| h (feet) | 0 | 2 | 5 | 10 |
|---|---|---|---|---|
| A(h) (sq ft) | 50.3 | 14.4 | 6.5 | 2.9 |
📐 Core Concept
Volume = \(\displaystyle\int_0^{10} A(h)\,dh\). A Riemann sum approximates this integral. A Left sum uses the left endpoint of each subinterval as the sample point. The three subintervals from the table are \([0,2]\), \([2,5]\), \([5,10]\).
⚡ Key Point — Left Riemann Sum Formula
For each subinterval \([a,b]\), the contribution is \(A(\text{left endpoint}) \times \Delta h\).
- Subinterval \([0,2]\): width \(= 2\), left endpoint \(h=0\) → \(A(0)=50.3\)
- Subinterval \([2,5]\): width \(= 3\), left endpoint \(h=2\) → \(A(2)=14.4\)
- Subinterval \([5,10]\): width \(= 5\), left endpoint \(h=5\) → \(A(5)=6.5\)
✅ Model Answer (a)
\(L = A(0)\cdot 2 + A(2)\cdot 3 + A(5)\cdot 5\)
\(= 50.3(2) + 14.4(3) + 6.5(5)\)
\(= 100.6 + 43.2 + 32.5\)
📐 Core Concept — Left Sum & Monotonicity
When a function is decreasing on an interval, the left endpoint gives the largest value in that interval. So the left Riemann sum rectangle is taller than the actual curve, producing an overestimate.
⚠️ Common Trap
Students say "overestimate" without linking it to the function being decreasing. You must state: "Because A is decreasing, left endpoints are larger than all other values in the subinterval, so each rectangle overestimates the actual area." — This reasoning is required for full credit.
✅ Model Answer (b)
The approximation is an overestimate.
Because A is continuous and decreasing, the value at the left endpoint of each subinterval is greater than or equal to A(h) for all h in that subinterval. Therefore each rectangular approximation lies above the graph of A, and the left Riemann sum overestimates the volume.
📐 Core Concept
Volume \(= \displaystyle\int_0^{10} f(h)\,dh\). This integral has no elementary antiderivative, so use the calculator.
🖩 TI-Nspire CX II — Numerical Integral
- Press menu → 4: Calculus → 3: Numerical Integral (nInt)
- OR type directly in Calculator app:
nInt(50.3/(e^(0.2x)+x), x, 0, 10) - Alternatively: press ∫ template, enter
50.3/(e^(0.2x)+x), lower=0, upper=10, variable=x - Press enter → Read the result
⚡ Make sure angle mode is Radian (irrelevant here, but good habit).
✅ Model Answer (c)
\(\displaystyle V = \int_0^{10} \frac{50.3}{e^{0.2h}+h}\,dh\)
📐 Core Concept — Chain Rule for Volume
The volume of water up to height h is \(\displaystyle V(h)=\int_0^h f(t)\,dt\). By the Fundamental Theorem of Calculus:
\(\dfrac{dV}{dh} = f(h)\)
Then by the chain rule: \(\dfrac{dV}{dt} = f(h)\cdot\dfrac{dh}{dt}\)
⚡ Key Point
- Compute \(f(5) = \dfrac{50.3}{e^{1}+5}\)
- Multiply by \(\dfrac{dh}{dt} = 0.26\)
🖩 TI-Nspire CX II — Evaluate f(5)
- In Calculator app type:
50.3/(e^(0.2·5)+5)→ press enter - Result ≈ 6.0 … then multiply by 0.26
✅ Model Answer (d)
\(\dfrac{dV}{dt} = f(5)\cdot\dfrac{dh}{dt}\)
\(f(5) = \dfrac{50.3}{e^{1}+5} \approx 6.0 \text{ sq ft}\)
\(\dfrac{dV}{dt} = 6.0 \times 0.26\)
📐 Core Concept — Polar Area Formula
Area bounded by \(r=f(\theta)\) between \(\theta=\alpha\) and \(\theta=\beta\):
Region R spans \(\theta\in[0, \pi/2]\) and is entirely bounded by \(r=f(\theta)\) and the x-axis (the origin).
⚠️ Critical Trap — Don't forget the ½ !
The polar area formula has a \(\frac{1}{2}\) factor. Forgetting it is the single most common polar area error.
