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}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})-(\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}
\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]
 
&= (\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{align}


</math>
</math>

Revision as of 02:44, 7 September 2022