Câu hỏi
16/12/2024 3Tính:
a)
b)
c)
d)
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Lời giải của Vua Trắc Nghiệm
a) Ta có \(\int\limits_0^3 {{{\left( {3x – 1} \right)}^2}} dx\)\( = \int\limits_0^3 {\left( {9{x^2} – 6x + 1} \right)} dx\)\( = 9\int\limits_0^3 {{x^2}} dx – 6\int\limits_0^3 x dx + \int\limits_0^3 {dx} \)
\( = \left. {3{x^3}} \right|_0^3 – \left. {3{x^2}} \right|_0^3 + \left. x \right|_0^3\)\( = 81 – 27 + 3 = 57\).
b) Ta có \(\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \sin x} \right)dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {dx} + \int\limits_0^{\frac{\pi }{2}} {\sin xdx} \)\( = \left. x \right|_0^{\frac{\pi }{2}} – \left. {\cos x} \right|_0^{\frac{\pi }{2}}\)\( = \frac{\pi }{2} + 1\).
c) Ta có \(\int\limits_0^1 {\left( {{e^{2x}} + 3{x^2}} \right)} dx\)\( = \int\limits_0^1 {{e^{2x}}} dx + 3\int\limits_0^1 {{x^2}} dx\)\( = \left. {\frac{{{e^{2x}}}}{2}} \right|_0^1 + \left. {{x^3}} \right|_0^1\)\( = \frac{{{e^2}}}{2} – \frac{1}{2} + 1 = \frac{{{e^2}}}{2} + \frac{1}{2}\).
d) Ta có \(\int\limits_{ – 1}^2 {\left| {2x + 1} \right|dx} \)\( = \int\limits_{ – 1}^{\frac{{ – 1}}{2}} {\left| {2x + 1} \right|dx + \int\limits_{\frac{{ – 1}}{2}}^2 {\left| {2x + 1} \right|dx} } \)\( = \int\limits_{ – 1}^{\frac{{ – 1}}{2}} {\left( { – 2x – 1} \right)dx + \int\limits_{\frac{{ – 1}}{2}}^2 {\left( {2x + 1} \right)dx} } \)
\[ = \left. {\left( { – {x^2} – x} \right)} \right|_{ – 1}^{\frac{{ – 1}}{2}} + \left. {\left( {{x^2} + x} \right)} \right|_{\frac{{ – 1}}{2}}^2\]\[ = \frac{1}{4} + 6 + \frac{1}{4} = \frac{{13}}{2}\].