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

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

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

Ta có: \[\int\limits_0^5 {f\left( t \right)dt = } \int\limits_0^5 {f\left( x \right)dx = 4} .\]

Có: \[\int\limits_0^5 {f\left( x \right)dx = \int\limits_0^4 {f\left( x \right)dx}  + \int\limits_4^5 {f\left( x \right)dx} } \]

 ⇒ \[\int\limits_4^5 {f\left( x \right)dx}  = \int\limits_0^5 {f\left( x \right)dx - \int\limits_0^4 {f\left( x \right)dx} } \] = 4 – (−2) = 6.

Vậy \[\int\limits_4^5 {f\left( x \right)dx}  = 6\].

a) Ta có: x² + x – 2 = 0 ⇔ (x + 2)(x – 1) = 0 ⇔ x = 1 hoặc x = −2.

Ta có: x² + x – 2 ≤ 0 với mọi x ∈ [−1; 1] và x² + x – 2 ≤ 0 với mọi x ∈ [1; 2].

Suy ra, \[\int\limits_{ - 1}^2 {\left| {{x^2} + x - 2} \right|} dx\]

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

         \[ = \left. { - \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} - 2x} \right)} \right|_{ - 1}^1 + \left. {\left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} - 2x} \right)} \right|_1^2 = \frac{{31}}{6}.\]

b) \[\int\limits_{ - 1}^1 {\left| {{e^x} - 1} \right|} dx\]

Ta có: e^x – 1 = 0 ⇔ x = 0.

Ta có e^x – 1 ≤ 0 với mọi x ∈ [−1; 0] và e^x – 1 ≥ 0 với mọi x ∈ [0; 1].

Từ đó, \[\int\limits_{ - 1}^1 {\left| {{e^x} - 1} \right|} dx = \int\limits_{ - 1}^0 {\left( {1 - {e^x}} \right)dx} + \int\limits_0^1 {\left( {{e^x} - 1} \right)dx} \]

                              \[ = \left. {\left( {x - {e^x}} \right)} \right|_{ - 1}^0 + \left. {\left( {{e^x} - x} \right)} \right|_0^1 = e + \frac{1}{e} - 2\].

Ta có: F(x) = \[\sqrt {4x + 1} \]

           \[F'\left( x \right) = \frac{2}{{\sqrt {4x + 1} }},x >  - \frac{1}{4}\].

Nhận thấy \[\frac{1}{{\sqrt {4x + 1} }} = \frac{{F'\left( x \right)}}{2}\].

Do đó \[\int\limits_0^1 {\frac{1}{{\sqrt {4x + 1} }}dx}  = \int\limits_0^1 {\frac{{F'\left( x \right)}}{2}dx} \]

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

                                \[ = \frac{1}{2}\left[ {F\left( 1 \right) - F\left( 0 \right)} \right] = \frac{1}{2}\left( {\sqrt 5  - 1} \right).\]

Theo giả thiết, ta có y = f(x) đi qua điểm (−1; 3) hay f(−1) = 3 và f'(x) = 3x2 – 4x + 1.

Ta có: f(2) – f(−1) = \[\int\limits_{ - 1}^2 {f'\left( x \right)dx}  = \int\limits_{ - 1}^2 {\left( {3{x^2} - 4x + 1} \right)dx} \]

                              \[ = \left. {\left( {{x^3} - 2{x^2} + x} \right)} \right|_{ - 1}^2 = 6\].

Suy ra f(2) – f(−1) = 6 hay f(2) – 3 = 6 suy ra f(2) = 9.

a) Hàm số f(x) liên tục trên khoảng (−∞; 1) và (1; +∞).

Ta có: \[\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} {x^2} = 1;{\rm{ }}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{x} = 1;{\rm{ }}f\left( x \right) = 1\].

Suy ra hàm số f(x) liên tục tại x = 1.

Vậy hàm số f(x) liên tục trên ℝ.

b) Ta có: \[\int\limits_{ - 1}^2 {f\left( x \right)dx}  = \int\limits_{ - 1}^1 {{x^2}dx}  + \int\limits_1^2 {\frac{1}{x}dx} \]

                                \[ = \left. {\frac{{{x^3}}}{3}} \right|_{ - 1}^1 + \left. {\ln \left| x \right|} \right|_{ - 1}^2 = \frac{2}{3} + \ln 2.\]

Ta có: \[\int\limits_0^3 {T'\left( t \right)dt}  = \int\limits_0^3 {\left( { - 140{e^{ - 2t}}} \right)dt} \]

                           \[ =  - 140\int\limits_0^3 {{{\left( {{e^{ - 2}}} \right)}^t}dt} \]

                                        \[ = \left. {\frac{{ - 140{e^{ - 2t}}}}{{\ln {e^{ - 2}}}}} \right|_0^3 = 70\left( {{e^{ - 6}} - 1} \right)\].

Theo đề, T(0) = 100℃.

Ta có: T(3) – T(0) = 70(e−6 – 1) ⇒ T(3) = 100 + 70(e−6 – 1) ≈ 30,2℃.

Vậy nhiệt độ của vật ở thời điểm 3 phút kể từ khi đặt vào môi trường là 30,2℃.

a) Kí hiệu h(t) là độ cao của vật (tính theo mét) tại thời điểm t (0 ≤ t ≤ 4).

Ta có:  h'(t) = v(t) và h(0) = 0.

Từ đó, \[h\left( 3 \right) - h\left( 0 \right) = \int\limits_0^3 {v\left( t \right)dt = \int\limits_0^3 {\left( {20 - 10t} \right)dt} } \]

                                 \[ = \left. {\left( {20t - 5{t^2}} \right)} \right|_0^3 = 15{\rm{ }}\left( m \right).\]

Suy ra h(3) = 15 + h(0) = 15 + 0 = 15 (m).

b) Quãng đường vật đi được trong 3 giây đầu là:

s = \[\int\limits_0^3 {\left| {v\left( t \right)} \right|dt = } \int\limits_0^3 {\left| {20 - 10t} \right|dt} \]

  \[ = \int\limits_0^2 {\left( {20 - 10t} \right)dt + \int\limits_2^3 {\left( {10t - 20} \right)} } dt\]

  \[ = \left. {\left( {20t - 5{t^2}} \right)} \right|_0^2 + \left. {\left( {5{t^2} - 20t} \right)} \right|_0^2\]  = 20 + 5 = 25 (m).