5.5 The Substitution Rule/54: Difference between revisions

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Line 7: Line 7:
\begin{align}
\begin{align}


u &=4+3x \\[2ex]
u &=x^2 \\[2ex]
du &= 3\,dx \\[2ex]
du &= 2x\,dx \\[2ex]
\frac{1}{3}du &= dx \\[2ex]
\frac{1}{2}du &= xdx \\[2ex]





Revision as of 19:20, 26 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 &= 2x\,dx \\[2ex] \frac{1}{2}du &= xdx \\[2ex] \end{align} }


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}^{7} \sqrt{4+3x}\,dx = \int_{4}^{25}\sqrt{u}\,du\\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} }