5.3 The Fundamental Theorem of Calculus/29

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< 5.3 The Fundamental Theorem of Calculus
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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}^{9}\frac{x-1}{\sqrt{x}}\,dx &= \int_{1}^{9}\left(\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}\right)dx = \int_{1}^{9}\left(\frac{x}{x^{1/2}}-\frac{1}{x^{1/2}}\right)dx = \int_{1}^{9}\left(x^{1/2} -x^{-1/2}\right)dx \\[2ex] &= \frac{2x^{3/2}}{3} - 2x^{1/2} \bigg|_{1}^{9} \\[2ex] &= \left[\frac{2(9)^{3/2}}{3}-2(9)^{1/2}\right]-\left[\frac{2(1)^{3/2}}{3}) - 2(1)^{1/2}\right] \\[2ex] &= \left[\frac{54}{3} - 6\right] - \left[\frac{2}{3}-2\right] = \frac{52}{3}-4 \\[2ex] &=\frac{40}{3} \end{align} }