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- https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Integral_Calculus/3%3A_L'Hopital's_Rule_and_Improper_Integrals/Numerical_IntegrationTo get the area of the i th rectangle we multiply the height by the base: A fourth method involves the trapezoidal rule which geometrically calculates the area of the trapezoid with base on the x-axis...To get the area of the i th rectangle we multiply the height by the base: A fourth method involves the trapezoidal rule which geometrically calculates the area of the trapezoid with base on the x-axis and heights f(xi) and f(x_{i+1}). The graphs show that the error is directly inked to the concavity of the integrand. Let n be even then using the even subscripted x values for the trapezoidal estimate and the midpoint estimate, gives
- https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/04%3A_Transcendental_Functions/4.03%3A_A_Hard_LimitBefore we can complete the calculation of the derivative of the sine, we need one other limit: lim This limit is just as hard as \sin x/x, but closely related to i...Before we can complete the calculation of the derivative of the sine, we need one other limit: \lim_{x\to0}{\cos x - 1\over x}. This limit is just as hard as \sin x/x, but closely related to it, so that we don't have to a similar calculation; instead we can do a bit of tricky algebra: {\cos x - 1\over x}={\cos x - 1\over x}{\cos x+1\over\cos x+1} ={\cos^2 x - 1\over x(\cos x+1)}={-\sin^2 x\over x(\cos x+1)}= -{\sin x\over x}{\sin x\over \cos x + 1}.
- https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Integral_Calculus/4%3A_Transcendental_Functions/4.8%3A_Integrals_Involving_Arctrig_Functions\begin{align*} \int \dfrac{dx}{x \sqrt{x^2-4}} &= \int \dfrac{x\,dx}{x^2 \sqrt{4 (x^4/4 -1)}} \\ &= \dfrac{1}{4} \int \dfrac{x\,dx}{(x^2/2)\sqrt{(x^2/2)^2-1}} \end{align*} \[\begin{align*} \int \...\begin{align*} \int \dfrac{dx}{x \sqrt{x^2-4}} &= \int \dfrac{x\,dx}{x^2 \sqrt{4 (x^4/4 -1)}} \\ &= \dfrac{1}{4} \int \dfrac{x\,dx}{(x^2/2)\sqrt{(x^2/2)^2-1}} \end{align*} \begin{align*} \int \dfrac{(2u-3)\, du}{u^2+1} &= \int \dfrac{2u\,du}{u^2+1}-3\int\dfrac{du}{u^2+1} \\ &= \ln \left| u^2+1 \right| -3\tan^{-1} u+C \\ &= \ln \left(\dfrac{ \left(\dfrac{x+3}{2} \right)^2}{4} +1\right) - 3 \tan^{-1}\dfrac{x+3}{2} + C \end{align*}.\nonumber
- https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/14%3A_Partial_Differentiation/14.04%3A_The_Chain_RuleNot surprisingly, the same chain rule that was formulated for a function on one variable also works for functions of more than two variables.
