From 7eb98a3eb318bb2896fcacc474e147493ce7a2be Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 27 Aug 2025 11:59:05 -0600 Subject: [PATCH] vault backup: 2025-08-27 11:59:05 --- education/math/MATH1220 (calc II)/Integral Review.md | 7 ++++++- 1 file changed, 6 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1220 (calc II)/Integral Review.md b/education/math/MATH1220 (calc II)/Integral Review.md index 6844735..9efb9bb 100644 --- a/education/math/MATH1220 (calc II)/Integral Review.md +++ b/education/math/MATH1220 (calc II)/Integral Review.md @@ -11,14 +11,19 @@ Define $$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x$$ - Or, the width of each interval times the interval index, plus the starting offset. Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivative of $f$ (i.e $F'(x) = f(x)$). + Then $\int_a^b f(x) dx = F(b) - F(a)$. +## Sum of an infinite series +The sum of an infinite series can be defined as follows: +$$ \sum_{i = 1}^n i = \frac{n(n+1)}{2}$$ + ## Examples 1. Find the area under the curve between 0 and 1 of the function $f(x) = x^2$ $$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$ 2. Find the Riemann Sum under the curve between -2 and 2 of the function $2x + 2$. $$ \int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$ -> Using the fact that $x_i = \Delta x + a$, $ +> Using the fact that $x_i = \Delta xi + a$, $f(x_i) = 2$ $$ = lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}$$ $$ = \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}$$ $$ = \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n} $$ \ No newline at end of file