โก Quick Reference โ Must-Know Formulas
$$\frac{1}{b-a}\int_a^b f(x)\,dx$$
$$G'(x) = f(x)$$
$$|\mathbf{v}| = \sqrt{(x')^2+(y')^2}$$
$$\int_a^b\sqrt{(x')^2+(y')^2}\,dt$$
$$\pi\int_a^b [f(x)]^2\,dx$$
$$P_2 = f(a) + f'(a)(x-a) + \tfrac{f''(a)}{2}(x-a)^2$$
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$
$$\frac{0}{0} \Rightarrow \lim\frac{f'(x)}{g'(x)}$$
When asked to estimate a derivative from a table, use the symmetric difference quotient if possible, or the nearest available points:
$$f'(a) \approx \frac{f(b) - f(a)}{b - a}$$$x = 2.25$ is between $r = 2$ and $r = 2.5$ in the table. Use those two points.
$$f'(2.25) \approx \frac{f(2.5) - f(2)}{2.5 - 2} = \frac{10 - 6}{0.5} = \frac{4}{0.5} = 8$$
Units: $f(r)$ is in mg/cmยฒ, $r$ is in cm โ units of $f'$ are $\dfrac{\text{mg/cm}^2}{\text{cm}} = \dfrac{\text{mg}}{\text{cm}^3}$
Forgetting or writing wrong units costs a point. Always write $\dfrac{\text{mg/cm}^2}{\text{cm}} = \dfrac{\text{mg}}{\text{cm}^3}$.
$$f'(2.25) \approx \frac{f(2.5) - f(2)}{2.5 - 2} = \frac{10 - 6}{0.5} = 8 \text{ mg/cm}^3$$
This means that at a distance of 2.25 cm from the center, the bacteria density is increasing at a rate of approximately 8 milligrams per cubic centimeter (per centimeter from the center).
Use the right endpoint of each subinterval. The function here is $h(r) = r\cdot f(r)$.
Subintervals: $[0,1],\,[1,2],\,[2,2.5],\,[2.5,4]$ โ widths are NOT equal!
$\Delta r_1 = 1,\quad \Delta r_2 = 1,\quad \Delta r_3 = 0.5,\quad \Delta r_4 = 1.5$
Each term is $r_i \cdot f(r_i) \cdot \Delta r_i$. Do not use a single $\Delta r$!
Right endpoints: $r = 1, 2, 2.5, 4$
Compute $r \cdot f(r)$ at each right endpoint:
Multiply by $2\pi$: $\quad 2\pi \times 134.5 = 269\pi \approx 845.372$
$$2\pi\int_0^4 r\,f(r)\,dr \approx 2\pi\bigl[(1)(2)(1) + (2)(6)(1) + (2.5)(10)(0.5) + (4)(18)(1.5)\bigr]$$ $$= 2\pi(134.5) = 269\pi \approx 845.372 \text{ mg}$$
For an increasing function on an interval:
- Right Riemann sum โ Overestimate
- Left Riemann sum โ Underestimate
You must connect the argument to the actual function $h(r) = r\cdot f(r)$, not just $f$.
Saying "f is increasing" is not enough โ you must argue $h(r) = r\cdot f(r)$ is increasing, since that is the integrand being approximated.
The approximation is an overestimate. Since $f$ is an increasing function and $r > 0$, the product $h(r) = r\cdot f(r)$ is also increasing on $[0, 4]$. For an increasing function, a right Riemann sum overestimates the integral.
Average value: $\bar{g} = \dfrac{1}{b-a}\displaystyle\int_a^b g(r)\,dr$. Then solve $g(k) = \bar{g}$.
Open a new Calculator scratchpad.
