ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Tích phân

Ta có:\[\mathop \smallint \limits_1^4 \left[ {f\left( x \right) + g\left( x \right)} \right]dx = \mathop \smallint \limits_1^4 f\left( x \right)dx + \mathop \smallint \limits_1^4 f\left( x \right)dx = 10\] nên A đúng.

\[\mathop \smallint \limits_1^4 f\left( x \right)dx = \mathop \smallint \limits_1^3 f\left( x \right)dx + \mathop \smallint \limits_3^4 f\left( x \right)dx \Rightarrow \mathop \smallint \limits_3^4 f\left( x \right)dx = \mathop \smallint \limits_1^4 f\left( x \right)dx - \mathop \smallint \limits_1^3 f\left( x \right)dx = 3 - \left( { - 2} \right) = 5\]nên C đúng, B sai.

\[\mathop \smallint \limits_1^4 \left[ {4f\left( x \right) - 2g\left( x \right)} \right]dx = 4\mathop \smallint \limits_1^4 f\left( x \right)dx - 2\mathop \smallint \limits_1^4 g\left( x \right)dx = - 2\]nên D đúng.

Ta có:\(I = \int\limits_1^1 {{x^5}dx = \frac{{{x^6}}}{6}} \left| {_1^2} \right. = \frac{{21}}{2}\)

Ta có:\[\int\limits_1^2 {\frac{{dx}}{{x + 3}}} = \ln \left| {x + 3} \right|\left| {_1^2} \right. = ln5 - ln4 = ln\frac{5}{4}\]

Do đó\[a = 5,b = 4\]

Khi đó:\[3a - b = 3.5 - 4 = 11 < 12\] nên A đúng.

\[a + 2b = 5 + 2.4 = 13\] nên B đúng.

\[a - b = 5 - 4 = 1 < 2\] nên C sai.

\[{a^2} + {b^2} = {5^2} + {4^2} = 41\] nên D đúng.

Ta có: \[2f\left( x \right) + xf\left( {\frac{1}{x}} \right) = x\] với\[x = \frac{1}{t}\]ta có \[2f\left( {\frac{1}{t}} \right) + \frac{1}{t}f\left( t \right) = \frac{1}{t}\]

\[ \Rightarrow f\left( {\frac{1}{t}} \right) = \frac{1}{2}\left( {\frac{1}{t} - \frac{1}{t}f\left( t \right)} \right)\]

\[ \Rightarrow f\left( {\frac{1}{x}} \right) = \frac{1}{2}\left( {\frac{1}{x} - \frac{1}{x}f\left( x \right)} \right)\]

Khi đó ta có

\[\begin{array}{*{20}{l}}{2f\left( x \right) + \frac{1}{2}x\left( {\frac{1}{x} - \frac{1}{x}f\left( x \right)} \right) = x}\\{ \Leftrightarrow 2f\left( x \right) + \frac{1}{2} - \frac{1}{2}f\left( x \right) = x}\\{ \Leftrightarrow \frac{3}{2}f\left( x \right) = x - \frac{1}{2}}\\{ \Leftrightarrow \frac{3}{2}\mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \mathop \smallint \limits_{\frac{1}{2}}^2 \left( {x - \frac{1}{2}} \right)dx}\\{ \Leftrightarrow \frac{3}{2}\mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \frac{9}{8} \Leftrightarrow \mathop \smallint \limits_{\frac{1}{2}}^2 f\left( x \right)dx = \frac{3}{4}}\end{array}\]

\[I = \mathop \smallint \limits_2^5 \frac{{dx}}{x} = ln|x|\left| {_2^5} \right. = ln5 - ln2 = ln\frac{5}{2}\]

Tính tích phân cho từng hàm số trong các đáp án:

+)\(\int\limits_0^\pi {cos3xdx = \frac{1}{3}sin3x} \left| {_0^\pi } \right. = 0\)

+)\(\int\limits_0^\pi {cos3xdx = - \frac{1}{3}sin3x} \left| {_0^\pi } \right. = \frac{2}{3}\)

+)\(\int\limits_0^\pi {cos\left( {\frac{x}{4} + \frac{\pi }{2}} \right)dx = 4sin\left( {\frac{x}{4} + \frac{\pi }{2}} \right)} \left| {_0^\pi } \right. = 2\left( {\sqrt 2 - 2} \right)\)

