5.3 The Fundamental Theorem of Calculus/29: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{1}^{9}\frac{x-1}{\sqrt{x}} dx = \int_{1}^{9}\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}} = \int_{1}^{9}\frac{x}{x^{1/2}}-\frac{1}{x^{1/2}} = \int_{1}^{9}x^{1/2} -x^{-1/2} = \int_{1}^{9}\frac{2x^{3/2}}{3} - 2x^{1/2} \bigg|_{1}^{9} \\[2ex] | \int_{1}^{9}\frac{x-1}{\sqrt{x}} dx &= \int_{1}^{9}\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}} = \int_{1}^{9}\frac{x}{x^{1/2}}-\frac{1}{x^{1/2}} = \int_{1}^{9}x^{1/2} -x^{-1/2} = \int_{1}^{9}\frac{2x^{3/2}}{3} - 2x^{1/2} \bigg|_{1}^{9} \\[2ex] | ||
&= (\frac{2(9)^{3/2}}{3})-2(9)^{1/2})-(\frac{2(1)^{3/2}}{3}) - 2(1)^{1/2}) = (\frac{54}{3} - 6) - (\frac{2}{3}-2) | &= (\frac{2(9)^{3/2}}{3})-2(9)^{1/2})-(\frac{2(1)^{3/2}}{3}) - 2(1)^{1/2}) = (\frac{54}{3} - 6) - (\frac{2}{3}-2) | ||
Revision as of 02:47, 7 September 2022
![{\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}dx&=\int _{1}^{9}{\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}=\int _{1}^{9}{\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}=\int _{1}^{9}x^{1/2}-x^{-1/2}=\int _{1}^{9}{\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=({\frac {2(9)^{3/2}}{3}})-2(9)^{1/2})-({\frac {2(1)^{3/2}}{3}})-2(1)^{1/2})=({\frac {54}{3}}-6)-({\frac {2}{3}}-2)&={\frac {52}{3}}-4={\frac {40}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b4cc0ef8be55242a85a78c7416e765e871a11ecc)