🖩 TI-Nspire CX II
- Calculator app:
nInt(0.5*(1+sin(x)*cos(2x))^2, x, 0, π/2) - Or use Integral template: \(\frac{1}{2}\int_0^{\pi/2}(1+\sin\theta\cos2\theta)^2\,d\theta\)
- Make sure mode is Radian: menu → Settings → Angle: Radian
✅ Model Answer (a)
\(\displaystyle A_R = \frac{1}{2}\int_0^{\pi/2}\bigl[1+\sin\theta\cos(2\theta)\bigr]^2\,d\theta\)
📐 Core Concept
First, find the intersection of \(f(\theta)\) and \(g(\theta)\) in \((0,\pi/2)\) — call it \(\theta^*\). Region S is bounded by \(f\) from \(0\) to \(\theta^*\), and then by \(g\) from \(\theta^*\) to \(\pi/2\) (or you express S as the total area between the curves + piece by piece). The ray \(\theta=k\) splits S into two pieces of equal area.
⚡ Strategy — Setting Up the Equation
- Find \(\theta^*\): solve \(f(\theta)=g(\theta)\) numerically → \(\theta^*\approx 1.1731\)
- Area of S = \(\frac{1}{2}\int_0^{\theta^*}[f]^2\,d\theta + \frac{1}{2}\int_{\theta^*}^{\pi/2}[g]^2\,d\theta\)
- Left piece up to k = Half of S
🖩 TI-Nspire CX II — Find intersection θ*
- In Graphs app: enter \(f(\theta)=1+\sin\theta\cos(2\theta)\) and \(g(\theta)=2\cos\theta\) in polar mode
- Or in Calculator:
solve(1+sin(x)*cos(2x)=2*cos(x), x) | 0<x<π/2 - \(\theta^*\approx 1.1731\) rad
✅ Model Answer (b)
Let \(\theta^*\approx 1.1731\) be where \(f(\theta)=g(\theta)\) in \((0,\pi/2)\). Then:
Area of \(S = \frac{1}{2}\displaystyle\int_0^{\theta^*}[f(\theta)]^2\,d\theta + \frac{1}{2}\int_{\theta^*}^{\pi/2}[g(\theta)]^2\,d\theta\)
The equation for \(k\) (where \(0<k<\theta^*\)):
📐 Core Concept — Distance in Polar Coordinates
Both points are on the same ray \(\theta\), so the distance between \((f(\theta),\theta)\) and \((g(\theta),\theta)\) is simply:
Average value: \(\displaystyle w_A = \frac{1}{\pi/2-0}\int_0^{\pi/2}w(\theta)\,d\theta\)
⚠️ Trap — Sign / Absolute Value
Check which curve is larger. For most of \([0,\pi/2]\), \(g\ge f\), but near \(\theta=\pi/2\) this can flip. Use absolute value or check intersection to avoid negative distances.
🖩 TI-Nspire CX II — Average Value
nInt(abs(2cos(x)-(1+sin(x)*cos(2x))), x, 0, π/2) / (π/2)
✅ Model Answer (c)
\(w(\theta) = |2\cos\theta - 1 - \sin\theta\cos(2\theta)|\)
\(\displaystyle w_A = \frac{2}{\pi}\int_0^{\pi/2}w(\theta)\,d\theta\)
⚡ Strategy
- Solve \(w(\theta)=w_A\approx 0.4338\) numerically using calculator
- To determine increasing/decreasing: check \(w'(\theta)\) at that value (numerically)
- Or observe the graph of \(w(\theta)\) to see the slope direction
🖩 TI-Nspire CX II
- In Graphs app, graph \(y = 2\cos(x)-(1+\sin(x)\cos(2x))\) and \(y=w_A\approx 0.4338\)
- Find intersection: menu → 6: Analyze Graph → 4: Intersection
- For derivative:
nDeriv(2cos(x)-(1+sin(x)*cos(2x)), x, θ_value)
✅ Model Answer (d)
Solving \(w(\theta)=0.4338\) on \([0,\pi/2]\) gives \(\theta\approx 0.9058\).
Checking \(w'(0.9058)\): since \(w'(\theta)<0\) at this point, the function is decreasing.
