Bài viết hướng dẫn phương pháp tính tích phân hàm chứa giá trị tuyệt đối, đây là dạng toán thường gặp trong chương trình Giải tích 12 chương 3.
1. Phương pháp tính tích phân hàm chứa giá trị tuyệt đối
Muốn tính tích phân \(I = \int_a^b | f(x)|dx\), ta thức hiện theo các bước sau:
+ Xét dấu hàm \(f(x)\) trên đoạn \([a;b]\) để mở dấu giá trị tuyệt đối.
+ Áp dụng công thức: \(\int_a^b | f(x)|dx\) \( = \int_a^c | f(x)|dx + \int_c^b | f(x)|dx.\)
2. Một số ví dụ minh họa
Ví dụ 1: Tính tích phân: \(I = \int_{ – 3}^3 {\left| {{x^2} – 1} \right|} dx.\)
Ta có: \(I = \int_{ – 3}^3 {\left| {{x^2} – 1} \right|} dx\) \( = \int_{ – 3}^{ – 1} {\left( {{x^2} – 1} \right)} dx\) \( + \int_{ – 1}^1 {\left( { – {x^2} + 1} \right)} dx\) \( + \int_1^3 {\left( {{x^2} – 1} \right)} dx\) \( = \left. {\left( {\frac{{{x^3}}}{3} – x} \right)} \right|_{ – 3}^{ – 1}\) \( + \left. {\left( { – \frac{{{x^3}}}{3} + x} \right)} \right|_{ – 1}^1\) \( + \left. {\left( {\frac{{{x^3}}}{3} – x} \right)} \right|_1^3\) \( = – \frac{1}{3} + 1 + 9 – 3 – \frac{1}{3} + 1\) \( – \frac{1}{3} + 1 + 9 – 3 – \frac{1}{3} + 1\) \( = \frac{{44}}{3}.\)
Vậy \(I = \int_{ – 3}^3 {\left| {{x^2} – 1} \right|} dx = \frac{{44}}{3}.\)
Ví dụ 2: Tính tích phân: \(I = \int_0^2 {\left| {{x^2} – 4x + 3} \right|} dx.\)
Ta có bảng xét dấu:
Nên \(I = \int_0^2 {\left| {{x^2} – 4x + 3} \right|} dx\) \( = \int_0^1 {\left( {{x^2} – 4x + 3} \right)} dx\) \( + \int_1^2 {\left( { – {x^2} + 4x – 3} \right)} dx\) \( = \left. {\left( {\frac{{{x^3}}}{3} – 2{x^2} + 3x} \right)} \right|_0^1\) \( + \left. {\left( { – \frac{{{x^3}}}{3} + 2{x^2} – 3x} \right)} \right|_1^2 = 2.\)
Vậy \(I = \int_0^2 {\left| {{x^2} – 4x + 3} \right|} dx = 2.\)
Ví dụ 3: Tính tích phân: \({I_{(m)}} = \int_0^1 {\left| {{x^2} – 2x + m} \right|} dx.\)
Đặt \(f(x) = {x^2} – 2x + m\) có \(\Delta’ = 1 – m.\)
+ Khi \(m \ge 1\) \( \Leftrightarrow \Delta’ = 1 – m \le 0\) \( \Rightarrow f(x) \ge 0\) \(\forall x \in R.\)
Do đó \({I_{(m)}} = \int_0^1 {\left| {{x^2} – 2x + m} \right|} dx\) \( = \int_0^1 {\left( {{x^2} – 2x + m} \right)} dx\) \( = \left. {\left( {\frac{{{x^3}}}{3} – {x^2} + mx} \right)} \right|_0^1\) \( = m – \frac{2}{3}.\)
+ Khi \(0 < m < 1\) thì \(\left\{ {\begin{array}{*{20}{l}}
{\Delta’ = 1 – m /> 0}\\
{f(0) = m /> 0}\\
{f(1) = m – 1 < 0}
\end{array}} \right.\)
Phương trình \(f(x) = m\) có hai nghiệm \({x_1} < {x_2}.\)
Do đó ta có \(0 < {x_1} < 1 < {x_2}\) với \({x_1},{x_2} = 1 \pm \sqrt {1 – m} .\)
Hay ta có:
