are the x-intercepts of y=4x-x^3 in the first quadrant (one will be zero, the other is easily found by factoring it)

luigi0210 · Answer

@Purespear you there?

anonymous · Answer

oh yeah thanks I got it now I was just trying the question out

agent0smith · Answer

The way i remember average value of a function (might help if you've done physics):

average velocity is displacement/change in time...  average value of a velocity function is therefore the area under the curve, divided by the change in time... so average value of any function is also the area under the curve, divided by the change in x.

anonymous · Answer

ohhh i think i got it is it b?