+)\(\int\limits_0^\pi {cos\left( {\frac{x}{4} + \frac{\pi }{2}} \right)dx = - 4sin\left( {\frac{x}{4} + \frac{\pi }{2}} \right)} \left| {_0^\pi } \right. = 2\sqrt 2 \)

Cách 1:

\[\begin{array}{l}I = \mathop \smallint \limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} \frac{{dx}}{{\sin x}}\\ = \mathop \smallint \limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} \frac{{\left( {co{s^2}\frac{x}{2} + si{n^2}\frac{x}{2}} \right)}}{{2sin\frac{x}{2}cos\frac{x}{2}}}dx\\ = \frac{1}{2}\mathop \smallint \limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} \left( {cot\frac{x}{2} + tan\frac{x}{2}} \right)dx\\ = \left[ {\ln \left| {sin\frac{x}{2}} \right| - \ln \left| {cos\frac{x}{2}} \right|} \right]\left| {_{\frac{\pi }{3}}^{\frac{\pi }{2}}} \right.\\ = \left[ {\ln \frac{{\sqrt 2 }}{2} - \ln \frac{{\sqrt 2 }}{2}} \right] - \left[ {\ln \frac{1}{2} - \ln \frac{{\sqrt 3 }}{2}} \right]\\ = \ln \sqrt 3 \end{array}\]

Cách 2:

Bước 1: Dùng máy tính như hình dưới, thu được giá trị 0,549306...

Bước 2: Lấy\[{e^{0,549306...}}\]cho kết quả \[1,732050808... \approx \sqrt 3 \]Chọn\[\frac{1}{2}\ln 3\]

Cách 3:

Thực hiện các phép tính sau trên máy tính (đến khi thu được kết quả bằng 0 thì ngưng)

Chọn \[\frac{1}{2}\ln 3\]

\(\int\limits_0^1 {\frac{1}{{{x^2} - x - 2}}} dx = \int\limits_0^1 {\frac{1}{{(x - 2)(x + 1)}}} dx = \frac{1}{3}\)

\(\int\limits_0^1 {\left[ {\frac{1}{{x - 2}} - \frac{1}{{x + 1}}} \right]} dx = \frac{1}{3}\left[ {\ln \left| {x - 2} \right| - \ln \left| {x + 1} \right|} \right]\left| {_0^1} \right. = - \frac{{2ln2}}{3}\)

Ta có :\(\int\limits_0^3 {x(x - 1)dx = \int\limits_0^3 {({x^2} - x)dx = \frac{{{x^3}}}{3}} } - \frac{{{x^2}}}{2}\left| {_0^3} \right. = 9 - \frac{9}{2} = \frac{9}{2}\)

+)\[\mathop \smallint \limits_0^{\ln \sqrt {10} } {e^{2x}}dx = \frac{{{e^{2x}}}}{2}\left| {_0^{\ln \sqrt {10} }} \right. = \frac{{{e^{2\ln \sqrt {10} }} - 1}}{2} = \frac{9}{2}\]

+)\[3\mathop \smallint \limits_0^{3\pi } \sin xdx = - 3cosx\left| {_0^{3\pi }} \right. = 6\]

+)\[\mathop \smallint \limits_0^2 \left( {{x^2} + x - 3} \right)dx = \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} - 3x} \right)\left| {_0^2} \right. = \frac{8}{3} + 2 - 6 = - \frac{4}{3}\]

+)\[\mathop \smallint \limits_0^\pi \cos (3x + \pi )dx = \frac{1}{3}sin(3x + \pi )\left| {_0^\pi } \right. = \frac{1}{3}(sin4\pi - sin\pi ) = 0\]

\[I = \mathop \smallint \limits_0^2 {x^3}dx = \frac{{{x^4}}}{4}\left| {_0^2} \right.\]và\[J = \int\limits_0^2 {xdx} = \frac{{{x^2}}}{2}\left| {_0^2} \right. = 2\]

Suy ra\[I.J = 8\]

