# Ap Calculus BC 2002 Form B Question 4?

## Apr 30, 2018

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

$g \left(6\right) = 5 + {\int}_{6}^{6} f \left(t\right) \mathrm{dt}$

We know that ${\int}_{a}^{a} \text{anything} \mathrm{dx} = 0$, therefore $g \left(6\right) = 5$.

$g ' \left(6\right) = \frac{d}{\mathrm{dx}} \left({\int}_{6}^{x} f \left(t\right) \mathrm{dt}\right) = f \left(6\right) = 3$

$g ' ' \left(6\right) = \frac{d}{\mathrm{dx}} \left(f \left(6\right)\right) = 0$ because the tangent is horizontal

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

This will be $\left[12 , 15\right]$ and $\left[- 3 , 0\right]$

c) Concavity is determined by the second derivative.

$g ' ' \left(x\right) = f ' \left(x\right)$

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 $\left(6 , 15\right)$

d) The trapezoidal approximation is simply done by drawing trapezoids on the graph and adding up their areas. $A = 3 \frac{- 1}{2} + 3 \frac{1}{2} + \left(1 + 3\right) \frac{3}{2} + \left(1 + 3\right) \frac{3}{2} + 3 \frac{1}{2} + 3 \frac{- 1}{2}$

$A = 6 + 6 = 12$

Thus the approximation for ${\int}_{- 3}^{15} f \left(x\right) \mathrm{dx}$ is $12$.

Hopefully this helps!