5.5 The Substitution Rule/59: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) | \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) | ||
&=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex] | &=\int_{1}^{\frac{1}{2}}e^u\,(-du) \\[2ex] | ||
&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex] | &=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex] | ||
&=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex] | &=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex] | ||
Revision as of 02:02, 6 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx}
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&={\frac {1}{x}}\\[2ex]du&=-{\frac {1}{x^{2}}}dx\\[2ex]-du&={\frac {1}{x^{2}}}dx\\[2ex]\end{aligned}}}
New upper limit:Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{2}}={\frac {1}{(2)}}}
New lower limit: Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1={\frac {1}{(1)}}}
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_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) &=\int_{1}^{\frac{1}{2}}e^u\,(-du) \\[2ex] &=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex] &=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex] &=e-\sqrt{e} \end {align} }