Tích phân

Ta có:

\(\begin{array}{l}\int\limits_1^b {\left( {2x - 6} \right)dx}  = 0 \Leftrightarrow \int\limits_1^b {2xdx}  - \int\limits_1^b {6dx}  = 0 \Leftrightarrow \left. {{x^2}} \right|_1^b - \left. {6x} \right|_1^b = 0\\ \Leftrightarrow {b^2} - 1 - 6b + 6 = 0 \Leftrightarrow {b^2} - 6b + 5 = 0 \Leftrightarrow \left[ \begin{array}{l}b = 1\\b = 5\end{array} \right.\end{array}\)

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

Ta có:

\(\int\limits_0^a {\left( {\cos x + \sin x} \right)dx} {\rm{\;}} = 0\) \( \Leftrightarrow \left. {\sin x} \right|_0^a - \left. {\cos x} \right|_0^a = 0\) \( \Leftrightarrow \sin a - \cos a + 1 = 0\)

\(\begin{array}{l} \Leftrightarrow \sin a - \cos a =  - 1\\ \Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin a - \dfrac{1}{{\sqrt 2 }}.\cos a =  - \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin a.\cos \dfrac{\pi }{4} - \cos a.\sin \dfrac{\pi }{4} =  - \dfrac{1}{{\sqrt 2 }}\end{array}\)

\( \Leftrightarrow \sin \left( {a - \dfrac{\pi }{4}} \right) = {\rm{\;}} - \dfrac{1}{{\sqrt 2 }}\)

\( \Leftrightarrow \sin \left( {a - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{{ - \pi }}{4}} \right)\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{a - \dfrac{\pi }{4} = {\rm{\;}} - \dfrac{\pi }{4} + k2\pi }\\{a - \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi }\end{array}} \right.\)

$\Leftrightarrow a = \dfrac{{3\pi }}{2}\left( {0 < a < 2\pi } \right)$

$\int\limits_2^3 {\dfrac{{{x^2} - x + 4}}{{x + 1}}dx}  = \int\limits_2^3 {\left( {x - 2 + \dfrac{6}{{x + 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} - 2x + 6\ln \left| {x + 1} \right|} \right)} \right|_2^3 = \dfrac{1}{2} + 6\ln \dfrac{4}{3}$.

Ta có: \(\int\limits_1^2 {\dfrac{{dx}}{{x + 3}}}  = \left. {\ln \left| {x + 3} \right|} \right|_1^2 = \ln 5 - \ln 4 = \ln \dfrac{5}{4}\)

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: \(\int\limits_{ - 1}^0 {\left( {x + 1 + \dfrac{2}{{x - 1}}} \right)dx}  = \left. {\left( {\dfrac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right|} \right)} \right|_{ - 1}^0 \)

$= \dfrac{1}{2} - 2\ln 2 \Rightarrow \left\{ \begin{array}{l}a = \dfrac{1}{2}\\b =  - 2\end{array} \right. \Rightarrow a + b =  - \dfrac{3}{2}$

Ta có: \(\int\limits_2^4 {2xdx}  = \left. {{x^2}} \right|_2^4 = 12\)

\(\int\limits_1^2 {\left[ {{a^2} + (4 - 4a)x + 4{x^3}} \right]dx = } \)\( = \left. {\left[ {{a^2}x + (2 - 2a){x^2} + {x^4}} \right]} \right|_1^2 = {a^2} - 6a + 21\)

\( \Rightarrow {a^2} - 6a + 21 = 12 \Leftrightarrow a = 3.\)

$\int\limits_0^x {\sin 2tdt}  = \dfrac{1}{2}\int\limits_0^x {\sin 2td(2t)}  =  - \dfrac{1}{2}\left. {\cos 2t} \right|_0^x =  - \dfrac{1}{2}\left( {\cos 2x - \cos 0} \right) =  - \dfrac{1}{2}\cos 2x + \dfrac{1}{2}$

Khi đó $ - \dfrac{1}{2}\cos 2x + \dfrac{1}{2} = 0 \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi  \Leftrightarrow x = k\pi \left( {k \in Z} \right)$

$\int\limits_0^x {\left( {\dfrac{1}{2}t + 2\left( {a + 1} \right)} \right)dt}  = \left. {\left( {\dfrac{{{t^2}}}{4} + 2(a + 1)t} \right)} \right|_0^x = \dfrac{{{x^2}}}{4} + 2(a + 1)x$

Bất phương trình: $\dfrac{{{x^2}}}{4} + 2(a + 1)x \ge  - 1 \Leftrightarrow {x^2} + 8(a + 1)x + 4 \ge 0$ đúng với mọi giá trị thực của $x$ khi và chỉ khi: $64{\left( {a + 1} \right)^2} - 16 \le 0 \Leftrightarrow {\left( {a + 1} \right)^2} \le \dfrac{1}{4} \Leftrightarrow  - \dfrac{1}{2} \le a + 1 \le \dfrac{1}{2} \Leftrightarrow  - \dfrac{3}{2} \le a \le  - \dfrac{1}{2}$