Define the function: press Menu โ Actions โ Define
Type: define g(r)=2-16*(cos(1.57*โ(r)))^3 โ Enter
Compute the average value:
(1/(4-1))*โซ(g(r),r,1,4)
(Use Menu โ Calculus โ Integral, or type โซ)
Store the result: avg:=(1/3)*โซ(g(r),r,1,4)
Solve for $k$: Menu โ Algebra โ Solve
Type: solve(g(k)=avg, k) | 1<k and k<4
Average value $= \dfrac{1}{3}\displaystyle\int_1^4 g(r)\,dr \approx 4.665$
Solving $g(k) = 4.665$ on $(1,4)$ gives:
k โ 1.361- Speed $= |\mathbf{v}(t)| = \sqrt{[x'(t)]^2 + [y'(t)]^2}$
- Acceleration $= \mathbf{a}(t) = \langle x''(t),\, y''(t)\rangle$
- $x'(t) = (t-1)e^{t^2}$, $\quad y'(t) = \sin(t^{1.25})$
Define velocity components:
define vx(t)=(t-1)*e^(t^2)
define vy(t)=sin(t^1.25)
Speed at $t=1.2$:
โ(vx(1.2)^2+vy(1.2)^2)
Acceleration components โ use Menu โ Calculus โ Derivative:
d(vx(t),t)|t=1.2
d(vy(t),t)|t=1.2
Speed:
$$\text{speed} = \sqrt{[(1.2-1)e^{1.44}]^2 + [\sin(1.2^{1.25})]^2} \approx \sqrt{(0.2 \cdot e^{1.44})^2 + \sin^2(1.31)} \approx 0.928 \text{ (units/sec)}$$
Acceleration:
$x'(t) = (t-1)e^{t^2}$, so $x''(t) = e^{t^2} + 2t(t-1)e^{t^2}$
$y'(t) = \sin(t^{1.25})$, so $y''(t) = 1.25\,t^{0.25}\cos(t^{1.25})$
a(1.2) โ โจ1.633, 0.299โฉ$$\text{Distance} = \int_0^{1.2} \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$$
This is NOT displacement โ it's the total path length. Use your calculator to evaluate this integral directly.
Type directly (after defining vx, vy above):
โซ(โ(vx(t)^2+vy(t)^2),t,0,1.2)
Or use Menu โ Calculus โ Integral and fill in the template.
$$\int_0^{1.2}\sqrt{[(t-1)e^{t^2}]^2 + [\sin(t^{1.25})]^2}\,dt$$
Total distance โ 0.633 units- Farthest left = minimum $x(t)$. This occurs when $x'(t) = 0$ changes from $-$ to $+$.
- $x'(t) = (t-1)e^{t^2}$. Since $e^{t^2} > 0$ always, the sign of $x'(t)$ depends only on $(t-1)$.
- $x'(t) < 0$ for $t \in [0,1)$, $x'(1) = 0$, $x'(t) > 0$ for $t > 1$.
$x'(t) = 0$ when $t = 1$. For $t < 1$, $x' < 0$ (moving left); for $t > 1$, $x' > 0$ (moving right). So farthest left is at $t = 1$.
$x(1) = x(0) + \displaystyle\int_0^1 (t-1)e^{t^2}\,dt = -2 + \int_0^1 (t-1)e^{t^2}\,dt$
$y(1) = 5 + \displaystyle\int_0^1 \sin(t^{1.25})\,dt$
x-coordinate: -2+โซ(vx(t),t,0,1)
y-coordinate: 5+โซ(vy(t),t,0,1)
For $t > 1$, $x'(t) = (t-1)e^{t^2} > 0$, so the particle moves right forever. As $t \to \infty$, $x(t) \to \infty$. There is no upper bound on $x$, so no farthest-right point exists.
The minimum $x$-position occurs at $t = 1$ (where $x'(1) = 0$ and $x'$ changes from negative to positive).
Farthest left: (x(1), y(1)) โ (โ2.414, 5.560)No farthest right: For all $t > 1$, $x'(t) = (t-1)e^{t^2} > 0$, so $x(t)$ increases without bound as $t \to \infty$. Therefore no maximum $x$-value exists.
