Tích phân (phương pháp đổi biến)

Đặt \(t = f\left( x \right) \Rightarrow dt = f'\left( x \right)dx\)

Đổi cận: \(\left\{ \begin{array}{l}x = a \Rightarrow t = f\left( a \right)\\x = b \Rightarrow t = f\left( b \right)\end{array} \right.\)

Khi đó \(I = \int\limits_{f\left( a \right)}^{f\left( b \right)} {{e^t}dt}  = 0\) (Vì \(f\left( a \right) = f\left( b \right)\))

Đặt \(2x - 1 = t \Rightarrow 2dx = dt\). Đổi cận \(\left\{ \begin{array}{l}x =  - 1 \Rightarrow t =  - 3\\x = 3 \Rightarrow t = 5\end{array} \right.\).

Khi đó: \(I = \int\limits_{ - 1}^3 {f\left( {2x - 1} \right)dx}  = \int\limits_{ - 3}^5 {f\left( t \right)\dfrac{1}{2}dt}  = \dfrac{1}{2}\int\limits_{ - 3}^5 {f\left( t \right)dt} \)

               \(\begin{array}{l} = \dfrac{1}{2}\left( {\int\limits_{ - 3}^{ - 2} {f\left( x \right)dx}  + \int\limits_{ - 2}^0 {f\left( x \right)dx}  + \int\limits_0^2 {f\left( x \right)dx}  + \int\limits_2^3 {f\left( x \right)dx}  + \int\limits_3^5 {f\left( x \right)dx} } \right)\\ = \dfrac{1}{2}\left( { - {S_{\Delta ABC}} + {S_{\Delta OCD}} + {S_{\Delta OED}} + {S_{\Delta EFI}} + {S_{IFHK}}} \right)\\ = \dfrac{1}{2}\left( { - \dfrac{1}{2} + 2 + 2 + \dfrac{1}{2} + 5} \right) = \dfrac{9}{2}.\end{array}\)

Bước 1: Biểu diễn \(\left( {6{x^2} + 2x} \right)f\left( {2{x^3} + {x^2} + 1} \right)\)

Khi \(x \ge 0\), ta có:

\(f\left( {2{x^3} + {x^2} + 1} \right) = x + 2\)\( \Leftrightarrow \left( {6{x^2} + 2x} \right)f\left( {2{x^3} + {x^2} + 1} \right)\)\( = \left( {6{x^2} + 2x} \right)(x + 2)(*)\)

Bước 2: Sử dụng phương pháp đổi biến số tính \(\int_1^4 f (x)dx\)

Lấy tích phân từ 0 đến 1 hai vế của \((*)\) ta được

\(\int_0^1 {\left( {6{x^2} + 2x} \right)} f\left( {2{x^3} + {x^2} + 1} \right)dx\)\( = \int_0^1 {\left( {6{x^2} + 2x} \right)\left( {x + 2} \right)dx} \)

\( \Leftrightarrow \int_0^1 f \left( {2{x^3} + {x^2} + 1} \right){\rm{d}}\left( {2{x^3} + {x^2} + 1} \right)\)\( = \dfrac{{49}}{6}\)

Đặt \(t = 2{x^3} + {x^2} + 1\)

Đổi cận:\(x = 0 \Rightarrow t = 1;x = 1 \Rightarrow t = 4\)

\( \Rightarrow \int_1^4 f (t)dt = \dfrac{{49}}{6} \Leftrightarrow \int_1^4 f (x)dx = \dfrac{{49}}{6}\)

Dựa vào các đáp án, xét:

$\int\limits_{ - 1}^2 {f(2x)dx} $$ = \dfrac{1}{2}\int\limits_{ - 1}^2 {f(2x)d(2x)}  $$= \dfrac{1}{2}\int\limits_{ - 2}^4 {f(x)dx } $$= 1$

$\begin{array}{l}\int\limits_{ - 3}^3 {f(x + 1)dx}  = \int\limits_{ - 3}^3 {f(x + 1)d(x + 1)} \\ = \int\limits_{ - 2}^4 {f(x)dx = 2} \end{array}$

$\int\limits_0^6 {\dfrac{1}{2}f(x - 2)dx}  = \int\limits_0^6 {\dfrac{1}{2}f(x - 2)d(x - 2)} $$ = \dfrac{1}{2}\int\limits_{ - 2}^4 {f(x)dx = 1}$

Do đó các đáp án B, C, D đều đúng, đáp án A sai.

