Question

The value of the integral ${\int }_1^2\sqrt{(2x+3)(3x^2+4)}dx$ cannot exceed

• A. $\sqrt{48}$
• B. $\sqrt{66}$
• C. $\sqrt{73}$
• D. none of these

Answer 1

The correct option is B $\sqrt{66}$

${\int }_1^2\sqrt{(2x+3)(3x^2+4)}dx\le \sqrt{{\int }_1^2(2x+3)dx.{\int }_1^2(3x^2+4)dx}$

$[\because \left|{\int }_a^{b}f\left(x\right).g\left(x\right)dx\right|\le \sqrt{{\int }_a^b{f}^2\left(x\right)dx.{\int }_a^b{g}^2\left(x\right)dx}]$

$=\sqrt{{[x^2+3x]}_1^2.{[x^3+4x]}_1^2}=\sqrt{6\times 11}=\sqrt{66}$