5.3 The Fundamental Theorem of Calculus/28: Difference between revisions

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\int_{0}^{1}\left(3+x\sqrt{x}\right)dx &= \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx  
\int_{0}^{1}\left(3+x\sqrt{x}\right)dx &= \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx  
= \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx  = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\
= \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx  = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\
&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\biggg|_{0}^{1} = 3x+\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{5/2}}{5}\bigg|_{0}^{1} \\
&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{5/2}}{5}\bigg|_{0}^{1} \\
&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\
&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\
&= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{5}   
&= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{5}   

Revision as of 21:43, 23 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\frac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{5/2}}{5}}{\bigg |}_{0}^{1}\\&=\left[3(1)+{\frac {2(1)^{5/2}}{5}}\right]-\left[3(0)+{\frac {2(0)^{5/2}}{5}}\right]\\&=3+{\frac {2}{5}}={\frac {15}{5}}+{\frac {2}{5}}={\frac {17}{5}}\end{aligned}}}