AP Calculus BC 6.11 Integrating Using Integration by Parts bc only - FRQs - Exam Style Questions
No-Calc Question
The function \(f\) is twice differentiable for all \(x\) with \(f(0)=0\).
Values of \(f’\), the derivative of \(f\), are given in the table for selected values of \(x\).
Values of \(f’\), the derivative of \(f\), are given in the table for selected values of \(x\).
| \(x\) | \(0\) | \(\pi\) | \(2\pi\) |
|---|---|---|---|
| \(f'(x)\) | \(5\) | \(6\) | \(0\) |
(a) For \(x\ge 0\), the function \(h\) is defined by \(h(x)=\displaystyle\int_{0}^{x}\sqrt{1+\big(f'(t)\big)^{2}}\,dt\).
Find the value of \(h'(\pi)\). Show the work that leads to your answer.
(b) What information does \(\displaystyle\int_{0}^{\pi}\sqrt{1+\big(f'(x)\big)^{2}}\,dx\) provide about the graph of \(f\)?
(c) Use Euler’s method, starting at \(x=0\) with two steps of equal size, to approximate \(f(2\pi)\).
Show the computations that lead to your answer.
(d) Find \(\displaystyle\int (t+5)\cos\!\left(\frac{t}{4}\right)dt\).
Show the work that leads to your answer.
Most-appropriate topic codes:
• TOPIC 6.4: The Fundamental Theorem of Calculus and Accumulation Functions (part a).
• TOPIC 8.13: The Arc Length of a Smooth, Planar Curve & Distance Traveled (BC) (part b).
• TOPIC 7.5: Approximating Solutions Using Euler’s Method (BC) (part c).
• TOPIC 6.11: Integrating Using Integration by Parts (BC) (part d).
• TOPIC 8.13: The Arc Length of a Smooth, Planar Curve & Distance Traveled (BC) (part b).
• TOPIC 7.5: Approximating Solutions Using Euler’s Method (BC) (part c).
• TOPIC 6.11: Integrating Using Integration by Parts (BC) (part d).
▶️ Answer/Explanation
(a) Compute \(h'(\pi)\)
By the FTC for accumulation functions, if \(h(x)=\displaystyle\int_{0}^{x}G(t)\,dt\) then \(h'(x)=G(x)\).Here \(G(t)=\sqrt{1+\big(f'(t)\big)^2}\).
Therefore \(h'(x)=\sqrt{1+\big(f'(x)\big)^2}\).
With \(f'(\pi)=6\): \(h'(\pi)=\sqrt{1+6^{2}}=\boxed{\sqrt{37}}.\)
(b) Meaning of \(\displaystyle\int_{0}^{\pi}\sqrt{1+\big(f'(x)\big)^{2}}\,dx\)
The integral \(\displaystyle\int_{a}^{b}\sqrt{1+\big(f'(x)\big)^{2}}\,dx\) equals the arc length of the curve \(y=f(x)\) from \(x=a\) to \(x=b\).Hence \(\displaystyle\int_{0}^{\pi}\sqrt{1+\big(f'(x)\big)^{2}}\,dx\) is the arc length of \(y=f(x)\) on \([0,\pi]\).
(c) Euler’s method (two steps to approximate \(f(2\pi)\))
Step size: \(\Delta x=\dfrac{2\pi-0}{2}=\pi\).Start \(x_{0}=0\), \(y_{0}=f(0)=0\).
Step 1 (to \(x_{1}=\pi\)): \(y_{1}\approx y_{0}+f'(x_{0})\Delta x = 0+5\,(\pi)=5\pi.\)
Step 2 (to \(x_{2}=2\pi\)): \(y_{2}\approx y_{1}+f'(x_{1})\Delta x = 5\pi+6\,(\pi)=11\pi.\)
Therefore \(f(2\pi)\approx \boxed{11\pi}\).
(d) Antiderivative \(\displaystyle\int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt\)
Use integration by parts: let \(u=t+5\), \(dv=\cos\!\left(\tfrac{t}{4}\right)dt\).Then \(du=dt\) and \(v=4\sin\!\left(\tfrac{t}{4}\right)\).
\(\displaystyle \int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt = u\,v-\int v\,du\)
\(\qquad=4(t+5)\sin\!\left(\tfrac{t}{4}\right)-\int 4\sin\!\left(\tfrac{t}{4}\right)dt\).
Since \(\displaystyle \int \sin\!\left(\tfrac{t}{4}\right)dt=-4\cos\!\left(\tfrac{t}{4}\right)\), the remaining integral is \(-16\cos\!\left(\tfrac{t}{4}\right)\).
Thus \[ \int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt = 4(t+5)\sin\!\left(\tfrac{t}{4}\right)+16\cos\!\left(\tfrac{t}{4}\right)+C. \] Final answer: \(\boxed{4(t+5)\sin(\tfrac{t}{4})+16\cos(\tfrac{t}{4})+C}\).