Phương pháp tự luận

\[\begin{array}{*{20}{c}}{I = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} dx = \mathop \smallint \limits_0^{2\pi } \left| {\sin \frac{x}{2} + \cos \frac{x}{2}} \right|dx = \sqrt 2 \mathop \smallint \limits_0^{2\pi } \left| {\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|dx}\\{ = \sqrt 2 \left[ {\mathop \smallint \limits_0^{\frac{{3\pi }}{2}} \sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)dx - \mathop \smallint \limits_{\frac{{3\pi }}{2}}^{2\pi } \sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)dx} \right] = 4\sqrt 2 }\end{array}\]

\(\begin{array}{l}\int\limits_{ - 1}^5 {\mid x2 - 2x - 3\mid dx = } \int\limits_{ - 1}^5 {|(x - 3)(x + 1)|dx = - } \int\limits_{ - 1}^3 {({x^2} - 2x - 3)dx + } \int\limits_3^5 {({x^2} - 2x - 3)dx} \\ = - \left( {\frac{{{x^3}}}{3} - {x^2} - 3x} \right)\left| {_{ - 1}^3} \right. + \left( {\frac{{{x^3}}}{3} - {x^2} - 3x} \right)\left| {_3^5} \right. = \frac{{64}}{3}\end{array}\)

\[\begin{array}{l}\mathop \smallint \limits_2^3 \frac{{{x^2} - x + 4}}{{x + 1}}dx = \mathop \smallint \limits_2^3 \left( {x - 2 + \frac{6}{{x + 1}}} \right)dx\\ = \left( {\frac{{{x^2}}}{2} - 2x + 6ln|x + 1|} \right)\left| {_2^3} \right. = \frac{1}{2} + 6ln\frac{4}{3}\end{array}\]

Ta có:

\[\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\]

\[ = \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]\]

\[\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\]

\[ = \left( {1 - \frac{1}{2}{{\sin }^2}2x} \right)\left( {1 - \frac{3}{4}{{\sin }^2}2x} \right) = 1 - \frac{5}{4}{\sin ^2}2x + \frac{3}{8}{\left( {{{\sin }^2}2x} \right)^2}\]

\[ = 1 - \frac{5}{4}.\frac{{1 - \cos 4x}}{2} + \frac{3}{8}.{\left( {\frac{{1 - \cos 4x}}{2}} \right)^2}\]

\[ = \frac{3}{8} + \frac{5}{8}\cos 4x + \frac{3}{{32}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right) = \frac{{15}}{{32}} + \frac{7}{{16}}\cos 4x + \frac{3}{{32}}{\cos ^2}4x\]

\[ = \frac{{15}}{{32}} + \frac{7}{{16}}\cos 4x + \frac{3}{{32}}.\frac{{1 + \cos 8x}}{2} = \frac{{33}}{{64}} + \frac{7}{{16}}\cos 4x + \frac{3}{{64}}\cos 8x\]

Do đó\[I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{33}}{{64}} + \frac{7}{{16}}\cos 4x + \frac{3}{{64}}\cos 8x} \right)dx\]

\[ = \frac{{33}}{{64}}x\left| {_0^{\frac{\pi }{2}}} \right. + \frac{7}{{64}}sin4x\left| {_0^{\frac{\pi }{2}}} \right. + \frac{3}{{512}}sin8x\left| {_0^{\frac{\pi }{2}}} \right. = = \frac{{33}}{{128}}\pi \]

Ta có\[f\left( x \right) = \smallint f'\left( x \right)dx = \smallint {\sin ^4}xdx\]

\[\begin{array}{*{20}{l}}{ = \smallint {{\left( {\frac{{1 - \cos 2x}}{2}} \right)}^2}dx}\\{ = \frac{1}{4}\smallint \left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx}\\{ = \frac{1}{4}\smallint \left( {1 - 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)dx}\\{ = \frac{1}{4}\left( {x - \sin 2x + \frac{1}{2}x + \frac{1}{2}.\frac{{\sin 4x}}{4}} \right) + C}\\{ = \frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}} + C}\end{array}\]

Theo bài ra ta có\[f\left( 0 \right) = 0 \Leftrightarrow C = 0 \Rightarrow f\left( x \right) = \frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}}\]

Vậy\[\mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( x \right)dx = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}}} \right)dx = \frac{{3{\pi ^2} - 16}}{{64}}\] (sử dụng MTCT).

