Tính tích phân sau: I=∫((x−1)^5+(2x+1)^10)dx

Question

Tính các tích phân sau:

\(I =\int _{0}^{1}\left(\left(x-1\right)^{5} +\left(2x+1\right)^{10} \right){\rm d}x .\)

Answer 1

\(I =\int _{0}^{1}\left(\left(x-1\right)^{5} +\left(2x+1\right)^{10} \right)dx \)

\(=\int _{0}^{1}\left(x-1\right)^{5} dx +\int _{0}^{1}\left(2x+1\right)^{10} dx \) 
+ Tính \(\int _{0}^{1}\left(x-1\right)^{5} {\rm d}x =\left. \frac{\left(x-1\right)^{6} }{6} \right|_{0}^{1} =-\frac{1}{6} \).

 + Tính \(\int _{0}^{1}\left(2x+1\right)^{10} {\rm d}x\)

\( =\frac{1}{2} .\int _{0}^{1}\left(2x+1\right)^{10} {\rm d}\left(2x+1\right)\)

\(=\left. \frac{\left(2x+1\right)^{11} }{22} \right|_{0}^{1} =\frac{1}{22} .\left(3^{11} -1\right)=\frac{88573}{11} .\)

 Vậy \( I =\int _{0}^{1}\left(\left(x-1\right)^{5} +\left(2x+1\right)^{10} \right){\rm d}x \)

\(=\int _{0}^{1}\left(x-1\right)^{5} {\rm d}x +\int _{0}^{1}\left(2x+1\right)^{10} {\rm d}x =\frac{531427}{66} .\)