vault backup: 2025-08-27 11:49:05
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@ -6,11 +6,14 @@ 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
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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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$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
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$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
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$$ \int_{-2}^2 2x + 2dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$
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