Tích phân \(\displaystyle\int\limits_{0}^{1}(3x+1)(x+3)\mathrm{\,d}x\) bằng
\(6\) | |
\(5\) | |
\(12\) | |
\(9\) |
Chọn phương án D.
\(\begin{aligned}
\displaystyle\int\limits_{0}^{1}(3x+1)(x+3)\mathrm{\,d}x&=\displaystyle\int\limits_{0}^{1}\left(3x^2+10x+3\right)\mathrm{\,d}x\\
&=\left(x^3+5x^2+3x\right)\bigg|_0^1\\
&=9.
\end{aligned}\)