Early Solutions to the 2026 AP Calculus AB FRQs | Step by Step

Part A: Approximate \(M'(7.5)\)

Task. Use the table to approximate the derivative at \(t=7.5\), then state its units.

1. Choose the centered secant interval.

   The interval \([5,10]\) is centered at \(7.5\). Its average rate of change is therefore the natural table-based approximation to the instantaneous rate \(M'(7.5)\).

   \[ M'(7.5)\approx\frac{M(10)-M(5)}{10-5}. \]

2. Substitute the table values.

   \[ M'(7.5)\approx\frac{16-7}{5}=\frac95=1.8. \]

3. Determine the units.

   \(M\) is measured in birds per day and \(t\) in days, so differentiating the rate with respect to time gives

   \[ \frac{\text{birds/day}}{\text{day}}=\text{birds per day}^2. \]

Part B(i): Midpoint Riemann-sum estimate

Task. Estimate the total number of male birds arriving from day \(0\) to day \(30\) with three midpoint subintervals.

1. Identify widths and midpoints.

   The subintervals are \([0,10]\), \([10,20]\), and \([20,30]\). Each width is \(10\), and the midpoints are \(5\), \(15\), and \(25\).

2. Multiply each width by the rate at its midpoint.

   \[ \begin{aligned} \int_0^{30}M(t)\,dt &\approx 10M(5)+10M(15)+10M(25)\\ &=10(7)+10(6)+10(2)\\ &=10(15)=150. \end{aligned} \]

3. Attach the accumulated-quantity unit.

   \((\text{birds/day})(\text{days})=\text{birds}\).

Part B(ii): Interpret the definite integral

Task. Explain the meaning of \(\int_0^{30}M(t)\,dt\) in context.

\(M(t)\) is an arrival rate. Integrating a rate over a time interval accumulates the total amount that arrived during that interval.

Part C: Total female birds from day 15 to day 45

Task. For \(15\le t\le45\), use \(F(t)=18+16\sin\left(\frac{\pi}{20}(t+15)\right)\) to find the number of arriving female birds.

1. Integrate the arrival rate over the requested time interval.

   \[ \begin{aligned} N &=\int_{15}^{45}\left[18+16\sin\left(\frac{\pi}{20}(t+15)\right)\right]dt\\ &=18(45-15)-\frac{320}{\pi} \left[\cos\left(\frac{\pi}{20}(t+15)\right)\right]_{15}^{45}. \end{aligned} \]

   The factor \(\frac{320}{\pi}\) comes from reversing the chain rule when integrating the sine term.

2. Evaluate the two angles.

   \[ \frac{\pi}{20}(45+15)=3\pi, \qquad \frac{\pi}{20}(15+15)=\frac{3\pi}{2}. \]

3. Simplify and round only at the end.

   \[ \begin{aligned} N &=540-\frac{320}{\pi}\left(\cos3\pi-\cos\frac{3\pi}{2}\right)\\ &=540-\frac{320}{\pi}(-1-0)\\ &=540+\frac{320}{\pi}\\ &\approx641.859. \end{aligned} \]

   A number of birds must be a whole number, so round to the nearest integer.

Part D: Show the two arrival rates are equal at some time

Task. Decide whether there is a time in \((15,20)\) when \(M(t)=F(t)\), and justify the conclusion.

1. Define a difference function.

   \[ D(t)=M(t)-F(t). \]

   Because \(M\) is differentiable and \(F\) is continuous, \(D\) is continuous on \([15,20]\). Continuity is required before applying the Intermediate Value Theorem.

2. Check the sign at \(t=15\).

   \[ \begin{aligned} F(15)&=18+16\sin\left(\frac{3\pi}{2}\right)=18-16=2,\\ D(15)&=M(15)-F(15)=6-2=4\gt0. \end{aligned} \]

3. Check the sign at \(t=20\).

   \[ \begin{aligned} F(20)&=18+16\sin\left(\frac{7\pi}{4}\right)=18-8\sqrt2,\\ D(20)&=5-(18-8\sqrt2)=8\sqrt2-13\approx-1.686\lt0. \end{aligned} \]

4. Apply the Intermediate Value Theorem.

   Since the continuous function \(D\) changes sign between \(15\) and \(20\), there is at least one \(c\) with \(15\lt c\lt20\) for which \(D(c)=0\). That equation means \(M(c)=F(c)\).

Part A: Area of region \(R\)

Task. Find the area under \(g(x)\) and above the \(x\)-axis from \(x=0\) to \(x=1\).