📐 Core Concept — FTC Part 2 (Net Change)
We know \(f(-2)=7\). Use this as the anchor and integrate \(f'\) (geometrically — count areas under the graph of \(f'\)) to reach \(-6\) and \(5\).
⚡ Geometric Areas from Graph of f′
- \(\displaystyle\int_{-6}^{-2}f'(x)\,dx\): Line segment from \((-6,2)\) to \((-1,0)\) → trapezoid area. From the graph: area \(= \frac{1}{2}(2+0)\cdot 4 = 4\) (triangle) and the segment from \(-2\) to \(-1\) contributes \(\frac{1}{2}(1)(?)…\) — read carefully from the graph. The segment goes from \((-6,2)\) to \((-1,0)\), so over \([-6,-2]\) (width 4): area ≈ \(\frac{1}{2}(2+\frac{2}{5})\cdot 4\). Use exact geometry from the graph.
- Semicircle on \([-1,1]\): area = \(\frac{1}{2}\pi(1)^2 = \frac{\pi}{2}\) (below x-axis → negative)
- Line from \((1,0)\) to \((3,2)\), then \((3,2)\) to \((5,0)\): triangles
⚠️ Critical — Signed Areas!
Areas where \(f'<0\) (below x-axis) are subtracted. The semicircle is below the x-axis → its contribution is \(-\frac{\pi}{2}\). Don't add it as a positive area!
✅ Model Answer (a)
For f(−6): Using \(f(-2)=f(-6)+\int_{-6}^{-2}f'(x)\,dx\)
From graph: \(\int_{-6}^{-2}f'(x)\,dx\). The graph of \(f'\) is linear from \((-6,2)\) to \((-1,0)\). Over \([-6,-2]\) this is a trapezoid: \(\frac{1}{2}(f'(-6)+f'(-2))\cdot 4 = \frac{1}{2}(2+\frac{2}{5})\cdot 4\).
More precisely from the graph: slope = \(\frac{0-2}{-1-(-6)}=\frac{-2}{5}\), so \(f'(-2)=2+\frac{-2}{5}(4)=2-\frac{8}{5}=\frac{2}{5}\). Trapezoid area \(=\frac{1}{2}(2+\frac{2}{5})\cdot 4 = \frac{1}{2}\cdot\frac{12}{5}\cdot 4 = \frac{24}{5}\).
\(7 = f(-6) + \frac{24}{5}\) → \(f(-6) = 7 - \frac{24}{5} = \frac{11}{5}\)
For f(5): \(f(5)=f(-2)+\int_{-2}^{5}f'(x)\,dx\)
Piece \([-2,-1]\): triangle, area \(=\frac{1}{2}\cdot 1\cdot\frac{2}{5}=\frac{1}{5}\) (positive)
Piece \([-1,1]\): semicircle below x-axis, area \(=-\frac{\pi}{2}\)
Piece \([1,3]\): triangle above x-axis, area \(=\frac{1}{2}(2)(2)=2\)
Piece \([3,5]\): triangle above x-axis, area \(=\frac{1}{2}(2)(2)=2\)
\(\int_{-2}^{5}f'(x)\,dx = \frac{1}{5} - \frac{\pi}{2} + 2 + 2 = \frac{21}{5}-\frac{\pi}{2}\)
📐 Core Concept
\(f\) is increasing where \(f'(x)>0\). Read directly from the graph of \(f'\).
✅ Model Answer (b)
From the graph, \(f'(x)>0\) on \((-6,-1)\) and \((1,5)\).
Therefore \(f\) is increasing on \(\mathbf{[-6,-1]}\) and \(\mathbf{[1,5]}\).
📐 Core Concept — Closed Interval Method
Evaluate f at all critical points (where \(f'=0\) or undefined) inside the interval AND at the endpoints. The smallest value is the absolute minimum.
