Задача 57861 Помогите...
Условие
Решение
[m] ∫ ^{0}_{-0,5}\frac{(2x+1)^{\frac{1}{2}}-6(2x+1)^{\frac{2}{3}}}{(2x+1)^{\frac{2}{3}}}=∫ ^{0}_{-0,5}\frac{(2x+1)^{\frac{1}{2}}}{(2x+1)^{\frac{2}{3}}}dx-6\cdot ∫ ^{0}_{-0,5}\frac{(2x+1)^{\frac{2}{3}}}{(2x+1)^{\frac{2}{3}}}dx=∫ ^{0}_{-0,5}(2x+1)^{\frac{1}{2}-\frac{2}{3}}dx-6\cdot ∫ ^{0}_{-0,5}dx=[/m]
так как [m] d(2x+1)=(2x+1)`dx=2dx[/m] ⇒ [m]dx=\frac{1}{2}d(2x+1)[/m]
[m]=\frac{1}{2}∫ ^{0}_{-0,5}(2x+1)^{-\frac{1}{6}}d(2x+1)-6\cdot ∫ ^{0}_{-0,5}dx=[/m]
[m]=\frac{1}{2}\cdot \frac{(2x+1)^{-\frac{1}{6}+1}}{-\frac{1}{6}+1}|^{0}_{-0,5}-6\cdot (x)|^{0}_{-0,5}=[/m]
[m]=\frac{1}{2}\cdot \frac{(2x+1)^{\frac{5}{6}}}{\frac{5}{6}}|^{0}_{-0,5}-6\cdot (x)|^{0}_{-0,5}=[/m]
[m]=0,6\cdot (1-0)-6\cdot (0-(-0,5))=0,6-3=-2,4[/m]