Xét \(\int\limits_0^1 {xf\left( {3x} \right)dx = 1} \).

Đặt \(t = 3x \Leftrightarrow dt = 3dx\). Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 3\end{array} \right.\).

\( \Rightarrow \int\limits_0^1 {xf\left( {3x} \right)dx}  = 1 \Leftrightarrow \int\limits_0^3 {\dfrac{t}{3}f\left( t \right)\dfrac{{dt}}{3}}  = 1\)\( \Leftrightarrow \dfrac{1}{9}\int\limits_0^3 {tf\left( t \right)dt}  = 1\)

\( \Leftrightarrow \int\limits_0^3 {tf\left( t \right)dt}  = 9 \Leftrightarrow 2\int\limits_0^3 {xf\left( x \right)dx}  = 18\)\( \Rightarrow \int\limits_0^3 {2xf\left( x \right)dx}  = 18\)

Ta có: \(\int\limits_0^3 {\left[ {{x^2}f'\left( x \right) + 2xf\left( x \right)} \right]dx} \)\( = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + \int\limits_0^3 {2xf\left( x \right)dx} \)

\(\begin{array}{l} \Leftrightarrow \int\limits_0^3 {\left( {{x^2}f\left( x \right)} \right)'dx}  = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow \left. {{x^2}f\left( x \right)} \right|_0^3 = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow 9f\left( 3 \right) = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow 9.1 = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow \int\limits_0^3 {{x^2}f'\left( x \right)dx}  = 9 - 18 =  - 9\end{array}\)

Đặt \(t = \sqrt[3]{{1 - x}}\)\( \Rightarrow {t^3} = 1 - x \Rightarrow 3{t^2}dt =  - dx \Leftrightarrow dx =  - 3{t^2}dt\)

Với \(x = 0 \Rightarrow t = 1\) ; \(x = 1 \Rightarrow t = 0\)

Khi đó \(I = \int\limits_1^0 {t.\left( { - 3{t^2}} \right)\,dt}  = 3\int\limits_0^1 {{t^3}dt} \)

Bước 1:

Đặt \(x = 4\sin t\)

Bước 2:

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow \sin t = 0 \Rightarrow t = 0\\x = \sqrt 8  \Rightarrow \sin t = \dfrac{{\sqrt 2 }}{2} \Rightarrow t = \dfrac{\pi }{4}\end{array} \right.\)

Bước 3:

Ta có: \(x = 4\sin t \Rightarrow dx = d\left( {4\sin t} \right)\)\( = \left( {4\sin t} \right)'dt = 4\cos tdt\)

Bước 4:

Khi đó ta có: \(I = 4\int\limits_0^{\dfrac{\pi }{4}} {\sqrt {16 - 16{{\sin }^2}t} \cos tdt} \)\( = 4\int\limits_0^{\dfrac{\pi }{4}} {4.{{\cos }^2}tdt} \)\( = 16\int\limits_0^{\dfrac{\pi }{4}} {{{\cos }^2}tdt} \) \( = 8\int\limits_0^{\dfrac{\pi }{4}} {\left( {1 + \cos 2t} \right)dt} \)  

Ta có: \(\int\limits_1^3 {f\left( x \right)dx = 4} .\)

Đặt \(2x + 1 = t \Rightarrow dt = 2dx \Rightarrow dx = \dfrac{1}{2}dt\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 1 \Rightarrow t = 3\end{array} \right.\)

\( \Rightarrow I = \int\limits_0^1 {f\left( {2x + 1} \right)dx}  = \dfrac{1}{2}\int\limits_1^3 {f\left( t \right)dt = } \dfrac{1}{2}.4 = 2.\)

Ta có

\(\begin{array}{l}I = \int\limits_0^2 {\left[ {f\left( {1 - 3x} \right) + 9} \right]dx} \\ \Leftrightarrow I = \int\limits_0^2 {f\left( {1 - 3x} \right)dx}  + \int\limits_0^2 {9dx} \end{array}\)

Đặt \({I_1} = \int\limits_0^2 {f\left( {1 - 3x} \right)dx} \).

