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

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\begin{align}
\begin{align}


\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\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 \\[0.6ex]


&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\tfrac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{\frac{5}{2}}}{5}\bigg|_{0}^{1} \\
&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\tfrac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{\frac{5}{2}}}{5}\bigg|_{0}^{1} \\

Revision as of 21:56, 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\\[0.6ex]&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\tfrac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{\frac {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}}}