math117: add improper integrals
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@ -641,3 +641,63 @@ For curves bounded by functions of $y$:
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$$L(y)=\int^b_a\sqrt{1+\left(\frac{dx}{dy}\right)^2\ dy}$$
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$$L(y)=\int^b_a\sqrt{1+\left(\frac{dx}{dy}\right)^2\ dy}$$
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### Solids of revolution
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### Solids of revolution
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### Improper integrals
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An improper integral is a definite integral where only one bound is defined:
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!!! example
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$\int_2^\infty$ or $\int_a^b$, where only $a$ is defined.
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These can be expanded into limits:
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$$\int_a^\infty f(x)\ dx = \lim_{t\to\infty}\int_a^t f(x)\ dx$$
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The integral converges to a value if the limit exists.
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$$\int_{-\infty}^a f(x)\ dx = \lim_{t\to-\infty}\int^a_tf(x)\ dx$$
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Discontinuities can be simply dodged. If there is a discontinuity:
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- at $b$: $\int_a^{b^-}f(x)\ dx$
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- at $a$: $\int_{a^+}^b f(x)\ dx$
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- at $a<c<b$: $\int_a^cf(x)\ dx + \int_c^bf(x)\ dx$
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Limits to both infinities must be broken up because they may not approach them at the same rate.
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$$\int^\infty_{-\infty}x\ dx = \int^0_{-\infty} x\ dx + \int^\infty_0 x\ dx$$
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## Polar form
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Please see [MATH 115: Linear Algebra#Polar form](/ce1/math115/#polar-form) for more information.
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Instead of $r$ and $\theta$, engineers use $\rho$ and $\phi$.
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For $\rho \geq 0$, these basic conversions go between the two forms:
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- $x=\rho\cos\phi$
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- $y=\rho\sin\phi$
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- $\phi=\sqrt{x^2+y^2}$
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- $\phi=\tan^{-1}\left(\frac{y}{x}\right) + 2k\pi,k\in\mathbb Z$
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Polar form allows for simpler representations such as $x^2+y^2=4 \iff \rho=2$
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Functions are described in the form $\rho=f(\phi)$, such as $\rho=\sin\phi+2$.
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### Area under curves
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From the axis to the curve:
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$$A=\int^\beta_\alpha\frac{1}{2}[f(\phi)]^2\ d\phi$$
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Between two curves:
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$$A=\int^\beta_\alpha\frac{1}{2}[f(\phi)^2-g(\phi)^2]\ d\phi$$
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Arc length:
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$$L=\int^\beta_\alpha\sqrt{f'(\phi)^2 + f(\phi)^2}\ d\phi = \int^\beta_\alpha\sqrt{\left(\frac{d\rho}{d\phi}\right)^2+\rho^2}\ d\phi$$
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## Complex numbers
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Please see [MATH 115: Linear Algebra#Complex Numbers](/ce1/math115/#complex-numbers) for more information.
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