Nên: \({I_{(m)}} = \int_0^1 {\left| {{x^2} – 2x + m} \right|} dx\) \( = \int_0^{{x_1}} {\left( {{x^2} – 2x + m} \right)} dx\) \( + \int_{{x_1}}^1 {\left( { – {x^2} + 2x – m} \right)} dx\) \( = \left. {\left( {\frac{{{x^3}}}{3} – {x^2} + mx} \right)} \right|_0^{{x_1}}\) \( + \left. {\left( { – \frac{{{x^3}}}{3} + {x^2} – mx} \right)} \right|_{{x_1}}^1\) \( = 2\left[ {\frac{{x_1^3}}{3} – x_1^2 + m{x_1}} \right] + \frac{2}{3} – m.\)
Thế \({x_1} = 1 – \sqrt {1 – m} \) vào ta có:
\({I_m} = \frac{2}{3}(1 – \sqrt {1 – m} )\)\(\left[ {{{(1 – \sqrt {1 – m} )}^2} – 3(1 – \sqrt {1 – m} ) + 3m} \right]\) \( + \frac{2}{3} – m\) \( = \frac{2}{3}(1 – \sqrt {1 – m} )(2m – 1 + \sqrt {1 – m} )\) \( + \frac{2}{3} – m.\)
+ Khi \(m \le 0\) thì \(\left\{ {\begin{array}{*{20}{l}}
{f(0) = m \le 0}\\
{f(1) = m – 1 \le 0}
\end{array}} \right.\)
Do đó ta có \({x_1} \le 0 < 1 < {x_2}\) \( \Rightarrow f(x) < 0\) \(\forall x \in [0;1].\)
Nên \({I_m} = \int_0^1 {\left( { – {x^2} + 2x – m} \right)} dx\) \( = \left. {\left( {\frac{{ – {x^3}}}{3} + {x^2} – mx} \right)} \right|_0^1\) \( = \frac{2}{3} – m.\)
Ví dụ 4: Tính tích phân: \(I = \int_0^2 {\left| {{x^2} – x} \right|} dx.\)
Ta có:
Do đó: \(I = \int_0^2 {\left| {{x^2} – x} \right|} dx\) \( = \int_0^1 {\left( { – {x^2} + x} \right)} dx\) \( + \int_1^2 {\left( {{x^2} – x} \right)} dx\) \( = \left. {\left( { – \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)} \right|_0^1\) \( + \left. {\left( {\frac{{{x^3}}}{3} – \frac{{{x^2}}}{2}} \right)} \right|_1^2 = 1.\)
Ví dụ 5: Tính tích phân: \(I(\alpha ) = \int_0^1 x |x – \alpha |dx.\)
+ Khi \(\alpha \le 0\) thì \(x – \alpha \ge 0\) \(\forall x \in [0;1].\)
Vậy \(I(\alpha ) = \int_0^1 x |x – \alpha |dx\) \( = \left. {\left( {\frac{{{x^3}}}{3} – \frac{{\alpha {x^2}}}{2}} \right)} \right|_0^1\) \( = \frac{1}{3} – \frac{\alpha }{2}.\)
+ Khi \(0 < \alpha < 1\), ta có:
Vậy \(I(\alpha ) = \int_0^\alpha x |x – \alpha |dx\) \( + \int_\alpha ^1 x |x – \alpha |dx\) \( = \int_0^\alpha {\left( { – {x^2} + \alpha x} \right)} dx\) \( + \int_\alpha ^1 {\left( {{x^2} – \alpha x} \right)} dx\) \( = \left. {\left( {\frac{{\alpha {x^2}}}{2} – \frac{{{x^3}}}{3}} \right)} \right|_0^\alpha \) \( + \left. {\left( {\frac{{{x^3}}}{3} – \frac{{\alpha {x^2}}}{2}} \right)} \right|_\alpha ^1\) \( = \frac{{{\alpha ^3}}}{3} – \frac{\alpha }{2} + \frac{1}{3}.\)
+ Khi \(\alpha \ge 1\) thì \(x – \alpha \le 0\) \(\forall x \in [0;1].\)
Vậy \(I(\alpha ) = \int_0^1 {\left( { – {x^2} + \alpha x} \right)} dx\) \( = \left. {\left( { – \frac{{{x^3}}}{3} + \frac{{\alpha {x^2}}}{2}} \right)} \right|_0^1\) \( = \frac{\alpha }{2} – \frac{1}{3}.\)