Đặt \(t = 1 - 3x \Rightarrow dt =  - 3dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 2 \Rightarrow t =  - 5\end{array} \right.\).

\( \Rightarrow {I_1} =  - \dfrac{1}{3}\int\limits_1^{ - 5} {f\left( t \right).dt}  = \dfrac{1}{3}\int\limits_{ - 5}^1 {f\left( x \right)dx}  = \dfrac{1}{3}.9 = 3\).

Đặt \({I_2} = \int\limits_0^2 {9dx}  = \left. {9x} \right|_0^2 = 18\).

Vậy \(I = {I_1} + {I_2} = 3 + 18 = 21.\)

Ta có: \(\int\limits_0^1 {f\left( x \right)} dx = 4\) và \(\int\limits_0^1 {f\left( {3x} \right)dx = 6.} \)

Đặt \(t = 3x \Rightarrow dt = 3dx \Rightarrow dx = \dfrac{1}{3}dt\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 3\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \int\limits_0^1 {f\left( {3x} \right)dx = 6} \\ \Leftrightarrow \int\limits_0^3 {\dfrac{1}{3}f\left( t \right)dt}  = 6\\ \Leftrightarrow \int\limits_0^3 {f\left( t \right)dt}  = 18\\ \Rightarrow \int\limits_0^3 {f\left( x \right)dx}  = 18\\ \Rightarrow \int\limits_1^3 {f\left( x \right)dx}  = \int\limits_0^3 {f\left( x \right)dx}  - \int\limits_0^1 {f\left( x \right)dx} \\ \Leftrightarrow \int\limits_1^3 {f\left( x \right)dx}  = 18 - 4 = 14.\end{array}\)

Đặt \(t = \sqrt {{x^2} + 9}  \Rightarrow {t^2} = {x^2} + 9\) \( \Rightarrow tdt = xdx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 3\\x = 4 \Rightarrow t = 5\end{array} \right.\).

Vậy \(I = \int\limits_0^4 {x\sqrt {{x^2} + 9} dx}  = \int\limits_3^5 {t.tdt}  = \int\limits_3^5 {{t^2}dt} \).

Đặt \(t = \sqrt {1 + 3\ln x}  \Rightarrow {t^2} = 1 + 3\ln x \Rightarrow 2tdt = \dfrac{3}{x}dx \Rightarrow \dfrac{1}{x}dx = \dfrac{2}{3}tdt\)

Đổi cận:  \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 1\\x = e \Rightarrow t = 2\end{array} \right.\).

Khi đó ta có: \(I = \int\limits_1^2 {t.\dfrac{2}{3}tdt}  = \dfrac{2}{3}\int\limits_1^2 {{t^2}dt} \)

Ta có: \(I = \int\limits_0^3 {\dfrac{x}{{1 + \sqrt {x + 1} }}dx} \)

Đặt \(t = \sqrt {x + 1} \) \( \Rightarrow {t^2} = x + 1 \Rightarrow dx = 2tdt\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 3 \Rightarrow t = 2\end{array} \right.\)

\( \Rightarrow I = \int\limits_0^2 {\dfrac{{{t^2} - 1}}{{1 + t}}2tdt} \) \( = 2\int\limits_0^2 {\dfrac{{\left( {t - 1} \right)\left( {t + 1} \right)}}{{t + 1}}tdt} \)\( = 2\int\limits_0^2 {t\left( {t - 1} \right)dt = 2\int\limits_0^2 {\left( {{t^2} - t} \right)dt} } \)

\( \Rightarrow f\left( t \right) = 2\left( {{t^2} - t} \right) = 2{t^2} - 2t.\)

Đặt \(t = {e^x} + 1 \Rightarrow dt = {e^x}dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 2\\x = \ln 2 \Rightarrow t = 3\end{array} \right.\).

