Giải SBT Toán 12 Chân trời sáng tạo Bài 2. Tích phân có đáp án

a) \[\int\limits_0^2 {\left( {3x - 2} \right)\left( {3x + 2} \right)dx} = \int\limits_0^2 {\left( {9{x^2} - 4} \right)dx} \]

                                        \[ = \left. {\left( {3{x^3} - 4x} \right)} \right|_0^2\]

                                        = (3.2³ – 4.2) – (3.0³ – 4.0) = 16.

b) \[\int\limits_1^2 {{t^2}\left( {5{t^2} - 2} \right)dt} = \int\limits_1^2 {\left( {5{t^4} - 2{t^2}} \right)dt} \]

                             \[ = \left. {\left( {{t^5} - \frac{2}{3}{t^3}} \right)} \right|_1^2\]

                             \[ = \left( {{2^5} - \frac{2}{3}{{.2}^3}} \right) - \left( {{1^5} - \frac{2}{3}{{.1}^3}} \right)\]

                                          \[ = \frac{{79}}{3}\].

c) \[\int\limits_{ - 1}^1 {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)dx} = \int\limits_{ - 1}^1 {\left( {{x^3} - 8} \right)dx} \]

                                               \[ = \left. {\left( {\frac{{{x^4}}}{4} - 8x} \right)} \right|_{ - 1}^1 = - 16\].

a) \[\int\limits_1^2 {\frac{{1 - 2x}}{{{x^2}}}dx} = \int\limits_1^2 {\left( {\frac{1}{{{x^2}}} - \frac{2}{x}} \right)dx} = \left. {\left( { - \frac{1}{x} - 2\ln \left| x \right|} \right)} \right|_1^2\]

                       \[ = \left( { - \frac{1}{2} - 2\ln 2} \right) - \left( { - 1 - 2\ln 1} \right) = \frac{1}{2} - 2\ln 2.\]

b)

 \[\int\limits_1^2 {{{\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)}^2}dx} = \int\limits_1^2 {\left( {x + \frac{1}{x} + 2} \right)dx} \]

                               \[ = \left. {\left( {\frac{{{x^2}}}{2} + \ln \left| x \right| + 2x} \right)} \right|_1^2 = \frac{7}{2} + \ln 2.\]

c) \[\int\limits_1^4 {\frac{{x - 4}}{{\sqrt x + 2}}dx} = \int\limits_1^4 {\frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x + 2}}} dx = \int\limits_1^4 {\left( {\sqrt x - 2} \right)dx} \]

                         \[ = \int\limits_1^4 {\left( {{x^{\frac{1}{2}}} - 2} \right)dx = \left. {\left( {\frac{2}{3}x\sqrt x - 2x} \right)} \right|_1^4 = - \frac{4}{3}.} \]

a) \[\int\limits_1^3 {{e^{x - 2}}dx}  = \int\limits_1^3 {\frac{{{e^x}}}{{{e^2}}}dx} \]

                  \[ = \left. {\frac{{{e^x}}}{{{e^2}}}} \right|_1^3 = \frac{{{e^3}}}{{{e^2}}} - \frac{e}{{{e^2}}} = e - \frac{1}{e}\].

b) \[\int\limits_0^1 {{{\left( {{2^x} - 1} \right)}^2}dx}  = \int\limits_0^1 {\left( {{4^x} - {{2.2}^x} + 1} \right)dx} \]

                           \[ = \left. {\left( {\frac{{{4^x}}}{{\ln 4}} - 2.\frac{{{2^x}}}{{\ln 2}} + x} \right)} \right|_0^1\]

                                       \[ = 1 - \frac{1}{{2\ln 2}}\].

c) \[\int\limits_0^1 {\frac{{{e^{2x}} - 1}}{{{e^x} + 1}}dx}  = \int\limits_0^1 {\frac{{\left( {{e^x} + 1} \right)\left( {{e^x} - 1} \right)}}{{{e^x} + 1}}dx} \]

                       \[ = \int\limits_0^1 {\left( {{e^x} - 1} \right)dx = \left. {\left( {{e^x} - x} \right)} \right|_0^1 = e - 2} \].

a)

\[\int\limits_0^\pi {\left( {2\cos x + 1} \right)dx} = \left. {\left( {2\sin x + x} \right)} \right|_0^\pi \]

                            \[ = \left( {2\sin \pi + \pi } \right) - \left( {2\sin 0 + 0} \right) = \pi .\]

b) \[\int\limits_0^\pi {\left( {1 + \cot x} \right){\rm{sinx}}dx} = \int\limits_0^\pi {\left( {1 + \frac{{\cos x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}} \right){\rm{sinx}}dx} \]

                                     \[ = \int\limits_0^\pi {\left( {\sin {\rm{x}} + \cos x} \right)dx} \]

                                     \[ = \left. {\left( { - \cos x + \sin {\rm{x}}} \right)} \right|_0^\pi = 2.\]

c)

\[\int\limits_0^{\frac{\pi }{4}} {{{\tan }^2}xdx} = \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx} \]

                  \[ = \left. {\left( {\tan x - x} \right)} \right|_0^{\frac{\pi }{4}}\]

                           \[ = \left( {\tan \frac{\pi }{4} - \frac{\pi }{4}} \right) - \left( {\tan 0 - 0} \right) = 1 - \frac{\pi }{4}\].

Ta có:

\[f\left( 4 \right) - f\left( 1 \right) = \int\limits_1^4 {f'\left( x \right)} dx = \int\limits_1^4 {\frac{{\sqrt x  - 1}}{x}} dx\]

                                       \[ = \int\limits_1^4 {\left( {\frac{1}{{\sqrt x }} - \frac{1}{x}} \right)dx} \]

                                                        \[ = \left. {\left( {2\sqrt x  - \ln x} \right)} \right|_1^4 = 2 - 2\ln 2.\]

Vậy f(4) – f(1) = 2 – 2ln2.

a) \[A = \int\limits_{ - 1}^2 {\left( {x - 4{x^2}} \right)dx + 4\int\limits_{ - 1}^2 {\left( {{x^2} - 1} \right)dx} } \]

        \[ = \int\limits_{ - 1}^2 {\left( {x - 4{x^2}} \right)dx + \int\limits_{ - 1}^2 {4\left( {{x^2} - 1} \right)dx} } \]

        \[ = \int\limits_{ - 1}^2 {\left( {x - 4{x^2} + 4{x^2} - 4} \right)dx = \int\limits_{ - 1}^2 {\left( {x - 4} \right)} } dx\]

        \[ = \left. {\left( {\frac{{{x^2}}}{2} - 4x} \right)} \right|_{ - 1}^2 =  - \frac{{21}}{2}\].

Vậy \[A =  - \frac{{21}}{2}\].

b) \[B = \int\limits_{ - 1}^0 {\left( {{x^3} - 6x} \right)dx}  + \int\limits_0^1 {\left( {{t^3} - 6t} \right)dt} \]

       \[ = \int\limits_{ - 1}^0 {\left( {{x^3} - 6x} \right)dx}  + \int\limits_0^1 {\left( {{x^3} - 6x} \right)dx} \]

       \[ = \int\limits_{ - 1}^1 {\left( {{x^3} - 6x} \right)dx}  = \left. {\left( {\frac{{{x^4}}}{4} - 3{x^2}} \right)} \right|_{ - 1}^1 = 0\].