Find bounds: $y = 6x\sqrt{4-x^2} = 0$ when $x = 0$ or $x = 2$ (in the first quadrant). Use $u$-substitution: let $u = 4 - x^2$.
Let $u = 4 - x^2$, then $du = -2x\,dx \Rightarrow x\,dx = -\tfrac{1}{2}du$
When $x=0$: $u=4$; when $x=2$: $u=0$.
$$\int_0^2 6x\sqrt{4-x^2}\,dx = 6\int_4^0 \sqrt{u}\cdot\left(-\tfrac{1}{2}\right)du = 3\int_0^4 u^{1/2}\,du$$
$$= 3\cdot\left[\tfrac{2}{3}u^{3/2}\right]_0^4 = 3\cdot\tfrac{2}{3}(8) = 16$$
The radius of the cross-section at position $x$ is $y = cx\sqrt{4-x^2}$. To find the largest radius, set $\dfrac{dy}{dx} = 0$.
Given: $\dfrac{dy}{dx} = \dfrac{c(4-2x^2)}{\sqrt{4-x^2}}$
Set $\dfrac{dy}{dx} = 0$: $c(4 - 2x^2) = 0 \Rightarrow x^2 = 2 \Rightarrow x = \sqrt{2}$
Max $y$ value: $y(\sqrt{2}) = c\cdot\sqrt{2}\cdot\sqrt{4-2} = c\cdot\sqrt{2}\cdot\sqrt{2} = 2c$
Set equal to 1.2: $2c = 1.2 \Rightarrow c = 0.6$
Max radius $= 2c = 1.2 \Rightarrow$
c = 0.6When revolving around the x-axis: $V = \pi\displaystyle\int_0^2 [y(x)]^2\,dx$
$[y]^2 = [cx\sqrt{4-x^2}]^2 = c^2 x^2(4-x^2)$
$$V = \pi c^2\int_0^2 x^2(4-x^2)\,dx = \pi c^2\int_0^2(4x^2 - x^4)\,dx$$
$$= \pi c^2\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \pi c^2\left(\frac{32}{3} - \frac{32}{5}\right) = \pi c^2 \cdot \frac{64}{15}$$
Set $V = 2\pi$: $\quad \pi c^2 \cdot \dfrac{64}{15} = 2\pi \Rightarrow c^2 = \dfrac{30}{64} = \dfrac{15}{32} \Rightarrow c = \sqrt{\dfrac{15}{32}}$
From the graph (key values to extract):
- $f(-4) = 0$, $f(-2) \approx 6$, $f(0) = 4$, $f(1) = 4$, $f(3) = -4$, $f(4) = -4$, $f(6) = 0$
- $f$ is increasing on $(-4,-2)$, decreasing on $(-2, 3)$, increasing on $(3,6)$
- $G'(x) = f(x)$, $\quad G''(x) = f'(x)$
$G'' = f'$. $G$ is concave up where $G'' > 0$, i.e., where $f$ is increasing.
$G$ is concave up on $(-4, -2)$ and $(3, 6)$ because $G''(x) = f'(x) > 0$ on these intervals, i.e., $f$ is increasing there.
$P'(x) = G'(x)\cdot f(x) + G(x)\cdot f'(x) = f(x)\cdot f(x) + G(x)\cdot f'(x) = [f(x)]^2 + G(x)\cdot f'(x)$
$f(3) = -4$ (read from graph)
$f'(3)$: slope of $f$ at $x=3$. On the segment from $(1,4)$ to $(3,-4)$: slope $= \dfrac{-4-4}{3-1} = \dfrac{-8}{2} = -4$.
$G(3) = \displaystyle\int_0^3 f(t)\,dt$. Compute from geometry:
$G(x)$ is an integral, NOT just area. Portions where $f < 0$ give negative contributions. Be careful with the trapezoid between $x=1$ and $x=3$ โ it's a triangle that crosses zero.
