Вычислите определенный интеграл с точностью 0.001
Разложим в ряд Тейлора: Посчитаем теперь определенный интеграл.
![\displaystyle \dfrac{1}{ \sqrt[3]{1+x^3} } =(1+x^3)^{1/3}=1- \frac{x^3}{3}+ \frac{2x^6}{9} - \frac{14x^9}{81}+O(x^{11})](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Cdfrac%7B1%7D%7B+%5Csqrt%5B3%5D%7B1%2Bx%5E3%7D+%7D+%3D%281%2Bx%5E3%29%5E%7B1%2F3%7D%3D1-+%5Cfrac%7Bx%5E3%7D%7B3%7D%2B+%5Cfrac%7B2x%5E6%7D%7B9%7D+-+%5Cfrac%7B14x%5E9%7D%7B81%7D%2BO%28x%5E%7B11%7D%29++ "\displaystyle \dfrac{1}{ \sqrt[3]{1+x^3} } =(1+x^3)^{1/3}=1- \frac{x^3}{3}+ \frac{2x^6}{9} - \frac{14x^9}{81}+O(x^{11})")
![\displaystyle \int\limits^{0.5}_0 \frac{dx}{ \sqrt[3]{1+x^3} } =\bigg(x- \frac{x^4}{12}+ \frac{2x^7}{63} - \frac{7x^{10}}{405}\bigg)\bigg|^{0.5}_0=\\ \\ \\ =0.5- \frac{0.5^4}{12}+ \frac{2\cdot0.5^7}{63} - \frac{7\cdot0.5^{10}}{405}\approx0.495](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint%5Climits%5E%7B0.5%7D_0+%5Cfrac%7Bdx%7D%7B+%5Csqrt%5B3%5D%7B1%2Bx%5E3%7D+%7D++%3D%5Cbigg%28x-+%5Cfrac%7Bx%5E4%7D%7B12%7D%2B+%5Cfrac%7B2x%5E7%7D%7B63%7D+-+%5Cfrac%7B7x%5E%7B10%7D%7D%7B405%7D%5Cbigg%29%5Cbigg%7C%5E%7B0.5%7D_0%3D%5C%5C+%5C%5C+%5C%5C+%3D0.5-+%5Cfrac%7B0.5%5E4%7D%7B12%7D%2B+%5Cfrac%7B2%5Ccdot0.5%5E7%7D%7B63%7D+-+%5Cfrac%7B7%5Ccdot0.5%5E%7B10%7D%7D%7B405%7D%5Capprox0.495 "\displaystyle \int\limits^{0.5}_0 \frac{dx}{ \sqrt[3]{1+x^3} } =\bigg(x- \frac{x^4}{12}+ \frac{2x^7}{63} - \frac{7x^{10}}{405}\bigg)\bigg|^{0.5}_0=\\ \\ \\ =0.5- \frac{0.5^4}{12}+ \frac{2\cdot0.5^7}{63} - \frac{7\cdot0.5^{10}}{405}\approx0.495")