⚡ Critical Points from Graph of f′
- \(f'(-1)=0\): local max (f increases before, decreases after) → f has a local max
- \(f'(1)=0\): local min (f decreases before, increases after) → f has a local min
- Endpoints: \(f(-6)\) and \(f(5)\)
✅ Model Answer (c)
Candidates: \(f(-6)=\frac{11}{5}\), \(f(-1)\), \(f(1)\), \(f(5)\)
\(f(-1) = f(-2) + \int_{-2}^{-1}f'(x)\,dx = 7 + \frac{1}{5} = \frac{36}{5}\)
\(f(1) = f(-1) + \int_{-1}^{1}f'(x)\,dx = \frac{36}{5} - \frac{\pi}{2} \approx 5.629\)
Comparing: \(f(-6)\approx 2.2\), \(f(1)\approx 5.629\), \(f(5)\approx 10.633\), \(f(-1)=7.2\)
📐 Core Concept
\(f''(x)\) is the slope (derivative) of \(f'(x)\). Read the slope of the graph of \(f'\) at those points.
⚠️ When f ′′ Does NOT Exist
If the graph of \(f'\) has a sharp corner at a point, then \(f'\) is not differentiable there → \(f''(x)\) does not exist. At \(x=3\), the graph of \(f'\) has a corner (the line segments from left and right have different slopes), so \(f''(3)\) does not exist.
✅ Model Answer (d)
f ′′(−5): The graph of \(f'\) is a straight line on \((-6,-1)\). Slope \(= \frac{0-2}{-1-(-6)}=\frac{-2}{5}\).
\(f''(-5) = -\dfrac{2}{5}\)
f ′′(3): The graph of \(f'\) has a corner at \(x=3\) (peak of two line segments with different slopes). The left-side slope and right-side slope of \(f'\) are different, so \(f'\) is not differentiable at \(x=3\).
📐 Core Concept — Linearization
The tangent line at \((0, H(0))\) is \(L(t) = H(0) + H'(0)\cdot t\).
Compute \(H'(0)\) by plugging into the differential equation.
✅ Model Answer (a)
\(H'(0) = -\frac{1}{4}(H(0)-27) = -\frac{1}{4}(91-27) = -\frac{1}{4}(64) = -16\)
Tangent line: \(L(t) = 91 - 16t\)
Approximation: \(L(3) = 91 - 16(3) = 91 - 48 = 43\)
📐 Core Concept — Concavity & Tangent Line
If \(H'' > 0\) (concave up), the graph lies above the tangent line → tangent line approximation is an underestimate. If \(H'' < 0\), it's an overestimate.
⚡ Compute H″
Differentiate \(\frac{dH}{dt} = -\frac{1}{4}(H-27)\) with respect to t:
\(\dfrac{d^2H}{dt^2} = -\dfrac{1}{4}\cdot\dfrac{dH}{dt} = -\dfrac{1}{4}\cdot\left(-\dfrac{1}{4}(H-27)\right) = \dfrac{1}{16}(H-27)\)
Since \(H(t) > 27\) for all \(t > 0\), we have \(H'' > 0\) → concave up.
✅ Model Answer (b)
\(\dfrac{d^2H}{dt^2} = \dfrac{1}{16}(H-27) > 0\) for \(H > 27\).
Since H is concave up, the graph of H lies above its tangent line.
📐 Core Concept — Separable ODE
The equation is \(\dfrac{dG}{dt} = -\dfrac{(G-27)^{3/2}}{64}\). Separate variables and integrate both sides.
⚡ Step-by-Step Solution
- 1 Separate: \(\dfrac{dG}{(G-27)^{3/2}} = -\dfrac{1}{64}\,dt\)
- 2 Integrate left: \(\displaystyle\int (G-27)^{-3/2}\,dG = \frac{(G-27)^{-1/2}}{-1/2} = -2(G-27)^{-1/2}\)
- 3 Integrate right: \(-\dfrac{t}{64} + C\)
- 4 Equation: \(-2(G-27)^{-1/2} = -\dfrac{t}{64} + C\)
- 5 Apply IC \(G(0)=91\): \(-2(64)^{-1/2}=-2\cdot\frac{1}{8}=-\frac{1}{4}\), so \(C=-\frac{1}{4}\)
- 6 Solve for G: \(-2(G-27)^{-1/2}=-\frac{t}{64}-\frac{1}{4}=-\frac{t+16}{64}\)
- 7 \((G-27)^{-1/2}=\dfrac{t+16}{128}\), so \((G-27)^{1/2}=\dfrac{128}{t+16}\)
- 8 \(G(t) = 27+\left(\dfrac{128}{t+16}\right)^2\)
⚠️ Don't Drop the Constant C!
