24 lines
1.3 KiB
Markdown
24 lines
1.3 KiB
Markdown
# Formal Definition
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Let $f$ be a continuous function on an interval $[a, b]$.
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Divide $[a, b]$ into $n$ equal parts of width $\Delta x = \dfrac{b-a}{n}$.
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Let $x_0, x_1, x_2, \cdots, x_n$ be the endpoints of this subdivision. $x_0 = a$ and $x_n = b$.
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Define $$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x$$
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- $\Delta x$ refers to the width of each sub-interval, or $\frac{b-a}{n}$.
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- $f(x_i)$ refers to the height of each subinterval, and can be found with the equation $x_i = \Delta xi + a$
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- Or, the width of each interval times the interval index, plus the starting offset.
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Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivative of $f$ (i.e $F'(x) = f(x)$).
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Then $\int_a^b f(x) dx = F(b) - F(a)$.
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## Examples
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1. Find the area under the curve between 0 and 1 of the function $f(x) = x^2$
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$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
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2. Find the Riemann Sum under the curve between -2 and 2 of the function $2x + 2$.
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$$ \int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$
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> Using the fact that $x_i = \Delta x + a$, $
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$$ = lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}$$
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$$ = \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}$$
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$$ = \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n} $$ |