## Calculus: Early Transcendentals 8th Edition

$\displaystyle\int\limits_1^9 \sqrt{x} dx=\dfrac{52}{3}$
$\displaystyle\int\limits_1^9 \sqrt{x} dx$ Rewrite the integrand like this: $\displaystyle\int\limits_1^9x^{1/2}dx$ Integrate and apply the second part of the fundamental theorem of calculus: $\displaystyle\int\limits_1^9x^{1/2}dx=\dfrac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}\Big|_1^9=\dfrac{1}{\frac{3}{2}}x^{3/2}\Big|_1^9=\dfrac{2}{3}x^{3/2}\Big|_1^9=...$ $...=\dfrac{2}{3}[(9)^{3/2}-(1)^{3/2}]=\dfrac{2}{3}[\sqrt{9^{3}}-\sqrt{1^{3}}]=\dfrac{2}{3}[\sqrt{729}-1]=...$ $...=\dfrac{2}{3}(27-1)=\dfrac{2}{3}(26)=\dfrac{52}{3}$