Giải SBT Toán 12 Tập 2 KNTT Bài 11. Nguyên hàm có đáp án

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

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

                  = \(\int {3\sqrt x dx + \int {\frac{2}{{\sqrt[3]{x}}}} } dx\)

                  = 2x\(\sqrt x \) + 3\(\sqrt[3]{{{x^2}}}\) + C.

Mà f(1) = 1 nên 2 + 3 + C = 1 hay C = −4.

Vậy f(x) = 2x\(\sqrt x \) + 3\(\sqrt[3]{{{x^2}}}\) − 4.

a) \(\int {\frac{{{{\left( {x + 2} \right)}^2}}}{{{x^4}}}dx} \) = \(\int {\frac{{{x^2} + 2x + 1}}{{{x^4}}}dx} \)

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

                          = \(\int {\frac{1}{{{x^2}}}dx + \int {\frac{2}{{{x^3}}}dx + \int {\frac{1}{{{x^4}}}dx} } } \)

                          = \( - \frac{1}{x}\) + 4.\(\frac{{{x^{ - 2}}}}{{ - 2}}\) + 4.\(\frac{{{x^{ - 3}}}}{{ - 3}}\) + C

                          = \( - \frac{1}{x}\) − \(\frac{2}{{{x^2}}}\) − \(\frac{4}{{3{x^3}}}\) + C.

b) \(\int {\sqrt x \left( {7{x^2} + 6} \right)dx} \) = \(\int {\left( {7{x^2}\sqrt x + 6\sqrt x } \right)} dx\)

                                 = \(\int {7{x^2}\sqrt x dx + \int {6\sqrt x } } dx\)

                                 = \(7\int {{x^{\frac{5}{2}}}} dx + 6\int {{x^{\frac{1}{2}}}dx} \)

                                 = 2\({x^{\frac{7}{2}}} + 4{x^{\frac{3}{2}}}\) + C

                                 = 2x³\(\sqrt x \) + 4x\(\sqrt x \) + C.

a) \(\int {\left( {3x + 4} \right)\sqrt[3]{x}} dx\) = \(\int {\left( {3x\sqrt[3]{x} + 4\sqrt[3]{x}} \right)} dx\)

                               = \(\int {3x\sqrt[3]{x}dx + \int {4\sqrt[3]{x}dx} } \)

                               = \(\int {3{x^{\frac{4}{3}}}dx} \) + \(\int {4{x^{\frac{1}{3}}}} dx\)

                               = \(\frac{9}{7}{x^2}\sqrt[3]{x} + 3x\sqrt[3]{x}\) + C

                               = \(\left( {\frac{9}{7}{x^2} + 3x} \right)\sqrt[3]{x}\) + C

b) \(\int {\frac{{{{\left( {2x + 3} \right)}^2}}}{{\sqrt x }}dx} \) = \(\int {\frac{{4{x^2} + 12x + 9}}{{\sqrt x }}dx} \)

                            = \(\int {\left( {4x\sqrt x + 12\sqrt x + \frac{9}{{\sqrt x }}} \right)dx} \)

                            = \(\int {4x\sqrt x dx + \int {12\sqrt x } } dx + \int {\frac{9}{{\sqrt x }}dx} \)

                            = \(4\int {{x^{\frac{3}{2}}}} dx + 12\int {{x^{\frac{1}{2}}}dx + 9\int {{x^{\frac{{ - 1}}{2}}}dx} } \)

                            = \(\frac{8}{5}{x^2}\sqrt x + 8x\sqrt x + 18\sqrt x + C\)

                            = \(\left( {\frac{8}{5}{x^2} + 8x + 18} \right)\sqrt x \) + C.

 

a) \(\int {\left( {2{e^x} + \frac{1}{{{3^x}}}} \right)} dx\) = \(\int {2{e^x}dx + \int {\frac{1}{{{3^x}}}dx} } \)

                              = \(\int {2{e^x}dx + \int {{3^{ - x}}dx} } \)

                              = 2ex − \(\frac{1}{{{3^x}.\ln 3}}\) + C.

b) \(\int {\left( {{x^2} + {2^x}} \right)} dx\) = \(\int {{x^2}dx + \int {{2^x}dx} } \)

                           = \(\frac{{{x^3}}}{3} + \frac{{{2^x}}}{{\ln 2}} + C\).

a) \(\int {{{\left( {{2^x} + {3^x}} \right)}^2}dx} \) = \(\int {\left( {{2^{2x}} + {{2.6}^x} + {3^{2x}}} \right)dx} \)

                            = \(\int {\left( {{4^x} + {{2.6}^x} + {9^x}} \right)} dx\)

                            = \(\int {{4^x}dx + \int {{{2.6}^x}dx + \int {{9^x}dx} } } \)

                            = \(\frac{{{4^x}}}{{\ln 4}} + \frac{{{{2.6}^x}}}{{\ln 6}} + \frac{{{9^x}}}{{\ln 9}}\)+ C.