Khi đó ta có

\(\begin{array}{l}\int\limits_0^{\ln 2} {\dfrac{{{e^{2x}}}}{{{e^x} + 1}}dx}  = \int\limits_0^{\ln 2} {\dfrac{{{e^x}}}{{{e^x} + 1}}.{e^x}dx}  = \int\limits_2^3 {\dfrac{{t - 1}}{t}dt} \\ = \int\limits_2^3 {\left( {1 - \dfrac{1}{t}} \right)dt}  = \left. {\left( {t - \ln t} \right)} \right|_2^3 = 1 - \ln 3 + \ln 2 = 1 + \ln \dfrac{2}{3}\\ \Rightarrow a = 1,\,\,b = 2,\,\,c = 3\end{array}\)

Vậy \(a - b + c = 1 - 2 + 3 = 2.\)

Lấy tích phân từ 0 đến 1 hai vế ta được:

\(\int\limits_0^1 {f\left( x \right)dx}  + \int\limits_0^1 {\left( {5x - 2} \right)f\left( {5{x^2} - 4x} \right)dx}  = \int\limits_0^1 {\left( {50{x^3} - 60{x^2} + 23x - 1} \right)dx} \)

Xét tích phân \(I = \int\limits_0^1 {\left( {5x - 2} \right)f\left( {5{x^2} - 4x} \right)dx} \).

Đặt \(t = 5{x^2} - 4x\) ta có \(dt = \left( {10x - 4} \right)dx \Leftrightarrow \left( {5x - 2} \right)dx = \dfrac{1}{2}dt\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 1\end{array} \right.\).

\( \Rightarrow I = \int\limits_0^1 {\dfrac{1}{2}f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^1 {f\left( x \right)dx} \)

Xét tích phân \(J = \int\limits_0^1 {\left( {50{x^3} - 60{x^2} + 23x - 1} \right)dx} \) ta có: \(J = \left. {\left( {\dfrac{{50{x^4}}}{4} - \dfrac{{60{x^3}}}{3} + \dfrac{{23{x^2}}}{2} - x} \right)} \right|_0^1 = 3\)

Khi đó ta có \(\int\limits_0^1 {f\left( x \right)dx}  + \dfrac{1}{2}\int\limits_0^1 {f\left( x \right)dx}  = 3\) \( \Leftrightarrow \dfrac{3}{2}\int\limits_0^1 {f\left( x \right)dx}  = 3\) \( \Leftrightarrow \int\limits_0^1 {f\left( x \right)dx}  = 2\).

Theo bài ra ta có

\(\begin{array}{l}f\left( {{x^2} + 3x + 1} \right) = x + 2\\ \Rightarrow f\left( {{x^2} + 3x + 1} \right)\left( {2x + 3} \right) = \left( {x + 2} \right)\left( {2x + 3} \right)\\ \Rightarrow \int\limits_0^1 {f\left( {{x^2} + 3x + 1} \right)\left( {2x + 3} \right)dx}  = \int\limits_0^1 {\left( {x + 2} \right)\left( {2x + 3} \right)dx} \\ \Rightarrow \int\limits_0^1 {f\left( {{x^2} + 3x + 1} \right)\left( {2x + 3} \right)dx}  = \dfrac{{61}}{6}\end{array}\)

Đặt \(t = {x^2} + 3x + 1 \Rightarrow dt = \left( {2x + 3} \right)dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 1 \Rightarrow t = 5\end{array} \right.\).

\( \Rightarrow \int\limits_0^1 {f\left( {{x^2} + 3x + 1} \right)\left( {2x + 3} \right)dx}  = \int\limits_1^5 {f\left( t \right)dt}  = \int\limits_1^5 {f\left( x \right)dx} \).

Vậy \(\int\limits_1^5 {f\left( x \right)dx}  = \dfrac{{61}}{6}\).

Đặt \(t = 3 - 2x \Rightarrow dt =  - 2dx\). Đổi cận \(\left\{ \begin{array}{l}x =  - 2 \Rightarrow t = 7\\x = 3 \Rightarrow t =  - 3\end{array} \right.\).

Khi đó ta có:

\(\begin{array}{l}I =  - \dfrac{1}{2}\int\limits_7^{ - 3} {f\left( {\left| t \right|} \right)dt}  = \dfrac{1}{2}\int\limits_{ - 3}^7 {f\left( {\left| t \right|} \right)dt} \\\,\,\,\, = \dfrac{1}{2}\left( {\int\limits_{ - 3}^0 {f\left( { - t} \right)dt}  + \int\limits_0^7 {f\left( t \right)dt} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( { - \int\limits_3^0 {f\left( x \right)dx}  + \int\limits_0^7 {f\left( x \right)dx} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {\int\limits_0^3 {f\left( x \right)dx}  + \int\limits_0^7 {f\left( x \right)dx} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {6 + 10} \right) = 8\end{array}\).

