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education/math/MATH1220 (calc II)/Integral Review.md
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@ -14,6 +14,11 @@ Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivati
Then $\int_a^b f(x) dx = F(b) - F(a)$. Then $\int_a^b f(x) dx = F(b) - F(a)$.
## Examples ## Examples
1. Find the area under the curve between 0 and 1 of the function $f(x) = x^2$
$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$ $$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
2. Find the Riemann Sum under the curve between -2 and 2 of the function $2x + 2$.
$$ \int_{-2}^2 2x + 2dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$ $$ \int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$
> Using the fact that $x_i = \Delta x + a$, $
$$ = lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}$$
$$ = \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}$$
$$ = \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n} $$