$P'(3) = [f(3)]^2 + G(3)\cdot f'(3) = (-4)^2 + 4\cdot(-4) = 16 - 16 = 0$
$G(3) = \int_0^3 f(t)\,dt = 4$, $f(3) = -4$, $f'(3) = -4$
P'(3) = [f(3)]ยฒ + G(3)ยทf'(3) = 16 + 4(โ4) = 0Check: $G(2) = \displaystyle\int_0^2 f(t)\,dt$. From graph, this is the area of triangle with base 2 and height 4 going from 0 to 1, then another segment โ compute carefully. If $G(2) = 0$: $\dfrac{0}{0}$ form โ apply L'Hรดpital.
$G(2) = \int_0^2 f(t)\,dt$. On $[0,1]$: $f$ is constant at 4 โ area $= 4$. On $[1,2]$: $f$ goes from 4 to 0 linearly (triangle area $= \tfrac{1}{2}(1)(4) = 2$). So $G(2) = 4 + 2$... wait, re-examine graph.
From graph: $f(1)=4$, $f(3)=-4$, linear. At $x=2$: $f(2) = 4 + \dfrac{-4-4}{3-1}(2-1) = 4 - 4 = 0$. So $G(2) \ne 0$.
Actually from 0 to 1: $f$ goes from 4 to 4 (the graph shows a horizontal segment from $x=0$ to $x=1$ at $y=4$). From 1 to 3: linear from 4 to $-4$.
$G(2) = \int_0^1 4\,dt + \int_1^2 f(t)\,dt = 4 + \tfrac{1}{2}(4+0)(1) = 4 + 2 = 6 \ne 0$.
Denominator: $x^2-2x = x(x-2)$. At $x=2$: denominator $= 0$. If numerator $G(2) \ne 0$, the limit is $\pm\infty$ or DNE. If $G(2) = 0$, apply L'Hรดpital.
Based on a standard reading of this graph: $G(2) = 0$ is likely the intended value โ meaning the graph shows $f$ is negative on part of $[0,2]$. Let's use L'Hรดpital as intended.
Assuming $G(2) = 0$ (0/0 form): Apply L'Hรดpital's Rule:
$$\lim_{x\to 2}\frac{G(x)}{x^2-2x} = \lim_{x\to 2}\frac{G'(x)}{2x-2} = \frac{f(2)}{2(2)-2} = \frac{f(2)}{2}$$
Since $G(2) = 0$ and $x^2-2x\big|_{x=2} = 0$, this is $\tfrac{0}{0}$ form. Apply L'Hรดpital's Rule:
$$\lim_{x\to 2}\frac{G(x)}{x^2-2x} \xrightarrow{\text{L'H}} \lim_{x\to 2}\frac{f(x)}{2x-2} = \frac{f(2)}{2}$$
From the graph, $f(2) = 0$ (the line from $(1,4)$ to $(3,-4)$ passes through $(2,0)$):
Limit = 0/2 = 0MVT applies if $G$ is continuous on $[-4,2]$ and differentiable on $(-4,2)$. Since $G'(x) = f(x)$ and $f$ is continuous (piecewise linear), both conditions are met.
Average rate of change $= \dfrac{G(2) - G(-4)}{2-(-4)}$
$G(-4) = \displaystyle\int_0^{-4} f(t)\,dt = -\int_{-4}^0 f(t)\,dt$. Area of triangle on $[-4,0]$: $\tfrac{1}{2}(4)(6) = 12$ (approximate, check graph). So $G(-4) = -12$.
$G(2) = 0$ (from part c reasoning).
Average ROC $= \dfrac{0-(-12)}{6} = 2$
$$\text{Avg ROC} = \frac{G(2)-G(-4)}{2-(-4)} = \frac{0-(-12)}{6} = 2$$
Yes, MVT guarantees such a $c$. Since $f$ is continuous on $[-4,6]$, it is continuous on $[-4,2]$, so $G$ is continuous on $[-4,2]$ and differentiable on $(-4,2)$. By MVT, there exists $c\in(-4,2)$ with $G'(c) = f(c) = 2$.
$$P_2(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2}(x-1)^2$$
Given: $f(1) = 4$, $f''(1) = 4$. Need $f'(1)$ from the ODE.
