Giải SBT Toán 12 Chân trời sáng tạo Bài 1. Nguyên hàm có đáp án

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

                        \[ = \int {{x^2}dx - \int {4xdx + \int {4dx} } } \]

                        \[ = \frac{{{x^3}}}{3} - 2{x^2} + 4x + C\].

b) \[\int {\left( {x - 1} \right)\left( {3x + 1} \right)dx}  = \int {\left( {3{x^2} - 2x - 1} \right)dx} \]

                                   \[ = \int {3{x^2}dx - \int {2xdx - \int {1dx} } } \]

                                   = x3 – x2 + x + C.

c) \[\int {\sqrt[3]{{{x^2}}}dx}  = \int {{x^{\frac{2}{3}}}dx = \frac{3}{5}{x^{\frac{5}{3}}} + C = \frac{3}{5}x\sqrt[3]{{{x^2}}}}  + C.\]

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

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

                        \[ = \int {\left( {{x^{ - \frac{1}{2}}} + {x^{\frac{1}{2}}} + {x^{\frac{3}{2}}}} \right)dx} \]

                        \[ = 2{x^{\frac{1}{2}}} - 2.\frac{2}{3}{x^{\frac{3}{2}}} + \frac{2}{5}{x^{\frac{5}{2}}} + C\]

                         \[ = 2\sqrt x  - \frac{4}{3}x\sqrt x  + \frac{2}{5}{x^2}\sqrt x  + C.\]

a) \[\int {\left( {{5^x} + 1} \right)\left( {{5^x} - 1} \right)dx}  = \int {\left( {{5^{2x}} - 1} \right)dx} \]

                                   \[ = \int {{5^{2x}}dx - \int {1dx = \int {{{25}^x}dx}  - \int {1dx} } } \]

                                   \[ = \frac{{{{25}^x}}}{{\ln 25}} - x + C = \frac{{{{25}^x}}}{{2\ln 5}} - x + C.\]

b) \[\int {{e^{ - 0,5x}}dx}  = \int {{{\left( {{e^{ - 0,5}}} \right)}^x}dx = \frac{{{{\left( {{e^{ - 0,5}}} \right)}^x}}}{{\ln {e^{ - 0,5}}}}}  + C\]

                   \[ = \frac{{{e^{ - 0,5x}}}}{{ - 0,5}} + C =  - 2{e^{ - 0,5x}} + C\].

c) \[\int {{2^{x - 1}}{{.5}^{2x + 1}}dx}  = \int {\frac{{{2^x}}}{2}{{.5}^{2x}}.5dx = \int {\frac{5}{2}{{.2}^x}{{.25}^x}dx} } \]

                        \[ = \frac{5}{2}\int {{{50}^x}dx = \frac{5}{2}.\frac{{{{50}^x}}}{{\ln 50}} + C.} \]

a)\[\int {\frac{{{{\cos }^2}x}}{{1 - \sin {\rm{x}}}}dx}  = \int {\frac{{1 - {{\sin }^2}x}}{{1 - \sin {\rm{x}}}}dx} \]

                        \[ = \int {\frac{{\left( {1 - \sin {\rm{x}}} \right)\left( {1 + \sin {\rm{x}}} \right)}}{{\left( {1 - \sin {\rm{x}}} \right)}}dx} \]

                        \[ = \int {\left( {1 + \sin {\rm{x}}} \right)dx}  = x - \cos x + C.\]

b) \[\int {\left( {1 + 3{{\sin }^2}\frac{x}{2}} \right)dx}  = \int {\left( {1 + 3.\frac{{1 - \cos x}}{2}} \right)dx} \]

                                \[ = \int {\left( {\frac{5}{2} - \frac{3}{2}\cos x} \right)dx} \]

                                \[ = \frac{5}{2}x - \frac{3}{2}\sin {\rm{x}} + C\].

c) \[\int {\frac{{2{{\cos }^3}x + 3}}{{{{\cos }^2}x}}dx}  = \int {\left( {2\cos x + \frac{3}{{{{\cos }^2}x}}} \right)dx} \]

                              \[ = \int {2\cos xdx + \int {\frac{3}{{{{\cos }^2}x}}dx} } \]

                                               \[ = 2\sin x + 3\tan x + C\]

a) \[f\left( x \right) = \int {f'\left( x \right)dx} \]

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

            = \[\frac{{{x^4}}}{2} - 2{x^2} + x + C.\]

Mà f(1) = 0 ⇒ \[\frac{1}{2} - 2 + 1 + C = 0\]⇒ \[C = \frac{1}{2}\].

