Learn & Review: Calculus 2 Final Exam -

Calculus II Final Exam Review Summary

This summary covers key calculus II concepts and problem-solving techniques, including integration by parts, trigonometric integrals, trigonometric substitution, partial fractions, improper integrals, numerical integration (Trapezoidal and Simpson's rules), arc length, and surface area of revolution.

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1. Integration by Parts

• Formula: $\int u , dv = uv - \int v , du$
• Application: Used for integrals where a product of functions is involved, particularly when one function can be easily differentiated and the other easily integrated.
• Example Problem: $\int x^2 \sin(x) , dx$
  • Step 1: Apply integration by parts once.
    • Let $u = x^2$, $dv = \sin(x) , dx$.
    • Then $du = 2x , dx$, $v = -\cos(x)$.
    • Integral becomes: $-x^2 \cos(x) - \int (-\cos(x))(2x , dx) = -x^2 \cos(x) + 2 \int x \cos(x) , dx$.
  • Step 2: Apply integration by parts again to $\int x \cos(x) , dx$.
    • Let $u = x$, $dv = \cos(x) , dx$.
    • Then $du = dx$, $v = \sin(x)$.
    • Integral becomes: $x \sin(x) - \int \sin(x) , dx = x \sin(x) - (-\cos(x)) = x \sin(x) + \cos(x)$.
  • Final Result: Substitute back: $-x^2 \cos(x) + 2(x \sin(x) + \cos(x)) + C = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C$.

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2. Trigonometric Integrals

• Technique: Use trigonometric identities and u-substitution.
• Key Identity: $\sin^2(x) + \cos^2(x) = 1$
• Strategy:
  • Separate one trigonometric function (e.g., $\cos(x) , dx$).
  • Convert the remaining trigonometric functions to the other type using identities (e.g., convert $\cos^4(x)$ to terms of $\sin(x)$).
  • Use u-substitution with $u$ being the trigonometric function that was separated.
• Example Problem: $\int \cos^5(x) \sin^4(x) , dx$
  • Rewrite as: $\int \cos^4(x) \sin^4(x) \cos(x) , dx$.
  • Use $\cos^2(x) = 1 - \sin^2(x)$: $\int (1 - \sin^2(x))^2 \sin^4(x) \cos(x) , dx$.
  • Let $u = \sin(x)$, $du = \cos(x) , dx$.
  • Integral becomes: $\int (1 - u^2)^2 u^4 , du$.
  • Expand and integrate using the power rule.
  • Substitute back $u = \sin(x)$.

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3. Trigonometric Substitution

• Purpose: Used for integrals involving expressions of the form $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$.
• Substitutions:
  • For $\sqrt{a^2 - x^2}$: Let $x = a \sin(\theta)$.
  • For $\sqrt{a^2 + x^2}$: Let $x = a \tan(\theta)$.
  • For $\sqrt{x^2 - a^2}$: Let $x = a \sec(\theta)$.
• Example Problem: $\int \frac{dx}{\sqrt{x^2 + 9}}$
  • Here $a^2 = 9$, so $a = 3$. Use $x = 3 \tan(\theta)$.
  • $dx = 3 \sec^2(\theta) , d\theta$.
  • $\sqrt{x^2 + 9} = \sqrt{(3 \tan(\theta))^2 + 9} = \sqrt{9 \tan^2(\theta) + 9} = \sqrt{9(\tan^2(\theta) + 1)} = \sqrt{9 \sec^2(\theta)} = 3 \sec(\theta)$.
  • Integral becomes: $\int \frac{3 \sec^2(\theta) , d\theta}{3 \sec(\theta)} = \int \sec(\theta) , d\theta$.
  • The integral of $\sec(\theta)$ is $\ln|\sec(\theta) + \tan(\theta)|$.
  • Convert back to x: Draw a right triangle where $\tan(\theta) = x/3$. The opposite side is $x$, adjacent is $3$, hypotenuse is $\sqrt{x^2 + 9}$.
    • $\sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{x^2 + 9}}{3}$.
  • Final Result: $\ln\left|\frac{\sqrt{x^2 + 9}}{3} + \frac{x}{3}\right| + C = \ln\left|\frac{\sqrt{x^2 + 9} + x}{3}\right| + C = \ln|\sqrt{x^2 + 9} + x| - \ln(3) + C$. Since $\ln(3)$ is a constant, it can be absorbed into $C$. So, $\ln|\sqrt{x^2 + 9} + x| + C$.

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4. Partial Fractions

• Purpose: Used to integrate rational functions where the degree of the numerator is less than the degree of the denominator.
• Steps:
  1. Factor the denominator completely.
  2. Decompose the rational function into a sum of simpler fractions based on the factors.
     • Linear factor $(ax+b)$: $\frac{A}{ax+b}$
     • Repeated linear factor $(ax+b)^n$: $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$
     • Quadratic factor $(ax^2+bx+c)$: $\frac{Ax+B}{ax^2+bx+c}$
  3. Solve for the unknown coefficients (A, B, C, etc.) by clearing the denominators and equating numerators, or by substituting strategic values of $x$.
  4. Integrate the resulting simpler fractions. The integral of $\frac{1}{ax+b}$ is $\frac{1}{a}\ln|ax+b|$.
• Example Problem: $\int \frac{x^2 + 12x + 12}{x^3 - 4x} , dx$
  • Factor denominator: $x(x^2 - 4) = x(x+2)(x-2)$.
  • Decomposition: $\frac{A}{x} + \frac{B}{x+2} + \frac{C}{x-2}$.
  • Multiply by $x(x+2)(x-2)$: $x^2 + 12x + 12 = A(x+2)(x-2) + Bx(x-2) + Cx(x+2)$.
  • Solve for coefficients:
    • Let $x=0$: $12 = A(2)(-2) \implies 12 = -4A \implies A = -3$.
    • Let $x=2$: $4 + 24 + 12 = C(2)(4) \implies 40 = 8C \implies C = 5$.
    • Let $x=-2$: $4 - 24 + 12 = B(-2)(-4) \implies -8 = 8B \implies B = -1$.
  • Integral becomes: $\int \left(\frac{-3}{x} + \frac{-1}{x+2} + \frac{5}{x-2}\right) , dx$.
  • Final Result: $-3 \ln|x| - \ln|x+2| + 5 \ln|x-2| + C$.

