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Exam Cram Sheet842 words

Exam Cram Sheet: Calculus I Integration Mastery

Integration

Exam Cram Sheet: Calculus I Integration Mastery

This guide provides an ultra-condensed review of the fundamental theorems, techniques, and applications of integration required for the Calculus I curriculum.

Topic Weighting

Based on typical Calculus I curricula, expect the following distribution in a final exam:

TopicEstimated WeightComplexity
The Fundamental Theorem of Calculus25%Medium
Integration Techniques (Substitution)30%High
Area & Volume Applications30%High
Riemann Sums & Definitions10%Medium
Physics & Physical Applications5%Low-Medium

Key Concepts Summary

1. The Fundamental Theorem of Calculus (FTC)

  • Part 1 (The Derivative of an Integral): ddx∫axf(t)dt=f(x)\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)dxd​∫ax​f(t)dt=f(x). This shows that differentiation and integration are inverse processes.
  • Part 2 (Evaluation Theorem): ∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x) dx = F(b) - F(a)∫ab​f(x)dx=F(b)−F(a), where F′(x)=f(x)F'(x) = f(x)F′(x)=f(x).

2. uuu-Substitution

Used when the integrand contains a function and its derivative (Chain Rule in reverse).

  • Choose u=g(x)u = g(x)u=g(x) such that du=g′(x)dxdu = g'(x)dxdu=g′(x)dx simplifies the integral.
  • Crucial: Change the limits of integration for definite integrals immediately after choosing uuu.

3. Visualizing Volume Methods

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Common Pitfalls

  • The Missing +C+C+C: Always include the constant of integration for indefinite integrals.
  • **Bounds in u−Sub:∗∗Forgettingtoconvertthexu-Sub:** Forgetting to convert the xu−Sub:∗∗Forgettingtoconvertthex-bounds to uuu-bounds. If u=x2u=x^2u=x2 and x∈[1,2]x \in [1,2]x∈[1,2], then u∈[1,4]u \in [1,4]u∈[1,4].
  • Washer Order: Subtracting (rinner−router)2(r_{inner} - r_{outer})^2(rinner​−router​)2 instead of (router2−rinner2)(r_{outer}^2 - r_{inner}^2)(router2​−rinner2​).
  • Variable Mismatch: Integrating with respect to xwhenthefunctionisdefinedintermsofyx when the function is defined in terms of yxwhenthefunctionisdefinedintermsofy (especially in area between curves).
  • Average Value vs. Net Change: Average value requires dividing by the interval: 1b−a∫abf(x)dx\frac{1}{b-a} \int_a^b f(x) dxb−a1​∫ab​f(x)dx. Net change is just ∫abf′(x)dx\int_a^b f'(x) dx∫ab​f′(x)dx.

Mnemonics / Memory Triggers

[!TIP] "S-C-U-B-A" for uuu-Substitution

  1. Select uuu.
  2. Calculate dududu.
  3. Update bounds (for definite integrals).
  4. Back-substitute uuu and dududu into the integral.
  5. Antidifferentiate.
  • **Integral of 1x:∗∗"Logisthenaturalchoicefortheinverseofx\frac{1}{x}:** "Log is the natural choice for the inverse of xx1​:∗∗"Logisthenaturalchoicefortheinverseofx." (Remember: ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C∫x1​dx=ln∣x∣+C).
  • Disk vs. Shell: "Disks are Dead-on" (axis is perpendicular to the radius), while "Shells are Shadows" (parallel to the axis).