Every separable ODE integration must include +C. Apply the initial condition immediately after integrating, before solving for G.
✅ Model Answer (c)
\(\displaystyle G(t) = 27 + \left(\frac{128}{t+16}\right)^2\)
\(G(3) = 27 + \left(\dfrac{128}{19}\right)^2 = 27 + \dfrac{16384}{361}\)
✅ Model Answer (a)
\(f(x) = \dfrac{3}{2x^2-7x+5}\)
\(f'(x) = \dfrac{-3(4x-7)}{(2x^2-7x+5)^2}\)
\(f'(3) = \dfrac{-3(12-7)}{(18-21+5)^2} = \dfrac{-15}{(2)^2} = \dfrac{-15}{4}\)
📐 Core Concept
Critical points: where \(f'(x)=0\) or undefined (but inside domain). Use the partial fractions form to differentiate more easily, or use the quotient rule result.
Using \(f'(x)=0\): numerator of \(f'\) = \(-3(4x-7)=0\) → \(x=\frac{7}{4}=1.75\). Check this is in \((1, 2.5)\): yes.
⚡ First or Second Derivative Test
At \(x=\frac{7}{4}\): check sign of \(f'\) on each side. For \(x<\frac{7}{4}\): \(4x-7<0\) so \(f'>0\). For \(x>\frac{7}{4}\): \(4x-7>0\) so \(f'<0\). Sign changes \(+\to-\) → relative maximum.
✅ Model Answer (b)
Setting \(f'(x)=0\): \(-3(4x-7)=0\) → \(x=\frac{7}{4}\)
\(f'>0\) for \(x<\frac{7}{4}\) and \(f'<0\) for \(x>\frac{7}{4}\)
📐 Core Concept — Improper Integral & Partial Fractions
Given: \(f(x) = \dfrac{1}{x-1}-\dfrac{2}{2x-5}\)
\(\displaystyle\int f(x)\,dx = \ln|x-1| - \ln|2x-5| + C = \ln\left|\frac{x-1}{2x-5}\right| + C\)
Evaluate as a limit: \(\displaystyle\lim_{b\to\infty}\left[\ln\left(\frac{x-1}{2x-5}\right)\right]_5^b\)
⚡ Key Limit
As \(x\to\infty\): \(\dfrac{x-1}{2x-5}\to\dfrac{1}{2}\). So the limit = \(\ln\tfrac{1}{2}\).
At \(x=5\): \(\ln\dfrac{4}{5}\).
✅ Model Answer (c)
\(\displaystyle\int_5^\infty f(x)\,dx = \lim_{b\to\infty}\left[\ln\left(\frac{x-1}{2x-5}\right)\right]_5^b\)
\(= \ln\frac{1}{2} - \ln\frac{4}{5} = \ln\frac{1}{2}\cdot\frac{5}{4} = \ln\frac{5}{8}\)
📐 Core Concept — Integral Test
Since \(f(x)\) is continuous, positive, and decreasing on \([5,\infty)\), the Integral Test applies: the series converges iff \(\int_5^\infty f(x)\,dx\) converges.
⚡ State ALL Integral Test Conditions
- \(f(x)\) is continuous on \([5,\infty)\) ✓ (no poles for \(x\ge 5\))
- \(f(x)\) is positive on \([5,\infty)\) ✓
- \(f(x)\) is decreasing on \([5,\infty)\) ✓ (since \(f'<0\) there)
You MUST state all three conditions for full credit.
✅ Model Answer (d)
Since \(f(x)\) is continuous, positive, and decreasing on \([5,\infty)\), the Integral Test applies.
From part (c), \(\displaystyle\int_5^\infty f(x)\,dx = \ln\tfrac{5}{8}\), which converges.
📐 Core Concept — Maclaurin Series
Plug in the given derivative values at \(x=0\).
⚡ Derivative Values (from given conditions)
- \(f(0)=0\), \(f'(0)=1\), \(f''(0)=(-1)^2\cdot 1!=1\)? Check: \((-1)^{n+1}(n-1)!\) at \(n=2\): \((-1)^3\cdot 1!=-1\). At \(n=3\): \((-1)^4\cdot 2!=2\). At \(n=4\): \((-1)^5\cdot 3!=-6\).
