Lời giải của Tự Học 365

Ta có : \(\int\limits_0^3 {x(x - 1)dx} \)\( = \int\limits_0^3 {\left( {{x^2} - x} \right)dx}  = \left. {\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2}} \right|_0^3\) \( = 9 - \dfrac{9}{2} = \dfrac{9}{2}\)

+) $\int\limits_0^{\ln \sqrt {10} } {{e^{2x}}dx}  = \left. {\dfrac{{{e^{2x}}}}{2}} \right|_0^{\ln \sqrt {10} } = \dfrac{{{e^{2\ln \sqrt {10} }} - 1}}{2} = \dfrac{9}{2}$,

+) $3\int\limits_0^{3\pi } {\sin xdx}  = \left. { - 3\cos x} \right|_0^{3\pi } = 6$,

+) $\int\limits_0^2 {\left( {{x^2} + x - 3} \right)dx}  = \left. {\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} - 3x} \right)} \right|_0^2 = \dfrac{8}{3} + 2 - 6 =  - \dfrac{4}{3}$,

+) $\int\limits_0^\pi  {\cos (3x + \pi )dx}  = \dfrac{1}{3}\left. {\sin (3x + \pi )} \right|_0^\pi  = \dfrac{1}{3}\left( {\sin 4\pi  - \sin \pi } \right) = 0$.

Vậy chọn $\int\limits_0^{\ln \sqrt {10} } {{e^{2x}}dx} $

Đáp án cần chọn là: c