Do hàm số\[f(x) = \sqrt {1 - \cos 2x} = \sqrt 2 \left| {\sin x} \right|\] là hàm liên tục và tuần hoàn với chu kì\[T = \pi \] nên ta có

\[\int\limits_0^T {f(x)dx = } \int\limits_T^{2T} {f(x)dx = } \int\limits_{2T}^{3T} {f(x)dx = ... = \int\limits_{(n - 1)T}^{nT} {f(x)dx} } \]

\( \Rightarrow \int\limits_0^{nT} {f(x)dx = } \int\limits_0^T {f(x)dx + \int\limits_T^{2T} {f(x)dx + \int\limits_{2T}^{3T} {f(x)dx + ... + \int\limits_{(n - 1)T}^{nT} {f(x)dx} = } } } n\int\limits_0^T {f(x)dx} \)

\( \Rightarrow \int\limits_0^{2017\pi } {\sqrt {1 - cos2x} } dx\)

\( = 2017\int\limits_0^\pi {\sqrt {1 - cos2x} } dx\)

\( = 2017\sqrt 2 \int\limits_0^\pi {sinxdx = 4034\sqrt 2 } \)

\[\begin{array}{*{20}{l}}{f(x) = A\sin \pi x + B \Rightarrow f'(x) = A\pi \cos \pi x}\\{f'(1) = 2 \Rightarrow A\pi \cos \pi = 2 \Rightarrow A = - \frac{2}{\pi }}\end{array}\]

\[\mathop \smallint \limits_0^2 f(x)dx = 4 \Rightarrow \mathop \smallint \limits_0^2 (A\sin \pi x + B)dx = 4\]

\[ \Rightarrow - \frac{A}{\pi }\cos 2\pi + 2B + \frac{A}{\pi }\cos 0 = 4 \Rightarrow B = 2\]

Ta có\[g\left( x \right) = 1 + 2\mathop \smallint \limits_0^x f\left( t \right)dt\]suy ra\(\left\{ {\begin{array}{*{20}{c}}{g(x) - 1 = 2\int\limits_0^2 {f(t)dt} }\\{g(0) = 1 + \int\limits_0^0 {f(t)dt} }\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{g\prime (x) = 2f(x) \Rightarrow f(x) = \frac{{g\prime (x)}}{2}}\\{g(0) = 1}\end{array}} \right.\)

Mà

\[g\left( x \right) \ge {\left[ {f\left( x \right)} \right]^3} \Leftrightarrow g\left( x \right) \ge {\left[ {\frac{{g'\left( x \right)}}{2}} \right]^3} \Leftrightarrow \sqrt[3]{{g\left( x \right)}} \ge \frac{{g'\left( x \right)}}{2} \Leftrightarrow \frac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}} \le 2\]

Với\[t \in \left[ {0;1} \right]\]Lấy tích phân hai vế ta được

\(\int\limits_0^t {\frac{{g\prime (x)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx} \)

\(\begin{array}{l} \Leftrightarrow \int\limits_0^t {{{[g(x)]}^{\frac{{ - 1}}{3}}}} d(g(x)) \le 2t\\ \Leftrightarrow 2t \ge 32{[g(x)]^{\frac{2}{3}}}\left| {_0^t} \right.\\ \Leftrightarrow \frac{4}{3}t \ge \sqrt[3]{{{g^2}(t)}} - \sqrt[3]{{{g^2}(0)}}\end{array}\)

Mà\[g\left( 0 \right) = 1\] nên\[\sqrt[3]{{{g^2}\left( t \right)}} \le \frac{4}{3}t + 1 \Rightarrow \sqrt[3]{{{g^2}\left( x \right)}} \le \frac{4}{3}x + 1\]

Từ đó ta có\[\mathop \smallint \limits_0^1 \sqrt[3]{{{g^2}\left( x \right)}}\,dx \le \mathop \smallint \limits_0^1 \left( {\frac{4}{3}x + 1} \right)dx\]

\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}(x)}}} dx \le \left( {\frac{2}{3}{x^2} + x} \right)\left| {_0^1} \right.\)

\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}(x)}}} dx \le \frac{5}{2}\)

Hay giá trị lớn nhất cần tìm là\[\frac{5}{3}.\]