Ví dụ 6: Cho \(f(x) = 3{x^3} – {x^2} – 4x + 1\) và \(g(x) = 2{x^3} + {x^2} – 3x – 1.\)
a) Giải bất phương trình \(f(x) \ge g(x).\)
b) Tính \(I = \int_{ – 1}^2 | f(x) – g(x)|dx.\)
a) Ta có: \(f(x) \ge g(x)\) \( \Leftrightarrow f(x) – g(x) \ge 0\) \( \Leftrightarrow {x^3} – 2x – x + 2 \ge 0\) \( \Leftrightarrow (x – 1)\left( {{x^2} – x – 2} \right) \ge 0\) \( \Leftrightarrow \left( {{x^2} – 1} \right)(x – 2) \ge 0\) \( \Leftrightarrow – 1 \le x \le 1\) hoặc \(x \ge 2.\)
b) Ta có: (dựa vào câu a, ta xác định được \(f(x) – g(x)\) âm, dương khi nào).
Vậy \(I = \int_{ – 1}^2 | f(x) – g(x)|dx\) \( = \int_{ – 1}^1 | f(x) – g(x)|dx\) \( + \int_1^2 | f(x) – g(x)|dx\) \( = \int\limits_{ – 1}^1 {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} \) \( – \int\limits_1^2 {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} \) \( = \int_{ – 1}^1 {\left( {{x^3} – 2{x^2} – x + 2} \right)} dx\) \( – \int_1^2 {\left( {{x^3} – 2{x^2} – x + 2} \right)} dx\) \( = \left. {\left( {\frac{{{x^4}}}{4} – \frac{{2{x^2}}}{3} – \frac{{{x^2}}}{2} + 2x} \right)} \right|_{ – 1}^1\) \( – \left. {\left( {\frac{{{x^4}}}{4} – \frac{{2{x^2}}}{3} – \frac{{{x^2}}}{2} + 2x} \right)} \right|_1^2 = \frac{{37}}{{12}}.\)
Ví dụ 7: Tính tích phân: \(I = \int_{ – \pi }^\pi {\sqrt {1 – \sin x} } dx.\)
Ta có: \(I = \int_{ – \pi }^\pi {\sqrt {{{\left( {\sin \frac{x}{2} – \cos \frac{x}{2}} \right)}^2}} } dx\) \( = \int_{ – \pi }^\pi {\left| {\sin \frac{x}{2} – \cos \frac{x}{2}} \right|} dx\) \( = \sqrt 2 \int_{ – \pi }^\pi {\left| {\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|} dx.\)
Đổi biến: đặt \(t = \frac{x}{2} + \frac{\pi }{4} \Rightarrow dt = \frac{{dx}}{2}.\)
Đổi cận: \(\left[ {\begin{array}{*{20}{l}}
{x = \pi }\\
{x = – \pi }
\end{array}} \right.\) \( \Rightarrow \left[ {\begin{array}{*{20}{l}}
{t = \frac{{3\pi }}{4}}\\
{t = – \frac{\pi }{4}}
\end{array}} \right.\)
Ta thấy: với \( – \frac{\pi }{4} \le t \le \frac{\pi }{2}\) thì \(\cos t \ge 0\), với \(\frac{\pi }{2} \le t \le \frac{{3\pi }}{4}\) thì \(\cos t < 0.\)
Suy ra: \(I = 2\sqrt 2 \int_{ – \frac{\pi }{4}}^{\frac{{3\pi }}{4}} | \cos t|dt\) \( = 2\sqrt 2 \int_{ – \frac{\pi }{4}}^{\frac{\pi }{2}} {\cos } tdt – 2\sqrt 2 \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{4}} { \cos tdt } \) \( = 2\sqrt 2 \sin \left. t \right|_{ – \frac{\pi }{4}}^{\frac{\pi }{2}} – 2\sqrt 2 \sin \left. t \right|_{\frac{\pi }{2}}^{\frac{{3\pi }}{4}} = 4\sqrt 2 .\)
Ví dụ 8: Tính tích phân: \(I = \int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} | \sin x|dx.\)
