Question

What is the average value of y for the part of the curve y=3x-x^2 which is in the first quadrant

Answer 1

The part of this parabola in the first quadrant spans from x=0 to x=3, which will be the bounds in your integral. The average value of a function is found by:\[\frac{1}{b-a}\int\limits_{a}^{b}f(x)dx\]

Answer 2

So,\[\frac{1}{3}\int\limits_{0}^{3}(3x-x^2)dx=\frac{1}{3}(\frac{3}{2}(3)^2-\frac{1}{3}(3)^3)\]\[=\frac{1}{3}(\frac{27}{2}-\frac{27}{3})=\frac{9}{2}-\frac{9}{3}=\frac{9}{6}=\frac{3}{2}=1.5\]

Answer 3

ok that makes sense. thank you!

Answer 4

no prob

Answer 5

I found the bounds by solving for the zeros.
x(3-x)=0

so, x=0 and x=3 are the zeros. also, the vertex is found by x=-b/(2a)=1.5 and f(1.5)=2.25. Since a=-1, the parabola opens downward. so the parabola starts out at zero and goes up to a maximum of y=2.25 and then goes back down and crosses the x-axis at x=3.