1. Write the area integral.

   \[ A_R=\int_0^1g(x)\,dx=\int_0^1\frac{14x+12}{x+12}\,dx. \]

2. Rewrite the rational function.

   This creates a constant term plus a logarithmic derivative that can be integrated directly.

   \[ \frac{14x+12}{x+12} =\frac{14(x+12)-156}{x+12} =14-\frac{156}{x+12}. \]

3. Integrate and evaluate.

   \[ \begin{aligned} A_R &=\int_0^1\left(14-\frac{156}{x+12}\right)dx\\ &=[14x-156\ln(x+12)]_0^1\\ &=14-156\ln\left(\frac{13}{12}\right)\\ &\approx1.513. \end{aligned} \]

Part B: Volume with rectangular cross sections

Task. Set up the volume integral when cross sections perpendicular to the \(x\)-axis are rectangles whose height is one-third of the base in region \(R\).

1. Express the base and height.

   For \(x\in[0,1]\), the vertical base segment runs from \(y=0\) to \(y=g(x)\), so

   \[ \text{base}=g(x),\qquad \text{height}=\frac13g(x). \]

2. Build the cross-sectional area.

   \[ A(x)=\bigl(g(x)\bigr)\left(\frac13g(x)\right)=\frac13\bigl(g(x)\bigr)^2. \]

3. Integrate the area from \(x=0\) to \(x=1\).

   \[ V=\int_0^1A(x)\,dx. \]

Part C: Area between \(f\) and \(g\)

Task. Find the shaded area between the two curves across both intersections.

1. Find the second intersection with a calculator.

   The curves meet at \(x=1\) and again at \(x=a\). Solve

   \[ 1.43^x+0.57=\frac{14x+12}{x+12}. \]
   \[ a\approx3.255816474,\qquad b=f(a)=g(a)\approx3.774392. \]

2. Identify the upper curve on each interval.

   • On \(0\le x\le1\), \(f(x)\ge g(x)\).
   • On \(1\le x\le a\), \(g(x)\ge f(x)\).

   The upper curve changes at \(x=1\), so one signed integral would cancel part of the geometric area.

3. Split the area integral at \(x=1\).

   \[ \begin{aligned} A &=\int_0^1\bigl(f(x)-g(x)\bigr)\,dx +\int_1^a\bigl(g(x)-f(x)\bigr)\,dx\\ &=\int_0^1\left(1.43^x+0.57-\frac{14x+12}{x+12}\right)dx\\ &\quad+\int_1^{3.255816474} \left(\frac{14x+12}{x+12}-1.43^x-0.57\right)dx. \end{aligned} \]

4. Evaluate both pieces with a calculator.

   \[ \int_0^1(f-g)\,dx\approx0.258873, \qquad \int_1^a(g-f)\,dx\approx0.372911. \]
   \[ A\approx0.258873+0.372911=0.631784. \]

Part D: Rotate region \(T\) about the \(y\)-axis

Task. Set up the volume integral for the solid formed by revolving region \(T\) around the \(y\)-axis.

1. Use horizontal slices.

   Horizontal slices rotate into disks. Their radius is the horizontal distance from the \(y\)-axis to \(x=h(y)\).

2. Find the lower \(y\)-bound.

   The curve meets the \(y\)-axis when \(x=h(y)=0\).

   \[ \frac{12y-12}{14-y}=0 \Longrightarrow 12y-12=0 \Longrightarrow y=1. \]

   The upper bound is \(y=3.5\).

3. Write the disk area and volume.

   \[ r(y)=h(y)=\frac{12y-12}{14-y}, \qquad A(y)=\pi[r(y)]^2. \]

Part A: Explain why the proposed slope field is incorrect

Task. Compare the displayed field with the sign and structure required by the differential equation.

1. Determine the required sign.

   Every displayed value has \(H\gt20\), so

   \[ H-20\gt0 \Longrightarrow -\frac1{15}(H-20)\lt0. \]

   The correct segments must therefore tilt downward from left to right.

2. Check how slopes vary across the field.

   The right-hand side depends only on \(H\), so every point on a common horizontal row should have the same slope. The proposed field has that row-by-row property, but the slopes have the wrong sign.

Part B: Find the tangent-line slope at \(t=0\)

Task. Calculate \(H'(0)\) from the differential equation and initial condition.

Substitute \(H(0)=75\):

Part C: Underestimate or overestimate?

Task. Decide whether the tangent-line approximation at \(t=0\) underestimates or overestimates \(H(5)\).

1. Use the second derivative to determine concavity.

   \[ H''(t)=\frac1{225}\bigl(H(t)-20\bigr). \]

   Because \(H(t)\gt20\) for \(t\gt0\), \(H''(t)\gt0\). Thus \(H\) is concave up.

