Question

(39)Приложения определенного интеграла. Длина дуги кривой\ Решить как можно подробнее

Answer 1

$y'=(x^\frac{3}{2})'=\frac{3}{2}x^\frac{1}{2}, \\ L= \int\limits^5_0 {\sqrt{1+(\frac{3}{2}x^\frac{1}{2})^2}} \, dx = \frac{1}{2}\int\limits^5_0 {\sqrt{4+9x}} \, dx =\frac{1}{18}\int\limits^5_0 {\sqrt{4+9x}} \, d(4+9x) = \\ = \frac{1}{18}\cdot\frac{2}{3}\sqrt{(4+9x)^3}|^5_0= \frac{1}{27}(\sqrt{(4+9\cdot5)^3}-\sqrt{(4+9\cdot0)^3}) = \\ = \frac{1}{27}(\sqrt{49^3}-\sqrt{4^3}) = \frac{1}{27}(\sqrt{7^6}-\sqrt{2^6}) = \frac{1}{27}(7^3-2^3) = \frac{335}{27} = 12 \frac{11}{27}$

Answer 2

$\mathfrak{L}=\int \limits_a^b\ \sqrt{1+(y')^2}\ dx= \Bigr| \ y'= \frac{3}{2}x^{ \frac{1}{2}}\ \Bigr| = \int \limits_0^5\ \sqrt{1+ \frac{9}{4}x }\ dx=\\ \\ =\frac{4}{9} \int \limits_0^5\ \sqrt{1+ \frac{9}{4}x }\ d(1+\frac{9}{4}x)=(\frac{4}{9}*\frac{2}{3}(1+\frac{9}{4}x)\sqrt{1+ \frac{9}{4}x }) \Big|_0^5 = \\ \\ =\frac{8}{27}(1+\frac{9}{4}*5)\sqrt{1+ \frac{9}{4}*5 }-\frac{8}{27}(1+\frac{9}{4}*0)\sqrt{1+ \frac{9}{4}*0 }=$

$=\frac{8}{27}*\frac{49}{4}*\frac{7}{2}-\frac{8}{27}=\frac{1}{27}(343-8)=\frac{335}{27}=12\frac{11}{27}$