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\(\bf \displaystyle \int\limits_{1}^{9}\ \sqrt{x}\cdot dx \iff\displaystyle \int\limits_{1}^{9}\ x^{\frac{1}{2}}\cdot dx \\\\\\ \cfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\implies \cfrac{x^{\frac{3}{2}}}{\frac{3}{2}}\implies \left. \cfrac{2\sqrt{x^3}}{3} \right]_{1}^{9}\)

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Use the fundamental theorem of calculus and by the way turn x into x^1/2 dx.

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