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

\(\begin{array}{l}\int\limits_0^T {f\left( x \right)dx = \int\limits_T^{2T} {f\left( x \right)dx} }  = \int\limits_{2T}^{3T} {f\left( x \right)dx} \\ = ... = \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{nT} {f\left( x \right)dx}  = \int\limits_0^T {f\left( x \right)dx}  \\+ \int\limits_T^{2T} {f\left( x \right)dx}+ \int\limits_{2T}^{3T} {f\left( x \right)dx}   + ... + \\ \int\limits_{\left( {n - 1} \right)T}^{nT} {f\left( x \right)dx}  = n\int\limits_{0}^{T} {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^{2017\pi } {\sqrt {1 - \cos 2x} dx}  \\= 2017\int\limits_0^\pi  {\sqrt {1 - \cos 2x} dx} \\ = 2017\sqrt 2 \int\limits_0^\pi  {\sin x dx = 4034\sqrt 2 } \end{array}\)

Ta có

\(I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\cos 2x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x + 3} \right)}^2}}}dx} \)

Đặt \(\sin x - \cos x + 3 = t \Leftrightarrow \left\{ \begin{array}{l}\left( {\cos x + \sin x} \right)dx = dt\\\cos x - \sin x = 3 - t\end{array} \right.\)

Đổi cận \(x = 0 \Rightarrow t = 2;\,x = \dfrac{\pi }{4} \Rightarrow t = 3\)

Suy ra \(I = \int\limits_2^3 {\dfrac{{\left( {3 - t} \right)dt}}{{{t^2}}} = \int\limits_2^3 {\left( {\dfrac{3}{{{t^2}}} - \dfrac{1}{t}} \right)dt}  = \left. {\left( { - \dfrac{3}{t} - \ln \left| t \right|} \right)} \right|_2^3}  = \dfrac{1}{2} + \ln 2 - \ln 3 = \dfrac{1}{2} + \ln \dfrac{2}{3}\)

Hay \(a = \dfrac{1}{2};b = \dfrac{2}{3} \Rightarrow 2a + 3b = 3.\)

\(\int\limits_0^2 {f(x)dx}  = \int\limits_0^1 {{x^2}dx}  + \int\limits_1^2 {\left( {2 - x} \right)} dx = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6}\) .

Ta có \(g\left( x \right) = 1 + 2\int\limits_0^x {f\left( t \right)dt} \)  suy ra \(\left\{ \begin{array}{l}g\left( x \right) - 1 = 2\int\limits_0^x {f\left( t \right)dt} \\g\left( 0 \right) = 1 + \int\limits_0^0 {f\left( t \right)dt} \end{array} \right. \)\(\Leftrightarrow \left\{ \begin{array}{l}g'\left( x \right) = 2f\left( x \right) \Rightarrow f\left( x \right) = \dfrac{{g'\left( x \right)}}{2}\\g\left( 0 \right) = 1\end{array} \right.\)

Mà

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

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

\(\begin{array}{l}\int\limits_0^t {\dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx} \\ \Leftrightarrow \int\limits_0^t {{{\left[ {g\left( x \right)} \right]}^{\dfrac{{ - 1}}{3}}}} d\left( {g\left( x \right)} \right) \le 2t\\ \Leftrightarrow 2t \ge \dfrac{3}{2}\left. {{{\left[ {g\left( x \right)} \right]}^{\dfrac{2}{3}}}} \right|_0^t \\\Leftrightarrow \dfrac{4}{3}t \ge \sqrt[3]{{{g^2}\left( t \right)}} - \sqrt[3]{{{g^2}\left( 0 \right)}}\end{array}\)

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

Từ đó ta có \(\int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \int\limits_0^1 {\left( {\dfrac{4}{3}x + 1} \right)dx} \)\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \left. {\left( {\dfrac{2}{3}{x^2} + x} \right)} \right|_0^1\\ \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \dfrac{5}{3}\)

Hay giá trị lớn nhất cần tìm là \(\dfrac{5}{3}.\)

Ta có:  \(2f\left( x \right) + xf\left( {\dfrac{1}{x}} \right) = x\), với \(x = \dfrac{1}{t}\) ta có \(2f\left( {\dfrac{1}{t}} \right) + \dfrac{1}{t}f\left( t \right) = \dfrac{1}{t}\) \( \Rightarrow f\left( {\dfrac{1}{t}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{t} - \dfrac{1}{t}f\left( t \right)} \right)\)

\( \Rightarrow f\left( {\dfrac{1}{x}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right)\) 