Vậy \[f\left( x \right) = \frac{{{x^4}}}{2} - 2{x^2} + x + \frac{1}{2}.\]

b) Ta có: \[f\left( x \right) = \int {f'\left( x \right)dx} \]

                        \[ = \int {\left( {5\cos x - \sin {\rm{x}}} \right)dx} \]

                        \[ = 5\sin x + \cos x + C\].

Mà \[f\left( {\frac{\pi }{2}} \right) = 1\] nên \[5\sin \left( {\frac{\pi }{2}} \right) + \cos \left( {\frac{\pi }{2}} \right) + C = 1\] hay \[5 + C = 1\] suy ra  C = −4.

Vậy f(x) = 5sinx + cosx – 4.

Theo giả thiết, hệ số góc của tiếp tuyến tại mỗi điểm (x; f(x)) là \[\frac{{1 - x}}{{{x^2}}}\] với x > 0 hay \[f'\left( x \right) = \frac{{1 - x}}{{{x^2}}}\] với x > 0 và f(1) = 2.

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

                    \[ = \int {\left( {\frac{1}{{{x^2}}} - \frac{1}{x}} \right)dx =  - \frac{1}{x} - \ln x + C.} \]

Mà f(1) = 2 nên −1 – ln1 + C = 2 hay C = 3.

Vậy  \[f\left( x \right) - \frac{1}{x} - \ln x + 3.\]

Ta có:

\[F'\left( x \right) = {\left[ {\ln \left( {\sqrt {{x^2} + 4}  - x} \right)} \right]^\prime }\]

          \[ = \frac{1}{{\sqrt {{x^2} + 4}  - x}}.{\left( {\sqrt {{x^2} + 4}  - x} \right)^\prime }\]

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

            \[ = \frac{1}{{\sqrt {{x^2} + 4}  - x}}.\frac{{x - \sqrt {{x^2} + 4} }}{{\sqrt {{x^2} + 4} }}\]

            \[ =  - \frac{1}{{\sqrt {{x^2} + 4} }}{\rm{ }}\left( {x \in \mathbb{R}} \right).\]

Suy ra \[\int {\frac{1}{{\sqrt {{x^2} + 4} }}} dx = \int {\left[ { - F'\left( x \right)} \right]dx =  - \int {F'\left( x \right)dx} } \]

                                = \[ - F\left( x \right) + C =  - \ln \left( {\sqrt {{x^2} + 4}  - x} \right) + C.\]

Vậy \[\int {\frac{1}{{\sqrt {{x^2} + 4} }}dx}  =  - \ln \left( {\sqrt {{x^2} + 4}  - x} \right) + C.\]

a) Ta có: \[x\left( t \right) = \int {v\left( t \right)dt = \int {\left( {8 - 0,4t} \right)dt} } \] = 8t – 0,2t² + C.

Ban đầu vật ở gốc tọa độ nên x(0) = 0, suy ra C = 0.

Vậy x(t) = 8t – 0,2t² với t ≥ 0.

b) Ta có: x(t) = 0 ⇒ 8t – 0,2t² = 0 ⇔ t = 0 hoặc t = 40.

a) \[P\left( t \right) = \int {P'\left( t \right)dt = \int {150\sqrt t dt = 150\int {{t^{\frac{1}{2}}}dt} } } \]

            \[ = 150.\frac{2}{3}.{t^{\frac{3}{2}}} + C = 100t\sqrt t  + C\].

Theo giả thiết, ta có P(0) = 1 000, suy ra C = 1 000.

Do đó, \[P\left( t \right) = 100t\sqrt t  + 1000\].