b) \(\int {{{\left( {{e^x} - {e^{ - x}}} \right)}^2}dx} \) = \(\int {\left( {{e^{2x}} - 2{e^x}.{e^{ - x}} + {e^{ - 2x}}} \right)} dx\)

                              = \(\int {{e^{2x}}dx + \int {{e^{ - 2x}}dx - \int {2dx} } } \)

                              = \(\frac{{{e^{2x}}}}{2} - \frac{{{e^{ - 2x}}}}{2} - 2x + C\).

a) \(\int {\left( {2\cos x + \frac{3}{{\sqrt x }}} \right)dx} \) = \(\int {2\cos xdx + \int {\frac{3}{{\sqrt x }}} } dx\)

                                     = 2sinx + 6\(\sqrt x \) + C.

b) \(\int {\left( {3\sqrt x - 4\sin x} \right)dx} \) = \(\int {3\sqrt x dx - \int {4\sin xdx} } \)

                                     = \(2x\sqrt x + 4\cos x + C.\)

a) \(\int {\left( {x + {{\sin }^2}\frac{x}{2}} \right)dx} \) = \(\int {xdx + \int {{{\sin }^2}\frac{x}{2}dx} } \)

                                 = \(\int {xdx + \int {\frac{{1 - \cos x}}{2}dx} } \)

                                 = \(\int {xdx + \int {\frac{1}{2}dx - \int {\frac{{\cos x}}{2}dx} } } \)

                                 = \(\frac{1}{2}{x^2} + \frac{1}{2}x - \frac{1}{2}\sin x + C\).

b) \(\int {{{\left( {2\tan x + \cot x} \right)}^2}dx} \) = \(\int {\left( {4{{\tan }^2}x + 4\tan x\cot x + {{\cot }^2}x} \right)dx} \)

                                       = \(\int {\left( {\frac{4}{{{{\cos }^2}x}} - 4 + 4 + \frac{1}{{{{\sin }^2}x}} - 1} \right)dx} \)

                                       = \(\int {\frac{4}{{{{\cos }^2}xdx}} + \int {\frac{1}{{{{\sin }^2}x}}dx - \int {1dx} } } \)

                                       = 4tanx – cotx – x + C.

Độ cao h(t) của viên đạn tại điểm t là:

h(t) = \(\int {\left( {150 - 9,8t} \right)dt} \) = 150t – 9,8\(\frac{{{t^2}}}{2}\)+ C = 150t – 4,9t² + C.

Thay t = 0 ta được h(0) = C = 0.

Vậy h(t) = 150t – 4,9t² (m).

a) Sau t = 3 giây, độ cao của viên đạn là:

h = h(3) = 150.3 – 4,9.3² = 405,9 (m).

b) Ta có: h(t) = 150t – 4,9t² (m).

               h'(t) = v(t) = 150 – 9,8t

               h'(t) = 0 ⇔ t = \(\frac{{150}}{{9,8}}\).

Ta có bảng xét dấu như sau:

Khi đó, viên đạn đạt độ cao lớn nhất tại thời điểm t_max = \(\frac{{150}}{{9,8}}\).

Như vậy h_max = 150t_max – 4,9\(t_{\max }^2\)≈ 1148,0 (m).

Ta có: F'(u) = f(u), với mọi u ∈ K.

\({\left[ {F\left( {u\left( x \right)} \right)} \right]^\prime }\) = \(F'\left( {u\left( x \right)} \right)\).u'(x) = \(f\left( {u\left( x \right)} \right).u'\left( x \right)\), với mọi x ∈ J.

Do đó, \(\int {f\left( {u(x)} \right).u'(x)dx} \) = \(F\left( {u\left( x \right)} \right)\) + C.

Áp dụng:

\(\int {{{\left( {2x + 1} \right)}^5}dx} \) = \({\int {\left( {2x + 1} \right)} ^5}\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx\)

                      = \(\frac{1}{2}\int {{{\left( {2x + 1} \right)}^5}{{\left( {2x + 1} \right)}^\prime }dx} \)

                      = \(\frac{1}{2}.\frac{{{{\left( {2x + 1} \right)}^6}}}{6} + C\)

                      = \(\frac{{{{\left( {2x + 1} \right)}^6}}}{{12}} + C\).

 \(\int {\frac{1}{{\sqrt {2x + 1} }}dx} \) = \(\int {\frac{1}{{\sqrt {2x + 1} }}.\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx} \)

                       = \(\int {\frac{1}{{2\sqrt {2x + 1} }}.{{\left( {2x + 1} \right)}^\prime }dx} \)

                       = \(\sqrt {2x + 1} + C.\)        

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

                      = \(\int {2dx - \int {\frac{3}{{x + 1}}dx} } \)

                      = 2x – 3ln\(\left| {x + 1} \right|\) + C.

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

                                 = \(\int {\left( {3 + 1 - \cos 2x} \right)dx} \)

                                 = \(\int {\left( {4 - \cos 2x} \right)dx} \)

                                 = 4x − \(\int {\cos 2x.\frac{{{{\left( {2x} \right)}^\prime }}}{2}dx} \)

                                 = 4x − \(\frac{{\sin 2x}}{2}\) + C.