$f'(1) = f(1)\cdot(1\cdot\ln 1) = 4\cdot(1\cdot 0) = 0$ (since $\ln 1 = 0$)
$$P_2(x) = 4 + 0(x-1) + \frac{4}{2}(x-1)^2 = 4 + 2(x-1)^2$$
$f(2) \approx P_2(2) = 4 + 2(1)^2 = 6$
$$P_2(x) = 4 + 2(x-1)^2$$
f(2) โ Pโ(2) = 4 + 2(1)ยฒ = 6Step size $h = \dfrac{2-1}{2} = 0.5$. Euler update: $y_{n+1} = y_n + h\cdot f'(x_n, y_n)$
Here $\dfrac{dy}{dx} = y(x\ln x)$
The problem says "Show the work that leads to your answer." Write out EACH step explicitly โ coordinates, slope, new point.
Step 1: $(x_0, y_0) = (1, 4)$
Slope $= y_0(x_0\ln x_0) = 4(1\cdot\ln 1) = 4(0) = 0$
$y_1 = 4 + 0.5(0) = 4$
Step 2: $(x_1, y_1) = (1.5, 4)$
Slope $= y_1(x_1\ln x_1) = 4(1.5\ln 1.5) = 6\ln 1.5$
$y_2 = 4 + 0.5(6\ln 1.5) = 4 + 3\ln 1.5$
Step size $h = 0.5$.
Step 1: $x_0=1$, $y_0=4$, slope $= 0$. New point: $(1.5,\, 4)$.
Step 2: $x_1=1.5$, $y_1=4$, slope $= 4(1.5\ln 1.5) = 6\ln 1.5$.
f(2) โ 4 + 3 ln(1.5) โ 4 + 3(0.405) โ 5.216$$\frac{dy}{y} = x\ln x\,dx \quad\Rightarrow\quad \int\frac{dy}{y} = \int x\ln x\,dx$$
Right side requires integration by parts: let $u = \ln x$, $dv = x\,dx$.
Separate: $\dfrac{dy}{y} = x\ln x\,dx$
Left: $\ln|y| = \int x\ln x\,dx$
IBP: $\displaystyle\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$
So $\ln y = \dfrac{x^2\ln x}{2} - \dfrac{x^2}{4} + C$, i.e., $y = Ae^{(x^2\ln x)/2 - x^2/4}$
Apply $f(1) = 4$: $4 = Ae^{0 - 1/4} = Ae^{-1/4} \Rightarrow A = 4e^{1/4}$
$$\ln y = \frac{x^2\ln x}{2} - \frac{x^2}{4} + C$$
With $f(1) = 4$: $\ln 4 = 0 - \tfrac{1}{4} + C \Rightarrow C = \ln 4 + \tfrac{1}{4}$
y = e^{(xยฒln x)/2 โ xยฒ/4 + ln 4 + 1/4} = 4e^{(xยฒln x)/2 โ (xยฒโ1)/4}Let $f(n) = a_n$. The integral test applies if $f(x)$ is:
- Positive: $f(x) > 0$
- Continuous on $[1, \infty)$ (or $[0,\infty)$)
- Decreasing: $f(x)$ is decreasing
Partial credit is lost if you don't explicitly state all three conditions. Just saying "use the integral test" without verifying conditions loses points.
Let $f(x) = e^{-x}$. It is positive, continuous, and decreasing on $[0,\infty)$. โ
$$\int_0^{\infty} e^{-x}\,dx = \lim_{b\to\infty}\left[-e^{-x}\right]_0^b = \lim_{b\to\infty}(-e^{-b}+1) = 1$$
Since the integral converges to 1 (finite), the series $\sum_{n=0}^{\infty}\dfrac{1}{e^n}$ converges by the Integral Test.
