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): ddxaxf(t)dt=f(x)\frac{d}{dx} \int_{a}^{x} 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), where F(x)=f(x)F'(x) = f(x).

2. uu-Substitution

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

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

3. Visualizing Volume Methods

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Common Pitfalls

  • The Missing +C+C: Always include the constant of integration for indefinite integrals.
  • **Bounds in uSub:Forgettingtoconvertthexu-Sub:** Forgetting to convert the x-bounds to uu-bounds. If u=x2u=x^2 and x[1,2]x \in [1,2], then u[1,4]u \in [1,4].
  • Washer Order: Subtracting (rinnerrouter)2(r_{inner} - r_{outer})^2 instead of (router2rinner2)(r_{outer}^2 - r_{inner}^2).
  • Variable Mismatch: Integrating with respect to xwhenthefunctionisdefinedintermsofyx when the function is defined in terms of y (especially in area between curves).
  • Average Value vs. Net Change: Average value requires dividing by the interval: 1baabf(x)dx\frac{1}{b-a} \int_a^b f(x) dx. Net change is just abf(x)dx\int_a^b f'(x) dx.

Mnemonics / Memory Triggers

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

  1. Select uu.
  2. Calculate dudu.
  3. Update bounds (for definite integrals).
  4. Back-substitute uu and dudu into the integral.
  5. Antidifferentiate.
  • **Integral of 1x:"Logisthenaturalchoicefortheinverseofx\frac{1}{x}:** "Log is the natural choice for the inverse of x." (Remember: 1xdx=lnx+C\int \frac{1}{x} 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) dxNote
xnx^nxn+1n+1+C\frac{x^{n+1}}{n+1} + Cn1n \neq -1
exe^xex+Ce^x + CThe "Do Nothing" Integral
1x\frac{1}{x}$\lnx
sin(x)\sin(x)cos(x)+C-\cos(x) + CDerivative of CC is negative
sec2(x)\sec^2(x)tan(x)+C\tan(x) + CStandard Trig
11x2\frac{1}{\sqrt{1-x^2}}arcsin(x)+C\arcsin(x) + CInverse Trig
11+x2\frac{1}{1+x^2}arctan(x)+C\arctan(x) + CVery common in exams

Applications

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

  1. Fundamental Theorem: Find ddx3x2sin(t)dt\frac{d}{dx} \int_{3}^{x^2} \sin(t) dt.
    (Hint: Use the Chain Rule)
  2. uu-Substitution: Evaluate xx2+1dx\int x \sqrt{x^2 + 1} dx.
  3. Area Between Curves: Find the area bounded by y=x2y = x^2 and y=x+2y = x+2.
  4. Volume by Disk: Rotate the region under y=xy = \sqrt{x} from x=0x=0 to x=4x=4 around the xx-axis.
  5. Net Change: A particle has velocity v(t)=3t26t.Findthetotaldistancetraveledbetweent=0v(t) = 3t^2 - 6t. Find the total distance traveled between t=0 and t=3t=3.
Click for Solutions
  1. Solution: 2xsin(x2)2x \cdot \sin(x^2) (Substitute x2x^2 and multiply by its derivative).
  2. Solution: 13(x2+1)3/2+C\frac{1}{3}(x^2+1)^{3/2} + C (Let u=x2+1u = x^2+1).
  3. Solution: 12(x+2x2)dx=4.5\int_{-1}^{2} (x+2 - x^2) dx = 4.5.
  4. Solution: π04(x)2dx=π04xdx=8π\pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx = 8\pi.
  5. Solution: 023t26tdt+233t26tdt\int_0^2 |3t^2 - 6t| dt + \int_2^3 |3t^2 - 6t| dt. Note the velocity changes sign at t=2t=2.

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