notes / updates | ap calculus bc
May 4, 2026 | Free Response Generator
For more
Free Response practice, feel free to use the following generator. Each question should take approximately 15 minutes.
After 2011, only Problems 1 and 2 are calculator active. Before 2011, problems 1, 2, and 3 were calculator active.
Free Response Generator
April 28, 2026 | Quiz Feedback.
Below find some notes. Graded work will be returned at our next class meeting.
P1.
\( y = \ln(1-x^2) \) on \([0,1/2] \)
Before setting up the arclength formula, rewrite the expression inside the root as a perfect square:
\( 1+ \left( \frac{dy}{dx}\right)^2 = 1 + \frac{4x^2}{(1-x^2)^2} = ... = \left( \frac{x^2+1}{1-x^2} \right)^2 \)
Perform polynomial division or rewrite the fraction before setting up partial fractions:
\( \frac{x^2+1}{1-x^2} = \frac{x^2-1+2}{1-x^2} = \frac{x^2-1}{1-x^2} + \frac{2}{1-x^2} = -1 + \frac{2}{1-x^2} = -1 + \frac{A}{1-x} + \frac{B}{1+x} \)
\( L = \int\limits_0^{1/2} -1 + \frac{2}{(1-x)(1+x)} dx = \int\limits_0^{1/2} -1 + \frac{1}{(1-x)} + \frac{1}{1+x} dx = ... = \ln(3)-\frac{1}{2} \)
P2.
Draw the region neatly before setting up the shell method. Note that the problem instructions should have included \( x=0 \) as the fourth (natural) bound
for a curve of the form: \( x = g(y) \).
I awarded credit for attempts that interpreted the region differently from what was intended.
\( V_{shell} =\int\limits_{-2}^2 2\pi(3-y)\frac{1}{1+y^2} dy =\int\limits_{-2}^2 6\pi\frac{1}{1+y^2} dy - \int\limits_{-2}^2 2\pi y\frac{1}{1+y^2} dy ... = 12\pi \arctan(2) \)
Alternatively, if you use the Washer Method, separate the setup into two integrals. The Shell Method is more convenient here, however.
A desmos illustration of the region is shown below:
April 18, 2026 | Quiz Feedback.
Below find some notes. Graded work will be returned on Monday April 20.
P1.
a) Draw neatly. Label the axes clearly with arrows and show the intersection point \( (\pi/3, 2 ) \).
\( A(R) = \int\limits_0^{\pi/3} \left(2- \sec x \right) dx = ... = \frac{2\pi}{3}-\ln|2+\sqrt{3}| \)
b) Washer method because there is a gap between the region and the rotational axis. Identify the radii:
\( R(x) = 3-\sec x \)
\( r(x) = 3-2 = 1 \)
\( V_{washer} = \pi \int\limits_0^{\pi/3} \left( (R(x))^2-(r(x))^2 \right) dx = ... = \pi \left( \frac{8\pi}{3} - 6\ln(2+\sqrt{3}) + \sqrt{3}\right) \)
c)
\( V_{washer} = \pi \int\limits_1^2 \left( 3^2 - (3-arcsec y)^2 \right) dy\)
OR:
\( V_{washer} = \pi \int\limits_1^2 \left( 3^2 - (3- \arccos(1/y))^2 \right) dy\)
P2.
a) Like in problem 1, be sure to label the axes and shade(!) the region.
\( A(R) = \int\limits_{-2}^0 \left(2^{-x} - 2^x \right) dx \left( \right) = ... = \frac{9}{4\ln2} \)
b) Washer method because there is a gap between the region and the rotational axis. Identify the radii:
\( R(x) = \left(2^{-x}\right)^2 = 2^{-2x} \)
\( r(x) = \left( 2^x\right)^2 = 2^{2x} \)
\( V_{washer} = \pi \int\limits_{-2}^0 \left( (R(x))^2-(r(x))^2 \right) dx = ... = \frac{225 \pi}{32 \ln 2} \)
c)
\( V_{washer} = \pi \int\limits_{-2}^0 \left( 2^{-x}-(-2) \right)^2 - \left( 2^x-(-2) \right)^2 dx\)
OR:
\( V_{washer} = \pi \int\limits_{-2}^0 \left( 2^{-x}+2 \right)^2 - \left( 2^x+2 \right)^2 dx\)
April 18, 2026 | Known Cross Sections - interactive applet
https://www.geogebra.org/m/XFgMaKTy
April 6, 2026 | Study Guide
In case you are interested, below I list the ISBN number for a recommended study guide for AP Calculus BC. Note that purchasing the study guide is not required.
