Tính I=∫0−1x2(x+1)3dx

Question

Tính \(I=\int _{-1}^{0}x^{2} \left(x+1\right)^{3} {\rm d}x .  \)

 A. \(I=-\frac{7}{70} \). B. \(I=-\frac{1}{60}\) . C.\( I=\frac{2}{15}\) . D. \(I=\frac{1}{60} .\)

Answer 1

Chọn D
\(I=\int _{-1}^{0}x^{2} \left(x+1\right)^{3} {\rm d}x =\int _{-1}^{0}x^{2} \left(x^{3} +3x^{2} +3x+1\right){\rm d}x .\)
\(=\int _{-1}^{0}\left({\rm x}^{5} {\rm +3x}^{4} {\rm +3x}^{3} {\rm +x}^{2} \right){\rm d}x=\left(\frac{1}{6} x^{6} +\frac{3}{5} x^{5} +\frac{3}{4} x^{4} +\frac{1}{3} x^{3} \right) \left|\begin{array}{c} {0} \\ {-1} \end{array}\right. =\frac{1}{60} .\)