5.5 The Substitution Rule/59
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Revision as of 01:58, 6 September 2022 by Elizabethh58413 (talk | contribs) (Created page with "<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> <br><br> <math> \begin{align} u &=\frac{1}{x} \\[2ex] du &=-\frac{1}{x^2}dx \\[2ex] -du &=\frac{1}{x^2}dx \\[2ex] \end{align} </math> New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> <br> New lower limit: <math> 1 = \frac{1}{(1)} <\math> <math> \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}...")
![{\displaystyle {\begin{aligned}u&={\frac {1}{x}}\\[2ex]du&=-{\frac {1}{x^{2}}}dx\\[2ex]-du&={\frac {1}{x^{2}}}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a2af755923cf83f08ed479cf396012097de2ef31)
New upper limit:Failed to parse (unknown function "\math"): {\displaystyle \frac{1}{2} = \frac{1}{(2)} <\math> <br> New lower limit: <math> 1 = \frac{1}{(1)} <\math> <math> \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} }