notes / updates | ap calculus bc

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February 23, 2026 | Snow Day - Office Hours
Office Hours - 12:30-1:30pm

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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 \( u = \tan^{-1}x, v' = x^{-2} \)

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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.)

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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: \( \frac{1}{2} \arcsin(2/3) \)

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January 22, 2026 | Mean Value Theorem for Integrals (Demo)
https://www.geogebra.org/m/Jan0wXhp

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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, \( f(x) = x^3 - 6x + 7\).

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