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5. Improper Integrals

• Definition: Integrals where either the interval of integration is infinite or the integrand has an infinite discontinuity within the interval.
• Convergence/Divergence: Determined by evaluating the integral using limits.
  • If the limit is a finite number, the integral converges.
  • If the limit is $\infty$, $-\infty$, or does not exist, the integral diverges.
• Example Problem: $\int_1^\infty \frac{1}{x^3} , dx$
  • Rewrite as a limit: $\lim_{a \to \infty} \int_1^a x^{-3} , dx$.
  • Find the antiderivative: $\int x^{-3} , dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$.
  • Evaluate the definite integral: $\lim_{a \to \infty} \left[-\frac{1}{2a^2} - \left(-\frac{1}{2(1)^2}\right)\right]$.
  • Calculate the limit: $\lim_{a \to \infty} \left(-\frac{1}{2a^2} + \frac{1}{2}\right) = 0 + \frac{1}{2} = \frac{1}{2}$.
  • Conclusion: Since the limit is a finite value ($\frac{1}{2}$), the integral converges.

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6. Numerical Integration

• Purpose: To approximate the value of a definite integral when an analytical solution is difficult or impossible.
• Trapezoidal Rule: Approximates the area under the curve using trapezoids.
  • Formula: $\int_a^b f(x) , dx \approx \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)]$
  • Where $\Delta x = \frac{b-a}{n}$ and $x_i = a + i \Delta x$.
  • Note: The first and last y-values are not multiplied by 2.
• Simpson's Rule: Approximates the area under the curve using parabolic segments. Generally more accurate than the Trapezoidal Rule for the same number of subintervals.
  • Formula: $\int_a^b f(x) , dx \approx \frac{\Delta x}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)]$
  • Requires an even number of subintervals ($n$).
  • Note: The coefficients follow the pattern 1, 4, 2, 4, 2, ..., 4, 1.

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7. Arc Length

• Purpose: To calculate the length of a curve between two points.
• Formula: For a function $y = f(x)$ from $x=a$ to $x=b$:
  • $L = \int_a^b \sqrt{1 + [f'(x)]^2} , dx$
• Example Problem: Arc length of $f(x) = \frac{2}{3}(x+4)^{3/2}$ from $x=0$ to $x=4$.
  • Find the derivative: $f'(x) = \frac{2}{3} \cdot \frac{3}{2} (x+4)^{1/2} \cdot 1 = \sqrt{x+4}$.
  • Square the derivative: $[f'(x)]^2 = (\sqrt{x+4})^2 = x+4$.
  • Set up the integral: $L = \int_0^4 \sqrt{1 + (x+4)} , dx = \int_0^4 \sqrt{x+5} , dx$.
  • Use u-substitution: Let $u = x+5$, $du = dx$.
    • New limits: When $x=0$, $u=5$. When $x=4$, $u=9$.
  • Integral becomes: $\int_5^9 \sqrt{u} , du = \int_5^9 u^{1/2} , du$.
  • Integrate: $\left[ \frac{2}{3} u^{3/2} \right]_5^9$.
  • Evaluate: $\frac{2}{3} (9^{3/2} - 5^{3/2}) = \frac{2}{3} (27 - 5\sqrt{5})$.

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8. Surface Area of Revolution

• Purpose: To calculate the surface area generated by rotating a curve around an axis.
• Formula: For a function $y = f(x)$ rotated around the x-axis from $x=a$ to $x=b$:
  • $SA = \int_a^b 2\pi f(x) \sqrt{1 + [f'(x)]^2} , dx$
• Example Problem: Surface area of $f(x) = x^3$ from $x=0$ to $x=2$ rotated about the x-axis.
  • Find the derivative: $f'(x) = 3x^2$.
  • Square the derivative: $[f'(x)]^2 = (3x^2)^2 = 9x^4$.
  • Set up the integral: $SA = \int_0^2 2\pi (x^3) \sqrt{1 + 9x^4} , dx$.
  • Use u-substitution: Let $u = 1 + 9x^4$, $du = 36x^3 , dx \implies x^3 , dx = \frac{du}{36}$.
    • New limits: When $x=0$, $u=1$. When $x=2$, $u=1 + 9(16) = 1 + 144 = 145$.
  • Integral becomes: $SA = \int_1^{145} 2\pi \sqrt{u} \left(\frac{du}{36}\right) = \frac{2\pi}{36} \int_1^{145} u^{1/2} , du = \frac{\pi}{18} \int_1^{145} u^{1/2} , du$.
  • Integrate: $\frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_1^{145}$.
  • Evaluate: $\frac{\pi}{18} \cdot \frac{2}{3} (145^{3/2} - 1^{3/2}) = \frac{\pi}{27} (145\sqrt{145} - 1)$.