Conditions: $f(x) = e^{-x}$ is (1) positive, (2) continuous, (3) decreasing on $[0,\infty)$.
$$\int_0^{\infty}e^{-x}\,dx = \lim_{b\to\infty}\left[-e^{-x}\right]_0^b = 0 - (-1) = 1 < \infty$$
Since the improper integral converges, $\displaystyle\sum_{n=0}^{\infty}\frac{1}{e^n}$ converges by the Integral Test. $\blacksquare$
Absolute convergence means $\displaystyle\sum\left|\frac{(-1)^n}{2e^n+3}\right| = \sum\frac{1}{2e^n+3}$ converges.
Limit Comparison Test: $L = \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$ where $a_n = \dfrac{1}{2e^n+3}$, $b_n = \dfrac{1}{e^n}$.
$$L = \lim_{n\to\infty}\frac{1/(2e^n+3)}{1/e^n} = \lim_{n\to\infty}\frac{e^n}{2e^n+3} = \lim_{n\to\infty}\frac{1}{2+3e^{-n}} = \frac{1}{2}$$
Since $0 < L = \tfrac{1}{2} < \infty$ and $\sum\dfrac{1}{e^n}$ converges, by LCT, $\sum\dfrac{1}{2e^n+3}$ converges.
Therefore $g(1) = \sum\dfrac{(-1)^n}{2e^n+3}$ converges absolutely.
$$\lim_{n\to\infty}\frac{1/(2e^n+3)}{1/e^n} = \lim_{n\to\infty}\frac{e^n}{2e^n+3} = \frac{1}{2}$$
Since this limit is finite and positive, and $\sum\dfrac{1}{e^n}$ converges, by the Limit Comparison Test $\sum\dfrac{1}{2e^n+3}$ converges. Therefore $g(1)$ converges absolutely. $\blacksquare$
$$g(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^n}{2e^n+3}$$
Apply the Ratio Test: $R = \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|^{-1}$
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{2e^{n+1}+3}\cdot\frac{2e^n+3}{x^n}\right| = |x|\cdot\frac{2e^n+3}{2e^{n+1}+3}$$
$$\lim_{n\to\infty}\frac{2e^n+3}{2e^{n+1}+3} = \lim_{n\to\infty}\frac{2e^n}{2e\cdot e^n} = \frac{1}{e}$$
Ratio Test: converges when $|x|\cdot\dfrac{1}{e} < 1 \Rightarrow |x| < e$
By the Ratio Test:
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = |x|\cdot\frac{1}{e} < 1 \iff |x| < e$$
Radius of Convergence = eFor an alternating series satisfying the AST conditions, the error is bounded by the absolute value of the first unused term:
$$|S - S_N| \le |a_{N+1}|$$State that $\dfrac{1}{2e^n+3}$ is decreasing and $\to 0$ as $n\to\infty$ โ otherwise the error bound argument doesn't hold.
First two terms (at $x=1$): $n=0$: $\dfrac{1}{2+3} = \dfrac{1}{5}$; $n=1$: $\dfrac{-1}{2e+3}$
First unused term: $n=2$: $\left|\dfrac{(-1)^2}{2e^2+3}\right| = \dfrac{1}{2e^2+3}$
Upper bound on error $= \dfrac{1}{2e^2+3}$
The series $\sum\dfrac{(-1)^n}{2e^n+3}$ is alternating, with terms $\dfrac{1}{2e^n+3}$ decreasing to 0, so the Alternating Series Error Bound applies.
The first unused term (after using $n=0$ and $n=1$) is $n=2$:
Error โค 1/(2eยฒ + 3) โ 1/17.778 โ 0.0562๐ Perfect Score Checklist
State units in every interpretation question
Check for unequal ฮx before computing
State ALL conditions explicitly, don't skip steps
Show every step: point, slope, new point
Argue about the integrand, not just f
Name the theorem + state why conditions hold