5.3 The Fundamental Theorem of Calculus/29: Difference between revisions
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<math> | <math> | ||
\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}\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} = (\frac{2(9)^{3/2}}{3})-2(9)^{1/2} | ||
</math> | </math> | ||
Revision as of 02:37, 7 September 2022