2. Relate concavity to the tangent line.

   Near the point of tangency, a tangent line to a concave-up graph lies below the graph. Therefore, the tangent-line value at \(t=5\) is less than the actual \(H(5)\).

3. Write the tangent line for reference.

   \[ L(t)=75-\frac{11}{3}t, \qquad L(5)=75-\frac{55}{3}=\frac{170}{3}. \]

Part D: Solve the differential equation

Task. Find the particular solution satisfying \(H(0)=75\).

1. Separate the variables.

   \[ \frac{dH}{dt}=-\frac1{15}(H-20) \quad\Longrightarrow\quad \frac1{H-20}\,dH=-\frac1{15}\,dt. \]

2. Integrate both sides.

   \[ \int\frac1{H-20}\,dH=\int-\frac1{15}\,dt, \qquad \ln|H-20|=-\frac{t}{15}+C. \]

3. Exponentiate.

   \[ |H-20|=e^Ce^{-t/15}. \]

   Since \(H(t)\gt20\), the difference \(H-20\) is positive. Absorb the positive constant into \(C_1\):

   \[ H-20=C_1e^{-t/15}. \]

4. Apply \(H(0)=75\).

   \[ 75-20=C_1e^0 \Longrightarrow C_1=55. \]
   \[ H(t)-20=55e^{-t/15}. \]

Part A: Find \(g'(2)\)

Task. For \(x\gt0\), \(g(x)=f(x)-\ln x\). Find \(g'(2)\).

1. Differentiate \(g\).

   \[ g'(x)=f'(x)-\frac1x. \]

2. Read \(f'(2)\) from the graph and substitute.

   \[ g'(2)=f'(2)-\frac12=1.5-0.5=1. \]

Part B: Point of inflection of \(f\)

Task. Find the point where the concavity of \(f\) changes on \(0\lt x\lt3\).

1. Connect \(f''\) to the behavior of \(f'\).

   \[ f''(x)=\bigl(f'(x)\bigr)'. \]

   \(f\) is concave up where \(f'\) increases and concave down where \(f'\) decreases.

2. Read the change from the graph.

   • \(f'\) increases on \((0,1)\), so \(f\) is concave up there.
   • \(f'\) decreases on \((1,3)\), so \(f\) is concave down there.

   The concavity changes at \(x=1\). Although \(f'\) changes from decreasing to increasing at \(x=3\), that endpoint is excluded from \(0\lt x\lt3\).

Part C: Increasing and concave down

Task. Find where \(f\) is both increasing and concave down.

1. Find where \(f\) is increasing.

   \(f\) increases where \(f'(x)\gt0\). From the graph, this occurs on \((-2,3)\) and \((3,4)\).

2. Find where \(f\) is concave down.

   \(f\) is concave down where \(f'\) is decreasing. From the graph, that occurs on \((-4,-3)\) and \((1,3)\).

3. Intersect the two conditions.

   The common interval is \((1,3)\).

Part D: Absolute extrema of \(f\)

Task. Given \(f(2)=3\), identify the absolute minimum and maximum of \(f\) on \([-4,4]\).

1. Use the sign of \(f'\) to locate candidates.

   \[ f'(x)\lt0\text{ on }(-4,-2), \qquad f'(x)\gt0\text{ on }(-2,3)\cup(3,4). \]

   Thus \(f\) decreases until \(x=-2\) and increases afterward. At \(x=3\), \(f'(3)=0\), but \(f'\) does not change sign. Therefore \(x=-2\) is the absolute-minimum candidate, and the absolute maximum must be at \(x=-4\) or \(x=4\).

2. Compare the endpoint values using accumulated change.

   \[ f(-4)=3-\int_{-4}^{2}f'(x)\,dx, \qquad f(4)=3+\int_{2}^{4}f'(x)\,dx. \]

   On \([-4,2]\), the positive area above the \(x\)-axis from \(-2\) to \(2\) exceeds the small negative area from \(-4\) to \(-2\). Therefore

   \[ \int_{-4}^{2}f'(x)\,dx\gt0 \quad\Longrightarrow\quad f(-4)\lt3. \]

   Also, \(f'(x)\ge0\) on \([2,4]\) and is positive except at \(x=3\), so

   \[ \int_2^4f'(x)\,dx\gt0 \quad\Longrightarrow\quad f(4)\gt3. \]

   Hence \(f(4)\gt f(-4)\).

Part A: Acceleration at \(t=1\)

Task. Find \(a(1)\).

