Задача 76571 ...
Условие
y = 1/3 √x^3 - √x, 9 ≤ x ≤ 36.
Решение
[m]y`=(\frac{1}{3}\cdot x^{\frac{3}{2}}-x^{\frac{1}{2}})`=\frac{1}{3}\cdot \frac{3}{2}\cdot x^{\frac{3}{2}-1}-\frac{1}{2}\cdot x^{\frac{1}{2}-1}=[/m]
[m]=\frac{1}{2}\cdot x^{\frac{1}{2}}-\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{1}{2}\cdot ( x^{\frac{1}{2}}- x^{-\frac{1}{2}})[/m]
[m]\sqrt{1+(y`)^2}=\sqrt{1+(\frac{1}{2}\cdot ( x^{\frac{1}{2}}- x^{-\frac{1}{2}}))^2}=\sqrt{1+\frac{1}{4}\cdot ( x-2+ x^{-1}}=\sqrt{\frac{1}{4}\cdot ( x^{\frac{1}{2}}+x^{-\frac{1}{2}})^2}=\frac{1}{2}(x^{\frac{1}{2}}+x^{-\frac{1}{2}})[/m]
[m]l= ∫ ^{36}_{9}\frac{1}{2}(x^{\frac{1}{2}}+x^{-\frac{1}{2}})=\frac{1}{2}\cdot (\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{1}{2}}}{\frac{1}{2}})|^{36}_{9}=\frac{1}{3}(\sqrt{36^3}-\sqrt{9^3})-(\sqrt{36}-\sqrt{9})=\frac{1}{3}(6^3-3^3)-(6-3)=60[/m]