Đặt \(t = \sin x \Rightarrow dt = \cos xdx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{2} \Rightarrow t = 1\end{array} \right.\), khi đó

\(\begin{array}{l}\int\limits_0^{\frac{\pi }{2}} {\dfrac{{\cos x}}{{{{\sin }^2}x - 5\sin x + 6}}dx}  = \int\limits_0^1 {\dfrac{{dt}}{{{t^2} - 5t + 6}}} \\ = \int\limits_0^1 {\dfrac{{dt}}{{\left( {t - 2} \right)\left( {t - 3} \right)}}}  = \int\limits_0^1 {\left( {\dfrac{1}{{t - 3}} - \dfrac{1}{{t - 2}}} \right)dt} \\ = \left. {\left( {\ln \left| {t - 3} \right| - \ln \left| {t - 2} \right|} \right)} \right|_0^1 = \left. {\ln \left| {\dfrac{{t - 3}}{{t - 2}}} \right|} \right|_0^1\\ = \ln 2 - \ln \dfrac{3}{2} = \ln \dfrac{4}{3}\end{array}\)

\( \Rightarrow a = 1,\,\,b = 3\).

Vậy \(a + b = 1 + 3 = 4\).

\(\begin{array}{l}\int\limits_0^1 {\dfrac{{xdx}}{{{{\left( {x + 2} \right)}^2}}} = \int\limits_0^1 {\dfrac{{x + 2}}{{{{\left( {x + 2} \right)}^2}}}dx}  - \int\limits_0^1 {\dfrac{2}{{{{\left( {x + 2} \right)}^2}}}dx} }  \\= \left. {\left( {\ln \left| {x + 2} \right| + \dfrac{2}{{x + 2}}} \right)} \right|_0^1\\ = \ln 3 + \dfrac{2}{3} - \ln 2 - 1 = \ln 3 - \ln 2 - \dfrac{1}{3}\\ \Rightarrow \left\{ \begin{array}{l}a =  - \dfrac{1}{3}\\b =  - 1\\c = 1\end{array} \right. \Rightarrow 3a + b + c = 3.\left( { - \dfrac{1}{3}} \right) - 1 + 1 =  - 1.\end{array}\)

\(I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{{{{\cos }^2}x + \sin x\cos x + 1}}{{{{\cos }^4}x + \sin x{{\cos }^3}x}}dx}  = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{{1 + \tan x + 1 + {{\tan }^2}x}}{{{{\cos }^2}x\left( {1 + \tan x} \right)}}dx}  = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{3}} {\dfrac{{{{\tan }^2}x + \tan x + 2}}{{{{\cos }^2}x\left( {1 + \tan x} \right)}}dx} \)

Đặt \(t = \tan x \Rightarrow dt = \dfrac{1}{{{{\cos }^2}x}}dx\). Đổi cận \(\left\{ \begin{array}{l}x = \dfrac{\pi }{4} \Rightarrow t = 1\\x = \dfrac{\pi }{3} \Rightarrow t = \sqrt 3 \end{array} \right.\).

\(\begin{array}{l} \Rightarrow I = \int\limits_1^{\sqrt 3 } {\dfrac{{{t^2} + t + 2}}{{t + 1}}dt}  = \int\limits_1^{\sqrt 3 } {\left( {t + \dfrac{2}{{t + 1}}} \right)dt} \\ = \left. {\dfrac{{{t^2}}}{2} + 2\ln \left| {t + 1} \right|} \right|_1^{\sqrt 3 } = \dfrac{3}{2} + 2\ln \left( {\sqrt 3  + 1} \right) - \dfrac{1}{2} - 2\ln 2 = 1 - 2\ln 2 + 2\ln \left( {1 + \sqrt 3 } \right)\\ \Rightarrow \left\{ \begin{array}{l}a = 1\\b =  - 2\\c = 2\end{array} \right. \Rightarrow abc = 1.\left( { - 2} \right).2 =  - 4\end{array}\)