1. Differentiate the relevant velocity piece.

   Because \(1\in[0,4]\), use the first piece:

   \[ a(t)=v'(t)=4t^3-24t^2+32t. \]

2. Evaluate at \(t=1\).

   \[ a(1)=4(1)^3-24(1)^2+32(1)=4-24+32=12. \]

Part B: Is the car speeding up or slowing down?

Task. Determine what happens to the speed at \(t=1\).

1. Check the velocity sign.

   \[ v(1)=1-8+16=9\gt0. \]

2. Compare it with the acceleration sign.

   \[ a(1)=12\gt0. \]

   Velocity and acceleration have the same sign, so the magnitude \(|v|\) is increasing.

Part C: Distance traveled from \(t=0\) to \(t=4\)

Task. Find the total distance traveled on \([0,4]\).

1. Start with the distance formula.

   \[ \text{distance}=\int_0^4|v(t)|\,dt. \]

2. Determine whether the absolute value is needed.

   \[ v(t)=t^4-8t^3+16t^2=t^2(t-4)^2\ge0 \quad\text{on }[0,4]. \]

   Because \(v(t)\) is nonnegative, \(|v(t)|=v(t)\) throughout this interval.

3. Integrate.

   \[ \begin{aligned} \text{distance} &=\int_0^4(t^4-8t^3+16t^2)\,dt\\ &=\left[\frac{t^5}{5}-2t^4+\frac{16}{3}t^3\right]_0^4\\ &=\frac{1024}{5}-512+\frac{1024}{3}\\ &=\frac{3072-7680+5120}{15}\\ &=\frac{512}{15}. \end{aligned} \]

Part D: Average velocity from \(t=6\) to \(t=12\)

Task. Find the average value of the velocity on \([6,12]\).

1. Use the average-value formula.

   \[ \bar v=\frac1{12-6}\int_6^{12}v(t)\,dt. \]

2. Substitute the third velocity piece and integrate.

   \[ \begin{aligned} \bar v &=\frac16\int_6^{12}\left[10\cos\left(\frac{\pi}{3}t\right)-10\right]dt\\ &=\frac16\left[\frac{30}{\pi}\sin\left(\frac{\pi}{3}t\right)-10t\right]_6^{12}. \end{aligned} \]

3. Evaluate the endpoint angles.

   \[ \begin{aligned} \bar v &=\frac16\left[ \left(\frac{30}{\pi}\sin4\pi-120\right) -\left(\frac{30}{\pi}\sin2\pi-60\right) \right]\\ &=\frac16(-120+60)=-10. \end{aligned} \]

Part A: Evaluate a limit

Task. Evaluate \(\displaystyle\lim_{x\to2}\frac{f(x)}{x}\).

Because \(f\) is twice differentiable, it is continuous. The denominator approaches \(2\ne0\), so direct substitution is valid:

Part B: Differentiate a composition

Task. If \(g(x)=f(f(x))\), find \(g'(2)\).

1. Apply the chain rule.

   \[ g'(x)=f'(f(x))\cdot f'(x). \]

2. Read each required value from the table.

   \(f(2)=3\), \(f'(3)=9\), and \(f'(2)=4\).

   \[ g'(2)=f'(f(2))f'(2)=f'(3)(4)=9\cdot4=36. \]

Part C: Recover \(h(2)\) from \(h'\)

Task. Given \(h(0)=10\) and \(h'(x)=f'(3x)\), find \(h(2)\).

1. Use accumulated change.

   \[ h(2)-h(0)=\int_0^2h'(x)\,dx=\int_0^2f'(3x)\,dx. \]

2. Substitute \(u=3x\).

   Since \(du=3\,dx\), \(dx=\frac13\,du\). The bounds change from \(x=0,2\) to \(u=0,6\).

   \[ h(2)=10+\frac13\int_0^6f'(u)\,du. \]

3. Apply the Fundamental Theorem of Calculus.

   \[ \begin{aligned} h(2) &=10+\frac13\bigl(f(6)-f(0)\bigr)\\ &=10+\frac13\bigl(5-(-1)\bigr)\\ &=10+2=12. \end{aligned} \]

Part D(i): Differentiate an integral with a variable upper bound

Task. If \(\displaystyle k(x)=\int_0^x t^2f(t)\,dt\), find \(k'(x)\).

By the Fundamental Theorem of Calculus, differentiate by evaluating the integrand at the upper limit:

Part D(ii): Find \(k''(3)\)

Task. Differentiate again and use the table.

1. Use the product rule.

   \[ k''(x)=2xf(x)+x^2f'(x). \]

2. Substitute \(x=3\), \(f(3)=8\), and \(f'(3)=9\).

   \[ k''(3)=2(3)(8)+(3)^2(9)=48+81=129. \]