ISBN-10 : 0593518225
ISBN-13 : 978-0593518229
April 6, 2026 | Quiz Notes (quick)
\( T(x) = -5(x-2)^2 - 3(x-2)^3 = f(2) + f'(2) (x-2) + \frac{f"(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!} (x-2)^3 \)
a) Since the linear term in the Taylor polynomial (\( f(2) + f'(2) (x-2) \)) is 0, that means:
\( f(2) = 0; \frac{f"(2)}{2!} = -5 \rightarrow f"(2)=-10 \)
Some of you took the derivative of the Taylor polynomial here without explaining the reasoning (link between the coefficients)
b) Since the linear term in the Taylor polynomial (\( f(2) + f'(2) (x-2) \)) is 0, that means:
\(f'(2) = 0 \), so yes, the function \(f(x) \) has a critical number at \(x=2 \).
\( f(2) \) is a relative maximum because the second derivative (see part a) is negative.
This is a direct conclusion of the Second Derivative Test for Local Extrema.
c) \( f(0) \approxeq T(0) = -20 + 24 = 4 \). No information about the derivatives of \(f(x) \) at \(x=0 \) is given,
so we do not know how the function behaves there.
April 2, 2026 | Maclaurin Series
Desmos demonstration - Maclaurin Series
March 30, 2026 | Notes on today's quiz
Some of the details below are simply hints or suggestions. Problem 3 includes a desmos demo.
P1.
Perform a Ratio Test for \( a_n = \frac{\pi^n (x-1)^{2n}}{(2n+1)!}\)
Some of us failed to update n to n+1 correctly. Also, be sure to use the absolute value notation with care.
\( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\pi^{n+1} (x-1)^{2(n+1)}}{(2(n+1)+1)!} \frac{(2n+1)!}{\pi^n (x-1)^{2n}} \right| = \left| \frac{\pi (x-1)^2}{(2n+3)(2n+2)} \right| \rightarrow 0 < 1 \) for all x as n approaches infinity.
Conclusion: the given power series converges for all real numbers. The domain is all reals.
P2.
Note that if \(f(x) = \frac{1}{1+2x}\), then \( f'(x) = \frac{-2}{(1+2x)^2}, f''(x) = \frac{8}{(1+2x)^2} \)
Start with the geometric series:
\( f(x) = \frac{1}{1+2x} = \frac{1}{1-(-2x)} = 1+ (-2x) + (-2x)^2 + ... + (-2x)^n + ... \)
as long as \( |-2x| < 1 \).
Complete the two-step term by term differentiation process to find:
\( \sum\limits_{n=2}^{\infty} (-1)^n n (n-1)2^n x^{n-2} \)
The radius is the same as the original function's radius of convergence: \( R = 1/2 \)
P3.
The first four nonzero terms are:
\( e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ... + \frac{u^n}{n!} + ... \)
Plug in \( u = -x^2 \):
\( e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + ... + \frac{(-1)^n x^{2n}}{n!} + ... \)
Multiply by \( x^3 \):
\( f(x) = x^3 e^{-x^2} = x^3 -x^5 + \frac{x^7}{2!} - \frac{x^9}{3!} + ... + \frac{(-1)^n x^{2n+3}}{n!} + ... \)
The anti-derivative is:
\( F(x) = C + \frac{x^4}{4} - \frac{x^6}{6} + \frac{x^8}{8*2!} - \frac{x^10}{10*3!} + ... + \frac{(-1)^n x^{2n+4}}{(2n+4)n!} + ... \)
Integrate the first four terms:
\( \int\limits_0^2 f(x) dx = F(2) - F(0) = F(2) \approxeq ... = \frac{-116}{15} \)
If you'd like to explore this integral with an interactive desmos illustration, click
here.
P4.
a) The radius of convergence of the series \( \sum (c_n + d_n) x^n \) is equal to the smaller of the two radii of convergence
of the series \( \sum c_n x^n \) and \( \sum d_n x^n \), which is 2. Otherwise the sum will not be defined.
b) The series \( \sum c_n x^{2n} \) will have a radius of convergence equal to \( \sqrt{R} \),
since we need to have \( |u| = | x^2 | < R \).
c) It is impossible to find a power series whose interval of convergence is \( [0, \infty) \).
Recall the three possibilities for intervals of convergence for power series: 1) center only 2) all reals 3) an open or closed or half-open interval whose midpoint is the center.
d) Because the new series is obtained using term by term integration, the series will also converge for \( |x| < 2 \).
February 23, 2026 | Snow Day - Office Hours
Office Hours - 12:30-1:30pm
February 17, 2026 | Notes on today's quiz
The details below are hints or suggestions.
P1.
\( \int_2^{\infty} \frac{4}{(x-1)(x+1)} dx \)
Solve the improper integral using partial fractions and the appropriate limit notation.
P2.
\(\int_{-4}^2 \frac{x}{\sqrt{16-x^2}} dx \)
Another improper integral. Solve using the appropriate limit notation and u-substitution: \( u = 16-x^2 \).
