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April 3, 2025 | Common Maclaurin Series with Desmos

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February 24, 2025 | Feedback on today's quiz.
P1.
Be sure to draw and shade the region. When drawing the graph, give a little more detail (for example, the vertical asymptotes).
$A = \int_2^{\infty} \frac{4}{x^2-1} dx = \lim\limits_{t\to \infty} \int_2^t \left( \frac{A}{x-1} + \frac{B}{x+1} \right) dx = ... = 2\ln 3$
Show work when calculating $A = 2, B=-2$.

P2.
Recognize the integral as improper due to the discontinuity. This means we NEED to use limits. Use u-substitution.
$\int_{-4}^2 \frac{x}{\sqrt{16-x^2}} dx = \lim\limits_{b\to -4^+} \frac{-1}{2} \int_b^2 \frac{-2x}{\sqrt{16-x^2}} dx = ... = -2 \sqrt{3}$

P3.
Rewrite the integral by completing the square:
$\int \frac{1}{\sqrt{(2y-1)^2-4}} dy$
Use trigonometric substitution with: $2y-1 = 2\sec\theta$.
Be sure to draw a right triangle. The final answer comes out to:
$\frac{1}{2} \ln \left| \frac{2y-1+\sqrt{4y^2-4y-3}}{2} \right| + C$

P4.
$\int_0^1 \ x \sqrt{2-\sqrt{1-x^2}}dx$

Option 1: Use u-substitution with $u = 1-x^2$, then w-substitution with $w = 2-\sqrt{u}$.
Option 2: Use trigonometric substitution with $x = \sin\theta$.
No matter which option you choose, be sure to update the bounds carefully whenever you switch variables. The final answer is: $\frac{2(8\sqrt{2}-7)}{15}$

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February 17, 2025 | Feedback on today's quiz.
P1.
$\int \frac{1}{\sqrt{x^2-a^2}} dx$
Use trigonometric substitution with: $x = a sec\theta$.
Be sure to draw a right triangle with all three sides labeled.
$dx = a \sec\theta \tan\theta d\theta$
After the substitution of the expressions for $x$ and $dx$, the integral reduces to:
$\int \sec\theta d\theta = \ln |\sec\theta + \tan\theta | + C = \ln\left| \frac{x +\sqrt{x^2-a^2}}{a} \right| + C$

P2.
One of your classmates pointed out that I was missing a $dx$ at the end of the integral for this problem.
$\int_0^{3\sqrt{3}/2} \left( \frac{x^3}{(4x^2+9)^{3/2}} \right) dx$
Option 1: $u = 4x^2 +9, du = 8 x dx, x^3 = x^2 x = \frac{u-9}{4} x$
Remember to switch the bounds of integration to $9 \leq u \leq 36$ when switching from $dx$ to $du$.


Option 2: Use trigonometric substitution with $x = \frac{3}{2} \tan\theta$
Be sure to draw a right triangle and update the bounds: $0 \leq \theta \leq \frac{\pi}{3}$
The final answer with either approach comes out to 3/32.

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January 15, 2025 | Problem 59 (Section 4.1)
The initial velocity is $v(0) = 16$ ft/s (positive because the object is moving up).
The initial position is $s(0) = 64$ ft.
a)
Find the velocity function: $v(t) = -32 t + 16$ by taking the anti-derivative of acceleration and using the initial value.
Find the position function $s(t) = -16t^2 + 16t + 64$ by taking the anti-derivative of velocity and using the initial value.
To find the time when the object hits the ground, set the position expression equal to zero, then solve.
b) Plug the t-value into the velocity function.

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January 13, 2025 | Feedback on Today's Quiz
P1.
We know for all four parts that $f'(x) = v(x)$.
a) $\int 2 v(x) dx = 2 \int v(x) dx = 2 f(x) + C$, so statement 1 is true.
b) The derivative of $f(2x)$ is $2 f'(2x) = 2 v(2x)$, so the second statement is false.
$f(x) = sin(x), v(x) = \cos(x)$ should work as an example.
c) $\int (v(x) + 1) dx = f(x) + x + K$, so the third statement is false.
d) If we use the same example as the one in part b), we can see that $(\sin x)^2$ is not an anti-derivative of $(\cos x)^2$. Therefore the fourth statement is false.

P2.
$a(t) = -32 \rightarrow \int (-32) dt = -32t + C = v(t)$
Since the initial velocity is zero, it follows that $C = 0$.
The velocity when the stone hits the ground is -120. The time at which this happens can be determined by solving:
$v(t) = -32t = -120 \rightarrow t=\frac{15}{4}$.
the position function is: $s(t) = \int v(t) dt = \int (-32 t) dt = -16t^2 + K$.
The position is zero when $t = \frac{15}{4}$, so find $K = 225$ feet.
The is also the height of the cliff since it is the value of the position function $s(t) = -16t^2 + 225$ when $t =0$

P3.
a. $\int \left[ \frac{1}{2\sqrt{1-x^2}} -\frac{3}{1+x^2} \right] dx = \frac{\arcsin x}{2} - 3\arctan x + C$

b. $\int \frac{\sec x + \cos x}{2 \cos x} dx = \int \left(\frac{1}{2}\sec^2x+\frac{1}{2} \right) dx = \frac{1}{2} \tan x + \frac{1}{2} x + C$

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