Given the graph of a differentiable function f (with a horizontal tangent and a linear portion) and the area of a bounded region, students evaluate integral-defined functions at several points using signed geometric area, find the critical points of an accumulation function, and compute values and derivatives of a function defined as the accumulation of f'.
Given the graph of a differentiable function f (with a horizontal tangent and a linear portion) and the area of a bounded region, students evaluate integral-defined functions at several points using signed geometric area, find the critical points of an accumulation function, and compute values and derivatives of a function defined as the accumulation of f'.
Read the complete question in the official College Board FRQ PDF ›
Original Tian2 solution. Each part identifies which rubric points are earned and why.
$g(x)=\int_0^x f(t)\,dt$ accumulates signed area under $f$ starting at $0$. **$g(-6)$:** $g(-6)=\int_0^{-6} f(t)\,dt=-\int_{-6}^0 f(t)\,dt$. The region $R$ between $f$ and the axes on $-6\le x\le 0$ lies in the second quadrant (so $f\ge 0$ there) and has area $12$, hence $\int_{-6}^0 f\,dt=12$ and $$g(-6)=-12.$$ **$g(4)$:** On $[0,7]$, $f$ is the line through $(0,2)$ with slope $-\tfrac12$, so $f(x)=2-\tfrac{x}{2}$ and $f(4)=0$. The region under $f$ on $[0,4]$ is a triangle with base $4$ and height $2$: $$g(4)=\int_0^4 f(t)\,dt=\tfrac12(4)(2)=4.$$ **$g(6)$:** Add the signed area on $[4,6]$, where $f$ is below the axis (a triangle with base $2$, height $1$, area $\tfrac12(2)(1)=1$, counted negative): $$g(6)=g(4)+\int_4^6 f(t)\,dt=4+\left(-\tfrac12(2)(1)\right)=4-1=3.$$
3 points, one for each value: g(-6)=-12 (signed area, region R has area 12 so the reversed-limit integral is -12); g(4)=4 (triangle 1/2*4*2); g(6)=3 (=4 minus the below-axis triangle 1/2*2*1). Per the SG special case, a response that instead defines the accumulation from -6 forfeits only the first point it would otherwise earn but remains eligible for the rest with consistent work.
By the Fundamental Theorem of Calculus, $g'(x)=f(x)$. A critical point of $g$ on $0\le x\le 6$ occurs where $g'(x)=f(x)=0$ (or is undefined; $f$ is continuous here). On $[0,7]$, $f(x)=2-\tfrac{x}{2}=0$ gives $x=4$. Since $f$ changes sign there (positive for $x<4$, negative for $x>4$), $g'$ changes sign and $$x=4$$ is the only critical point of $g$ on $0\le x\le 6$.
2 points: 1 for invoking the FTC g'(x)=f(x); 1 for the answer x=4 with a reason (g'=f=0 there, i.e. f crosses zero at x=4 on the interval). Listing x=4 without noting g'=f, or including spurious values where f is nonzero, does not earn the second point.
$h(x)=\int_{-6}^x f'(t)\,dt$. By the Fundamental Theorem of Calculus this accumulation of $f'$ recovers $f$ up to the lower-limit value: $$h(x)=\int_{-6}^x f'(t)\,dt=f(x)-f(-6).$$ **$h(6)$:** With $f(6)=2-\tfrac{6}{2}=-1$ and $f(-6)=\tfrac12$ (read from the graph), $$h(6)=f(6)-f(-6)=-1-\tfrac12=-\tfrac{3}{2}.$$ **$h'(6)$:** Differentiate $h(x)=f(x)-f(-6)$ (the second term is constant), so $h'(x)=f'(x)$. On $[0,7]$, $f$ is the line of slope $-\tfrac12$, hence $$h'(6)=f'(6)=-\tfrac12.$$ **$h''(6)$:** $h''(x)=f''(x)$. Since $f$ is linear near $x=6$, $f''(6)=0$, so $$h''(6)=0.$$
4 points: 1 for using the FTC to write h(x)=f(x)-f(-6) (or evaluating int_{-6}^6 f' as f(6)-f(-6)); 1 for h(6)=-3/2; 1 for h'(6)=f'(6)=-1/2; 1 for h''(6)=f''(6)=0 (f is linear near x=6). A value of h(6) without the FTC step, or h'/h'' computed from the wrong relationship, loses the corresponding points.
Our worked solutions and practice questions are original instructional content created by Tian2 AP. They are aligned to the concepts and skills described in College Board’s Course and Exam Description and are not reproductions of, or affiliated with, College Board’s official materials.
