AP Calculus BC 6.6 Applying Properties of Definite Integrals - FRQs - Exam Style Questions
No-Calc Question
The graph of the differentiable function \(f\), shown for \(-6\le x\le 7\), has a horizontal tangent at \(x=-2\) and is linear for \(0\le x\le 7\).
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 \(12\).
(a) The function \(g\) is defined by \(g(x)=\displaystyle\int_{0}^{x} f(t)\,dt\). Find the values of \(g(-6)\), \(g(4)\), and \(g(6)\).
(b) For the function \(g\) defined in part (a), find all values of \(x\) in the interval \(0\le x\le 6\) at which the graph of \(g\) has a critical point. Give a reason for your answer.
(c) The function \(h\) is defined by \(h(x)=\displaystyle\int_{-6}^{x} f'(t)\,dt\). Find the values of \(h(6)\), \(h'(6)\), and \(h”(6)\). Show the work that leads to your answers.
Most-appropriate topic codes:
• TOPIC 6.6: Applying Properties of Definite Integrals (part a) | • TOPIC 6.4: The Fundamental Theorem of Calculus and Accumulation Functions (part b) | • TOPIC 6.7: The Fundamental Theorem of Calculus and Definite Integrals (part c)
▶️ Answer/Explanation
(a) Values of \(g(x)=\displaystyle\int_{0}^{x} f(t)\,dt\)
Use areas and properties of definite integrals.Region \(R\) on \([-6,0]\) has area \(12\) above the \(x\)-axis, so \(\displaystyle \int_{-6}^{0} f(t)\,dt=12\).
\[ g(-6)=\int_{0}^{-6} f(t)\,dt =-\int_{-6}^{0} f(t)\,dt =-12. \]
On \(0\le x\le 6\) the graph is a line through \((0,2)\) and \((6,-1)\).
Slope \(m=\dfrac{-1-2}{6-0}=-\dfrac{3}{6}=-\dfrac12\), so \(f(x)=2-\dfrac{x}{2}\) for \(0\le x\le 6\).
The \(x\)-intercept is at \(x=4\).
\[ g(4)=\int_{0}^{4} f(t)\,dt =\text{(area of triangle base 4, height 2)} =\tfrac12\cdot 4\cdot 2 =4. \]
\[ g(6)=\int_{0}^{6} f(t)\,dt =\underbrace{\int_{0}^{4} f(t)\,dt}_{\text{+ area }=4} +\underbrace{\int_{4}^{6} f(t)\,dt}_{\text{− area }=\,-1} =4-1 =\boxed{3}. \]
(b) Critical points of \(g\) on \(0\le x\le 6\)
By the FTC, \(g'(x)=f(x)\).Critical points occur where \(g'(x)=0\) (and \(f\) is defined).
From the graph, \(f(x)=0\) only at \(x=4\) on \([0,6]\).
Before \(x=4\): \(f(x)>0\Rightarrow g'(x)>0\) (increasing).
After \(x=4\): \(f(x)<0\Rightarrow g'(x)<0\) (decreasing).
Therefore \(x=\boxed{4}\) is the only critical point, and \(g\) has a relative maximum there.
(c) Values for \(h(x)=\displaystyle\int_{-6}^{x} f'(t)\,dt\)
Use the FTC and the fact that the integral of a derivative gives a net change.\[ h(6)=\int_{-6}^{6} f'(t)\,dt =f(6)-f(-6) =(-1)-(0.5) =\boxed{-1.5}. \]
\[ h'(x)=f'(x)\quad\Rightarrow\quad h'(6)=f'(6)=\text{slope of the line on }[0,6] =\boxed{-\tfrac12}. \]
\[ h”(x)=f”(x)\quad\Rightarrow\quad h”(6)=\boxed{0} \] because \(f\) is linear on \(0\le x\le 7\).