5.5 The Substitution Rule/54: Difference between revisions

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&= \int_{0}^{\pi} \left(\frac{1}{2}du\right)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex]
&= \int_{0}^{\pi} \left(\frac{1}{2}du\right)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex]
&= \frac{1}{2}\cdot\sin{(u)}\bigg|_{0}^{\pi} \\[2ex]
&= \frac{1}{2}\sin{(u)}\bigg|_{0}^{\pi} \\[2ex]
&= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex]
&= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex]
&= 0
&= 0

Latest revision as of 22:50, 28 August 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\sqrt{\pi}} x\cos{(x^2)}\,dx }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &=x^2 \\[2ex] du &= 2xdx \\[2ex] \frac{1}{2}du &= xdx \\[2ex] \end{align} }


New upper limit:
New lower limit: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 0 = (0)^2}


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{0}^{\sqrt{\pi}} x\cos{(x^2)}\,dx &= \int_{0}^{\sqrt{\pi}} (xdx)\cos{(x^2)} \\[2ex] &= \int_{0}^{\pi} \left(\frac{1}{2}du\right)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[2ex] &= \frac{1}{2}\sin{(u)}\bigg|_{0}^{\pi} \\[2ex] &= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex] &= 0 \end{align} }