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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 $x_i = \Delta xi + a$, $f(x_i) = 2$ 
 $$ = lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}$$
 $$ = \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}$$
 $$ = \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n} $$