Question

Ap Calculus BC 2002 Form B Question 4?

Answer 1

a) We're going to have to recall the FTC here.

`g(6) = 5 + int_6^6 f(t) dt`

We know that `int_a^a "anything" dx = 0`, therefore `g(6) = 5`.

`g'(6) = d/dx(int_6^x f(t) dt) = f(6) = 3`

`g''(6) = d/dx(f(6)) = 0` because the tangent is horizontal

b) Since `g'(x) = f(x)`, we seek to find the intervals where `f(x)`, or `g'(x)` is negative.

This will be `[12, 15]` and `[-3, 0]`

c) Concavity is determined by the second derivative.

`g''(x) = f'(x)`

We are looking for places on the given graph where the tangent line's slopes are negative, or when the graph is decreasing (since we seek the intervals where `g` is CONCAVE DOWN.

This will occur on `(6, 15)`

d) The trapezoidal approximation is simply done by drawing trapezoids on the graph and adding up their areas.

`A = 3(-1)/2 + 3(1)/2 + (1 + 3)(3)/2 + (1 + 3)(3)/2 + 3(1)/2 + 3(-1)/2`

`A = 6 + 6 = 12`

Thus the approximation for `int_(-3)^15 f(x) dx` is `12`.

Hopefully this helps!