I+really+like+this+question!



a) ~ g(6) = 5 because the integral of f(t) between 6 and 6 is zero ~g'(x) = f(x), so f'(6) = 3 ~g(x) = f'(x), so g(6) = 0 because at a horizontal tangent the derivative is zero

b) g is decreasing on the intervals [-3,0] and [12,15] because the area under the curve {the integral of f(x)} is decreasing. This can be seen because f(x) in negative on these integrals.

c) When g(x) is negative, the graph of g is concave down. g(x) = f'(x). To see where f'(x) is negative, look at the graph. Where the slope is negative, g''(x) is negative. Therefore, g is concave down on the interval [6,15].I like to think that where f ' (x) is negative, f (x) is decreasing -- i.e. over the OPEN interval (6, 15).

d) 3[1 + 2*3 + 2*1+ 2*0 - 1] = 24  I think you need to fix this -- remember that the "outside term" is (delta x)/2 and that f(-3) = -1 **i miss-read. this is the integral between 3 and 15.

the approximate integral between -3 and 15: 3[-1+2*1+2*3+2*1-1]=24

Deb Levy: If it's a trapezoidal approximation, wouldn't it be 3(1/2)[f(-3)+2f(0)+2f(3)+2f(6)+2f(9)+2f(12)+f(15)]? I think that would work out to an integral of 12 instead of 24.