Euler's Method
Apply Euler's Method to approximate the solution to an IVP numerically: yₙ₊₁ = yₙ + h·f(xₙ,yₙ). Understand that smaller step sizes h give more accurate approximations and identify when the method overestimates or underestimates.
A lesson of AP Calculus AB, in Differential Equations, about 10 minutes.
Advanced Integration Techniques
By the end of this lesson, you will be able to apply integration by parts, use partial fractions, evaluate improper integrals, and apply Euler's method.
Integration by Parts
The formula:
$$\int u, dv = uv - \int v, du$$
This is the "undo" of the product rule for differentiation.
LIATE rule for choosing $u$: choose $u$ as the first in this order:
- Logarithmic functions
- Inverse trig functions
- Algebraic (polynomial)
- Trigonometric functions
- Exponential functions
(What's left over becomes $dv$.)
Example 1: $\displaystyle\int x e^x, dx$
Let $u = x$ (algebraic), $dv = e^x, dx$
$du = dx$, $v = e^x$
$= xe^x - \displaystyle\int e^x, dx = xe^x - e^x + C = e^x(x - 1) + C$
Example 2: $\displaystyle\int x\ln x, dx$
Let $u = \ln x$ (logarithmic — LIATE says L first), $dv = x, dx$
$du = \dfrac{1}{x}, dx$, $v = \dfrac{x^2}{2}$
$= \dfrac{x^2}{2}\ln x - \displaystyle\int \dfrac{x^2}{2} \cdot \dfrac{1}{x}, dx = \dfrac{x^2}{2}\ln x - \dfrac{1}{2}\int x, dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$
Example 3 (repeated): $\displaystyle\int e^x \sin x, dx$
Let $u = \sin x$, $dv = e^x, dx$ → $du = \cos x, dx$, $v = e^x$
$= e^x\sin x - \displaystyle\int e^x \cos x, dx$
Apply IBP again: $u = \cos x$, $dv = e^x, dx$
$= e^x\sin x - \left[e^x\cos x + \displaystyle\int e^x\sin x, dx\right]$
Let $I = \displaystyle\int e^x\sin x, dx$:
$I = e^x\sin x - e^x\cos x - I$
$2I = e^x(\sin x - \cos x)$
$I = \dfrac{e^x(\sin x - \cos x)}{2} + C$
Partial Fraction Decomposition
Purpose: integrate rational functions $\dfrac{P(x)}{Q(x)}$ by rewriting as simpler fractions.
When to use: numerator degree < denominator degree; denominator factors.
Case 1: Distinct linear factors
$\dfrac{3x + 1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$
Multiply both sides by $(x-1)(x+2)$: $3x + 1 = A(x+2) + B(x-1)$
$x = 1$: $4 = 3A \Rightarrow A = 4/3$
$x = -2$: $-5 = -3B \Rightarrow B = 5/3$
$\displaystyle\int \dfrac{3x+1}{(x-1)(x+2)}, dx = \dfrac{4}{3}\ln|x-1| + \dfrac{5}{3}\ln|x+2| + C$