Ta có: \(I = \int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} | \sin x|dx\) \( = \int_{ – \frac{\pi }{2}}^0 {( – \sin x)} dx + \int_0^{\frac{\pi }{2}} {\sin } xdx\) \( = \cos \left. x \right|_{ – \frac{\pi }{2}}^0 + \left. {( – \cos x)} \right|_0^{\frac{\pi }{2}}\) \( = 1 + 1 = 2.\)
Ví dụ 9: Tính \(I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} | \sin 2x|dx.\)
Đặt \(t = 2x \Rightarrow dt = 2dx.\)
Đổi cận \(\left[ {\begin{array}{*{20}{l}}
{x = \frac{{3\pi }}{4}}\\
{x = \frac{\pi }{4}}
\end{array}} \right.\) \( \Rightarrow \left[ {\begin{array}{*{20}{l}}
{t = \frac{{3\pi }}{2}}\\
{t = \frac{\pi }{2}}
\end{array}} \right.\)
Do đó: \(I = \frac{1}{2}\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} | \sin t|dt\) \( = \frac{1}{2}\int_{\frac{\pi }{2}}^\pi | \sin t|dt + \frac{1}{2}\int_\pi ^{\frac{{3\pi }}{2}} | \sin t|dt\) \( = \frac{1}{2}\int_{\frac{\pi }{2}}^\pi {\sin t} dt – \frac{1}{2}\int_\pi ^{\frac{{3\pi }}{2}} {\sin } tdt\) (vì \(\frac{\pi }{2} \le t \le \pi \) thì \(\sin t \ge 0\), \(\frac{\pi }{2} \le t \le \frac{{3\pi }}{2}\) thì \(\sin t \le 0\)).
\(I = – \frac{1}{2}\cos \left. t \right|_{\frac{\pi }{2}}^\pi + \frac{1}{2}\cos \left. t \right|_\pi ^{\frac{{3\pi }}{2}} = 1.\)
Ví dụ 10: Tính tích phân: \(I = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\sqrt {{{\tan }^2}x + {{\cot }^2}x – 2} } dx.\)
Ta có: \(\sqrt {{{\tan }^2}x + {{\cot }^2}x – 2} \) \( = \sqrt {{{(\tan x + \cot x)}^2}} \) \( = |\tan x – \cot x|\) \( = \left| {\frac{{\sin x}}{{\cos x}} – \frac{{\cos x}}{{\sin x}}} \right|\) \( = \left| {\frac{{{{\sin }^2}x – {{\cos }^2}x}}{{\sin x\cos x}}} \right|\) \( = \left| {\frac{{{{\cos }^2}x – {{\sin }^2}x}}{{\sin x\cos x}}} \right|\) \( = 2\left| {\frac{{\cos 2x}}{{\sin 2x}}} \right|.\)
Ta có: \(\frac{\pi }{6} \le x \le \frac{\pi }{3}\) \( \Rightarrow \frac{\pi }{3} \le 2x \le \frac{{2\pi }}{3}.\)
Do đó: \(\sin 2x \ge 0\), \(\left\{ \begin{array}{l}
\cos 2x \le 0\:{\rm{khi}}\:x \in \left[ {\frac{\pi }{4};\frac{\pi }{3}} \right]\\
\cos 2x \ge 0\:{\rm{khi}}\:x \in \left[ {\frac{\pi }{6};\frac{\pi }{4}} \right]
\end{array} \right.\)
Vậy \(I = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} 2 \left| {\frac{{\cos 2x}}{{\sin 2x}}} \right|dx\) \( + \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} 2 \left| {\frac{{\cos 2x}}{{\sin 2x}}} \right|dx\) \( = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} 2 \frac{{\cos 2x}}{{\sin 2x}}dx – \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} 2 \frac{{\cos 2x}}{{\sin 