P3.
\( \int \frac{x^2}{x^6+3x^3+2} dx \)
Solving using u-substitution: \( u = x^3 \), followed by partial fractions.
P4.
\(\int_1^{\sqrt{3}} \frac{\sqrt{1+x^2}}{x^2} dx \)
Use trigonometric substitution. Be sure to update the bounds, draw a triangle, and use
the identiy from precalculus: \( \csc^2\theta = 1 + \cot^2 \theta \)
P5.
\( \int_0^3 \frac{dx}{x^2-6x+5} \)
An improper integral. Break it into integrals and evaluate the first of the two using partial fractions.
The first integral is divergent, therefore there is no need to compute the second integral.
P6.
\( \int_0^1 x \sqrt{2-\sqrt{1-x^2}} dx \)
Trig substitution \( x=\sin\theta \) followed by u-substitution: \( u=2-\cos\theta \).
Be careful with the bounds.
P7.
\( \int \sqrt{1+e^x} dx \)
u-substitution works here. Set u equal to the integrand.
P8.
\( \int \frac{\tan^{-1}x}{x^2} dx \)
Use integration by parts with \( u = \tan^{-1}x, v' = x^{-2} \)
February 13, 2026 | Notes on today's quiz
P1.
The area of the semi-circle is given by the integral of the equation of the upper-half of the circle.
Feel free to use symmetry to rewrite as:
\( A = \int_{-m}^m \sqrt{m^2-x^2} dx = 2 \int_0^m \sqrt{m^2-x^2} dx \)
Draw a triangle for the trigonometric substitution: \( x = m \sin\theta \) and update the bounds:
\( x = 0 \rightarrow \theta = 0 \)
\( x = m \rightarrow \theta = \pi/2 \)
You'll need to use the cosine double angle formula in order to evaluate the definite integral.
P2.
From Algebra 2, \( A^3 + B^3 = (A+B) (A^2 - AB + B^2) \) We reviewed this at the start of the school year.
The factored form makes it easy to break \( \frac{1}{x^3+8} \) into partial fractions:
\( \frac{1}{(x+2)(x^2-2x+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2-2x+4}\)
Find: \( A = \frac{1}{12}, B = \frac{-1}{12}, C=\frac{1}{3} \), and then integrate. Break up the second fraction so that
you use u-substitution and arctangent. (If we are not reading the textbook, maybe we should start.)
February 10, 2026 | Notes on today's quiz
P1.
\( A = \int_3^5 \frac{x^3}{\sqrt{x^2-4}} dx \)
Draw a right triangle with one acute angle \(\theta \), the adjacent side 2, and hypotenuse \(x\).
\( x = 2\sec\theta \rightarrow dx = 2 \sec\theta \tan\theta d\theta \)
The bounds of integration are:
\( x = 3 \rightarrow \theta = \arccos(2/3) \text{ or } \tan\theta = \sqrt{5}/2\)
\( x = 5 \rightarrow \theta = \arccos(2/5) \text{ or } \tan\theta = \sqrt{21}/2\)
The final answer for area is: \( \frac{33\sqrt{21}-17\sqrt{5}}{3}\)
P2.
\( y_{avg} = \frac{1}{1-0} \int_0^1 f(x) dx = \int_0^1 \frac{1}{\sqrt{9-4x^2}} dx \)
Proceed with trig substitution: \( x = \frac{3\sin\theta}{2} \rightarrow dx = \frac{3}{2} \cos(\theta) d\theta \)
Be sure to update the bounds:
\( x=0 \rightarrow \theta = 0 \)
\( x=1 \rightarrow \theta = \arcsin(2/3) = m \)
The integral becomes: \( \int_0^1 \frac{1}{\sqrt{9-4x^2}} dx = \int_0^m \frac{1}{3 \cos\theta} \frac{3}{2} \cos(\theta) d\theta \)
The final answer is: \( \frac{1}{2} \arcsin(2/3) \)
January 22, 2026 | Mean Value Theorem for Integrals (Demo)
https://www.geogebra.org/m/Jan0wXhp
January 15, 2026 | Feedback on today's quiz
We know that \( f''(x) = 6x\). From the fact that the tangent line equation is: \( y = 5 - 3x \) at \(x=1 \),
we also know that \( f(1) = 2, f'(1) = -3 \)
Integrate: \( f'(x) = \int f''(x) dx = \int 6x dx = 3x^2 + C_1 \)
Plug in \( x=1, f'(1) = -3 \), to find \( C_1 = -6 \)
Integrate again: \( f(x) = \int f'(x) dx = \int (3x^2 -6) dx = x^3 -6x + C_2 \)
Plug in \(x=1, f(1) = 2 \), to find \( C_2 = 7 \)
Therefore, \( f(x) = x^3 - 6x + 7\).