Question

12．若f（x）=x²+3${∫}_{0}^{1}$f（x）dx，則${∫}_{0}^{1}$f（x）dx=（　�。�

• A． 4
• B． -6
• C． -$\frac{1}{6}$
• D． $\frac{16}{3}$

Answer 1

分析 兩邊取定積分，即可得到關(guān)于${∫}_{0}^{1}$f（x）dx的方程解得即可．

解答 解：f（x）=x²+3${∫}_{0}^{1}$f（x）dx，兩邊取定積分得到，
${∫}_{0}^{1}$f（x）dx=${∫}_{0}^{1}$（x²+3${∫}_{0}^{1}$f（x）dx）dx，
∴${∫}_{0}^{1}$f（x）dx=${∫}_{0}^{1}$x²dx+${∫}_{0}^{1}$（3${∫}_{0}^{1}$f（x）dx）dx=$\frac{1}{3}{x}^{3}$|${\;}_{0}^{1}$+3${∫}_{0}^{1}$f（x）dx•x|${\;}_{0}^{1}$=$\frac{1}{3}$+3${∫}_{0}^{1}$f（x）dx，
∴2${∫}_{0}^{1}$f（x）=-$\frac{1}{3}$，
∴${∫}_{0}^{1}$f（x）=-$\frac{1}{6}$．
故選：C．

點評 本題考查了定積分的計算；解答本題的關(guān)鍵是兩邊取定積分，屬于基礎(chǔ)題．