✅ Model Answer (a)
Term for \(n=1\): \(\dfrac{f'(0)}{1!}x = x\)
Term for \(n=2\): \(\dfrac{f''(0)}{2!}x^2 = \dfrac{-1}{2}x^2 = -\dfrac{x^2}{2}\)
Term for \(n=3\): \(\dfrac{2}{6}x^3 = \dfrac{x^3}{3}\)
Term for \(n=4\): \(\dfrac{-6}{24}x^4 = -\dfrac{x^4}{4}\)
First four terms: \(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots\)
📐 Core Concept — Absolute vs Conditional Convergence
At \(x=1\): series = \(\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\)
Absolute convergence test: \(\displaystyle\sum\left|\frac{(-1)^{n+1}}{n}\right|=\sum\frac{1}{n}\) — this is the harmonic series, which diverges.
Conditional convergence: The alternating series \(\sum\frac{(-1)^{n+1}}{n}\) satisfies AST conditions (terms decrease to 0) → converges.
✅ Model Answer (b)
At \(x=1\): the series becomes \(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\).
The series of absolute values is \(\sum\frac{1}{n}\) (harmonic series), which diverges — so the series is not absolutely convergent.
By the Alternating Series Test (terms \(\frac{1}{n}\to 0\) and are decreasing), the alternating series converges.
📐 Core Concept — Integrate Term-by-Term
If \(f(x)=\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\), then integrate from 0 to x term by term:
\(g(x)=\displaystyle\int_0^x f(t)\,dt = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\cdot\frac{x^{n+1}}{n+1}\)
✅ Model Answer (c)
\(g(x) = \dfrac{x^2}{1\cdot 2} - \dfrac{x^3}{2\cdot 3} + \dfrac{x^4}{3\cdot 4} - \dfrac{x^5}{4\cdot 5}+\cdots\)
\(= \dfrac{x^2}{2} - \dfrac{x^3}{6} + \dfrac{x^4}{12} - \dfrac{x^5}{20}+\cdots\)
📐 Core Concept — Alternating Series Error Bound
For an alternating series satisfying AST conditions, the error from using the first \(n\) terms is bounded by the absolute value of the first omitted term:
⚡ Strategy
\(P_4\) is the degree-4 Taylor polynomial for g. The first omitted term in the series for g is the \(n=4\) term (degree 5 term):
First omitted term at \(x=\frac{1}{2}\): \(\dfrac{(1/2)^5}{4\cdot 5} = \dfrac{1/32}{20} = \dfrac{1}{640}\)
Since \(\dfrac{1}{640} < \dfrac{1}{500}\), the bound is established. ✓
⚠️ State AST Conditions Explicitly
You must verify that the series is alternating, the terms decrease in absolute value, and the terms approach 0. Don't just cite the bound without verifying conditions.
✅ Model Answer (d)
The series for \(g(x)\) is an alternating series with terms decreasing in magnitude to 0 on \([-1,1]\). By the Alternating Series Error Bound, the error is bounded by the absolute value of the first omitted term.
The \(P_4\) polynomial captures terms through degree 4 (i.e., \(n=1,2,3\) in the general term since max degree is 4). The first omitted term corresponds to degree 5:
\(\left|a_{\text{next}}\right| = \dfrac{(1/2)^5}{4\cdot 5} = \dfrac{1}{640}\)
Since \(\dfrac{1}{640} < \dfrac{1}{500}\):
nInt(expr, x, a, b)
Or: menu → 4:Calculus → 3:Numerical Integral
nDeriv(expr, x, value)
Or: menu → 4:Calculus → 2:Numerical Derivative
solve(eq, x) | a < x < b
menu → 3:Algebra → 1:Solve
Graphs app → menu → 3:Graph Type → 2:Polar
Set angle mode to Radian!
menu → 4:Settings → Angle: Radian
Always check this before any trig integration!
Graphs app → menu → 6:Analyze Graph → 4:Intersection
Drag boundaries around intersection point
Always include units — cubic feet, sq ft/min, etc.
Justify every answer — state the theorem or test used
State all conditions for Integral Test, AST, etc.
Don't round mid-problem — keep decimals full until final answer
3 decimal places — AP rounds to 3 decimal places for numerical answers
Check IVP solutions — verify your ODE solution satisfies the initial condition