A. so find the area and divide by the width. g(x) is just the average value of f(x) over the interval.
A = ∫[0,3√π] x sin(x^3) dx = 0.334558
I hope you have some good numeric integration tools handy.
B. Again, just do the integral. Consider the area as a bunch of strips of width dx, whose height is the space between the two curves.
∫[0,m] mx - x^2 dx = 1/2 m x^2 - 1/3 x^3 [0,m] = 1/2 m^3 - 1/3 m^3 = 1/6 m^3
so, now you want
m^3/6 = 8
m^3 = 48
m = 2∛6
Consider the function f(x) = xsin(x^3) on the interval [0, 3√π]
A. Find a constant function (g(x) = constant) on the interval [0, 3√π] such that the area under the graph of g is the same as the area under the graph of f on the interval [0, 3√π].
B. Consider the curves y = x^2 and y = mx, where m is some positive constant m is, the two curves enclose a region in the first quadrant.
Without using a calculator, find the positive constant m such that the area of the region bounded by the curves y = m^2 and y = mx is equal to 8.
1 answer
1 answer