Formula / Equation Sheet

Basic Integrals

FunctionIntegral ∫f(x)dx\int f(x) dx∫f(x)dxNote
xnx^nxnxn+1n+1+C\frac{x^{n+1}}{n+1} + Cn+1xn+1​+Cn≠−1n \neq -1n=−1
exe^xexex+Ce^x + Cex+CThe "Do Nothing" Integral
1x\frac{1}{x}x1​$\lnx
sin⁡(x)\sin(x)sin(x)−cos⁡(x)+C-\cos(x) + C−cos(x)+CDerivative of CCC is negative
sec⁡2(x)\sec^2(x)sec2(x)tan⁡(x)+C\tan(x) + Ctan(x)+CStandard Trig
11−x2\frac{1}{\sqrt{1-x^2}}1−x2​1​arcsin⁡(x)+C\arcsin(x) + Carcsin(x)+CInverse Trig
11+x2\frac{1}{1+x^2}1+x21​arctan⁡(x)+C\arctan(x) + Carctan(x)+CVery common in exams

Applications

  • Arc Length: L=∫ab1+[f′(x)]2dxL = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dxL=∫ab​1+[f′(x)]2​dx
  • Work: W=∫abF(x)dxW = \int_{a}^{b} F(x) dxW=∫ab​F(x)dx
  • Hydrostatic Force: F=∫abρg(depth)(width)dyF = \int_{a}^{b} \rho g (depth) (width) dyF=∫ab​ρg(depth)(width)dy
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Practice Set

  1. Fundamental Theorem: Find ddx∫3x2sin⁡(t)dt\frac{d}{dx} \int_{3}^{x^2} \sin(t) dtdxd​∫3x2​sin(t)dt.
    (Hint: Use the Chain Rule)
  2. uuu-Substitution: Evaluate ∫xx2+1dx\int x \sqrt{x^2 + 1} dx∫xx2+1​dx.
  3. Area Between Curves: Find the area bounded by y=x2y = x^2y=x2 and y=x+2y = x+2y=x+2.
  4. Volume by Disk: Rotate the region under y=xy = \sqrt{x}y=x​ from x=0x=0x=0 to x=4x=4x=4 around the xxx-axis.
  5. Net Change: A particle has velocity v(t)=3t2−6t.Findthetotaldistancetraveledbetweent=0v(t) = 3t^2 - 6t. Find the total distance traveled between t=0v(t)=3t2−6t.Findthetotaldistancetraveledbetweent=0 and t=3t=3t=3.
▶Click for Solutions
  1. Solution: 2x⋅sin⁡(x2)2x \cdot \sin(x^2)2x⋅sin(x2) (Substitute x2x^2x2 and multiply by its derivative).
  2. Solution: 13(x2+1)3/2+C\frac{1}{3}(x^2+1)^{3/2} + C31​(x2+1)3/2+C (Let u=x2+1u = x^2+1u=x2+1).
  3. Solution: ∫−12(x+2−x2)dx=4.5\int_{-1}^{2} (x+2 - x^2) dx = 4.5∫−12​(x+2−x2)dx=4.5.
  4. Solution: π∫04(x)2dx=π∫04xdx=8π\pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx = 8\piπ∫04​(x​)2dx=π∫04​xdx=8π.
  5. Solution: ∫02∣3t2−6t∣dt+∫23∣3t2−6t∣dt\int_0^2 |3t^2 - 6t| dt + \int_2^3 |3t^2 - 6t| dt∫02​∣3t2−6t∣dt+∫23​∣3t2−6t∣dt. Note the velocity changes sign at t=2t=2t=2.
All Calculus I: Single-Variable Differential Calculus Study Resources

Related Notes

  • Curriculum Overview: Integration Formulas and the Net Change Theorem785 words
  • Curriculum Overview: Mastering Antiderivatives685 words
  • Exam Cram: Application of Derivatives685 words
  • Exam Cram: Applications of Integration780 words
  • Curriculum Overview: Applied Optimization Problems745 words
  • Curriculum Overview: Approximating Areas685 words
  • A Preview of Calculus: Curriculum Overview685 words
  • Curriculum Overview: Arc Length of a Curve and Surface Area685 words
  • Curriculum Overview: Mastery of Areas between Curves685 words
  • Master Curriculum Overview: Basic Classes of Functions785 words
  • Calculus I: Single-Variable Differential Calculus — Curriculum Overview745 words
  • Briefing Doc: The Fundamentals and Applications of Integration685 words

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