vault backup: 2025-03-25 09:06:37
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@ -35,3 +35,12 @@ $$ \int f(x) dx $$
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| $\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}$ | $\int \dfrac{1}{\sqrt{1+x^2}}dx = \tan^{-1}x + C$ |
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| $\dfrac{d}{dx} k f(x) = k f'(x)$ | $\int k*f(x)dx = k\int f(x)dx$ |
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| $\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)$ | $\int (f(x) \pm g(x))dx = \int f(x) dx \pm \int g(x) dx$ |
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# Definite Integrals
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Let $f$ be a continuous function on the interval $[a, b]$. Divide $[a, b]$ into $n$ equal parts of width $\Delta x = \dfrac{b - a}{n}$ . Let $x_0, x_1, x_2, \cdots, x_3$ be the endpoints of the subdivision.
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The definite integral of $f(x)$ with respect to $x$ from $x = a$ to $x = b$ can be denoted:
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$$ \int_{a}^b f(x) dx $$
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And __can__ be defined as:
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$$ \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i = 1}^n f(x_i)\Delta x$$
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