Khi đó ta có

\(\begin{array}{l}2f\left( x \right) + \dfrac{1}{2}x\left( {\dfrac{1}{x} - \dfrac{1}{x}f\left( x \right)} \right) = x\\ \Leftrightarrow 2f\left( x \right) + \dfrac{1}{2} - \dfrac{1}{2}f\left( x \right) = x\\ \Leftrightarrow \dfrac{3}{2}f\left( x \right) = x - \dfrac{1}{2}\\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \int\limits_{\frac{1}{2}}^2 {\left( {x - \frac{1}{2}} \right)dx} \\ \Leftrightarrow \dfrac{3}{2}\int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{9}{8} \Leftrightarrow \int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx}  = \dfrac{3}{4}\end{array}\)

Theo đề bài ta có

 \(\begin{array}{l}\left\{ \begin{array}{l}f\left( 1 \right) = 10\\f\left( 2 \right) = 20\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 + a + b + c = 10\\{2^3} + {2^2}.a + 2b + c = 20\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + b + c = 9\\4a + 2b + c = 12\end{array} \right. \Rightarrow 3a + b = 3\end{array}\)

\(\begin{array}{l} \Rightarrow \int\limits_0^3 {f'\left( x \right)dx}  = \left. {f\left( x \right)} \right|_0^3 = f\left( 3 \right) - f\left( 0 \right)\\ = {3^3} + {3^2}.a + 3b + c - c = 27 + 9a + 3b\\ = 27 + 3\left( {3a + b} \right) = 27 + 3.3 = 36.\end{array}\)

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

\(\begin{array}{l} = \int {{{\left( {\dfrac{{1 - \cos 2x}}{2}} \right)}^2}dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx} \\ = \dfrac{1}{4}\int {\left( {1 - 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)dx} \\ = \dfrac{1}{4}\left( {x - \sin 2x + \dfrac{1}{2}x + \dfrac{1}{2}.\dfrac{{\sin 4x}}{4}} \right) + C\\ = \dfrac{{3x}}{8} - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + C\end{array}\) 

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

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

Bước 1:

Theo bài ra ta có:

\(2xf'\left( x \right) = f\left( x \right) + {x^2}\)

\( \Leftrightarrow \sqrt {2x} f'\left( x \right) - \dfrac{1}{{\sqrt {2x} }}f\left( x \right) = \dfrac{{{x^2}}}{{\sqrt {2x} }}\)

\( \Leftrightarrow \dfrac{{f'\left( x \right)\sqrt {2x} {\rm{\;}} - \dfrac{1}{{\sqrt {2x} }}f\left( x \right)}}{{2x}} = \dfrac{{{x^2}}}{{2x\sqrt {2x} }}\)

\( \Leftrightarrow \dfrac{{f'\left( x \right)\sqrt {2x} {\rm{\;}} - f\left( x \right){{\left( {\sqrt {2x} } \right)}^\prime }}}{{2x}} = \dfrac{x}{{2\sqrt {2x} }}\)

\( \Leftrightarrow {\left( {\dfrac{{f\left( x \right)}}{{\sqrt {2x} }}} \right)^\prime } = \dfrac{1}{{2\sqrt 2 }}\sqrt x \)

Bước 2:

Lấy nguyên hàm hai vế ta được:

\(\begin{array}{l}\int {\left( {\dfrac{{f\left( x \right)}}{{\sqrt {2x} }}} \right)'dx}  = \int {\dfrac{1}{{2\sqrt 2 }}\sqrt x dx} \\ \Leftrightarrow \dfrac{{f\left( x \right)}}{{\sqrt {2x} }} = \dfrac{1}{{2\sqrt 2 }}.\dfrac{2}{3}x\sqrt x  + C\\ \Leftrightarrow f\left( x \right) = \dfrac{1}{{3\sqrt 2 }}\sqrt {2x} .x\sqrt x  + C\sqrt {2x} \\ \Leftrightarrow f\left( x \right) = \dfrac{1}{3}{x^2} + C\sqrt {2x} \end{array}\)

Bước 3:

Ta lại có \(f\left( 1 \right) = \dfrac{1}{3} + C\sqrt 2  = 2 \Leftrightarrow C = \dfrac{{5\sqrt 2 }}{6}\) \( \Rightarrow f\left( x \right) = \dfrac{1}{3}{x^2} + \dfrac{{5\sqrt 2 }}{6}\sqrt {2x}  = \dfrac{1}{3}{x^2} + \dfrac{5}{3}\sqrt x \)

Bước 4:

Vậy \(\int\limits_1^4 {f\left( x \right)dx}  = \int\limits_1^4 {\left( {\dfrac{1}{3}{x^2} + \dfrac{5}{3}\sqrt x } \right)dx}  \)\(= \dfrac{{133}}{9}\).