2x}}dx\) \( = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} 2 \frac{{d(\sin 2x)}}{{\sin 2x}} – \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} 2 \frac{{d(\sin 2x)}}{{\sin 2x}}\) \( = \ln \left. {|\sin 2x|} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} – \left. {\ln |\sin 2x|} \right|_{\frac{\pi }{4}}^{\frac{\pi }{3}}\) \( = \left( {\ln 1 – \ln \frac{{\sqrt 3 }}{2}} \right) – \left( {\ln \frac{{\sqrt 3 }}{2} – \ln 1} \right)\) \( = – 2\ln \frac{{\sqrt 3 }}{2}.\)
Ví dụ 11: Tính tích phân: \(I = \int_0^\pi {\sqrt {1 + \cos 2x} } dx.\)
Ta có: \(I = \int_0^\pi {\sqrt {1 + \cos 2x} } dx\) \( = \int_0^\pi {\sqrt {2{{\cos }^2}x} } dx\) \( = \int_0^\pi {\sqrt 2 } |\cos x|dx\) \( = \sqrt 2 \int_0^{\frac{\pi }{2}} {\cos } xdx – \sqrt 2 \int_{\frac{\pi }{2}}^\pi {\cos } xdx\) \( = \sqrt 2 \sin \left. x \right|_0^{\frac{\pi }{2}} – \sqrt 2 \sin \left. x \right|_{\frac{\pi }{2}}^\pi \) \( = 2\sqrt 2 .\)
Ví dụ 12: Tính tích phân: \(I = \int_0^\pi | \cos x|\sqrt {\sin x} dx.\)
Ta có: \(I = \int_0^\pi | \cos x|\sqrt {\sin x} dx\) \( + \int_{\frac{\pi }{2}}^\pi | \cos x|\sqrt {\sin x} dx\) \( = \int_0^{\frac{\pi }{2}} {\cos } x.{(\sin x)^{\frac{1}{2}}}dx\) \( – \int_{\frac{\pi }{2}}^\pi {\cos } x.{(\sin x)^{\frac{1}{2}}}dx\) \( = \int_0^{\frac{\pi }{2}} {{{(\sin x)}^{\frac{1}{2}}}} d(\sin x)\) \( – \int_{\frac{\pi }{2}}^\pi {{{(\sin x)}^{\frac{1}{2}}}} d(\sin x)\) \( = \frac{2}{3}\left. {{{(\sin x)}^{\frac{3}{2}}}} \right|_0^{\frac{\pi }{2}} – \frac{2}{3}\left. {{{(\sin x)}^{\frac{3}{2}}}} \right|_{\frac{\pi }{2}}^\pi \) \( = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}.\)
Ví dụ 13: Tính tích phân: \(I = \int_{ – 1}^1 {\frac{{|x|dx}}{{{x^4} – {x^2} – 12}}} .\)
Vì hàm số \(f(x) = \frac{{|x|}}{{{x^4} – {x^2} – 12}}\) là hàm số chẵn, liên tục trong \([ – 1;1].\)
Suy ra: \(I = \int_{ – 1}^1 {\frac{{|x|dx}}{{{x^4} – {x^2} – 12}}} \) \( = 2\int_0^1 {\frac{{|x|dx}}{{{x^4} – {x^2} – 12}}} \) \( = 2\int_0^1 {\frac{{xdx}}{{{x^4} – {x^2} – 12}}} .\)
Đặt \(t = {x^2} \Rightarrow dt = 2xdx.\)
Đổi cận \(\left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = 0}
\end{array}} \right.\) \( \Rightarrow \left[ {\begin{array}{*{20}{l}}
{t = 1}\\
{t = 0}
\end{array}} \right.\)
Vậy \(I = \int_0^1 {\frac{{dt}}{{{t^2} – t – 12}}} \) \( = \int_0^1 {\frac{{dt}}{{(t – 4)(t + 3)}}} \) \( = \frac{1}{7}\int_0^1 {\left( {\frac{1}{{t – 4}} – \frac{1}{{t + 3}}} \right)} dt\) \( = \frac{1}{7}\ln \left. {\left| {\frac{{t – 4}}{{t + 3}}} \right|} \right|_0^1\) \( = \frac{2}{7}\ln \frac{3}{4}.\)
