A.\[\mathop \smallint \limits_0^2 \left( {{x^2} + x - 3} \right)dx\]
B. \[3\mathop \smallint \limits_0^{3\pi } \sin xdx\]
C. \[\mathop \smallint \limits_0^{\ln \sqrt {10} } {e^{2x}}dx\]
D. \[\mathop \smallint \limits_0^\pi \cos (3x + \pi )dx\]
Ta có :\(\int\limits_0^3 {x(x - 1)dx = \int\limits_0^3 {({x^2} - x)dx = \frac{{{x^3}}}{3}} } - \frac{{{x^2}}}{2}\left| {_0^3} \right. = 9 - \frac{9}{2} = \frac{9}{2}\)
+)\[\mathop \smallint \limits_0^{\ln \sqrt {10} } {e^{2x}}dx = \frac{{{e^{2x}}}}{2}\left| {_0^{\ln \sqrt {10} }} \right. = \frac{{{e^{2\ln \sqrt {10} }} - 1}}{2} = \frac{9}{2}\]
+)\[3\mathop \smallint \limits_0^{3\pi } \sin xdx = - 3cosx\left| {_0^{3\pi }} \right. = 6\]
+)\[\mathop \smallint \limits_0^2 \left( {{x^2} + x - 3} \right)dx = \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} - 3x} \right)\left| {_0^2} \right. = \frac{8}{3} + 2 - 6 = - \frac{4}{3}\]
+)\[\mathop \smallint \limits_0^\pi \cos (3x + \pi )dx = \frac{1}{3}sin(3x + \pi )\left| {_0^\pi } \right. = \frac{1}{3}(sin4\pi - sin\pi ) = 0\]
Đáp án cần chọn là: C
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