Practice AP Calculus BC questions on evaluating definite integrals with geometry, interval properties, and linearity when function information is given.
The graph of the differentiable function f, shown for −6≤x≤7, has a horizontal tangent at x=-2 and is linear for 0≤x≤7. Let R be the region in the second quadrant bounded by the graph of f, the vertical line x=-6, and the x - and y-axes. Region R has area 12.
The function g is defined by g(x)=∫0xf(t)dt. Find the values of g(-6), g(4), and g(6).
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g(−6)=∫0−6f(t)dt=−∫−60f(t)dt=−12
g(-6)
1 point
g(4)=∫04f(t)dt=21⋅4⋅2=4
g(4)
1 point
g(6)=∫06f(t)dt=21⋅4⋅2−21⋅2⋅1=3
g(6)
1 point
Scoring notes:
- Supporting work is not required for any of these values. However, any supporting work that is
shown must be correct to earn the corresponding point.
- Special case: A response that explicitly presents g(x)=∫−6xf(t)dt does not earn the first point it
would have otherwise earned. The response is eligible for all subsequent points for correct answers,
or for consistent answers with supporting work.
○ Note: ∫−6−6f(t)dt=0,∫−64f(t)dt=16,∫−66f(t)dt=15
- Labeled values may be presented in any order. Unlabeled values are read from left to right and
from top to bottom as g(-6), g(4), and g(6), respectively. A response that presents only 1 or 2
values must label them to earn any points.
Total for part (a) 3 points