Cho hàm số \(f(x) = \left\{ \begin{array}{l} 2{x^2} - 1\,\,\,\,\,\,{\rm{khi}}\,\,x < 0\\ x - 1\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,0 \le x \le 2\\ 5 - 2x\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x > 2\, \end{array} \right.\). Tính tích phân \(\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{f\left( 2-7\tan x \right)}\frac{1}{{{\cos }^{2}}x}\text{d}x\).
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Lời giải:
Báo saiXét \(I=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{f\left( 2-7\tan x \right)\frac{1}{{{\cos }^{2}}x}\text{d}x}\)
Đặt \(2-7\tan x=t\Rightarrow \frac{1}{{{\cos }^{2}}x}\text{d}x=-\frac{1}{7}\text{d}t\)
Với \(x=-\frac{\pi }{4}\)\(\Rightarrow \)\(t=9\)
\(x=\frac{\pi }{4}\)\(\Rightarrow \)\(t=-5\)
\(\Rightarrow I=\frac{1}{7}\int\limits_{-5}^{9}{f\left( t \right)\text{d}t}=\frac{1}{7}\int\limits_{-5}^{9}{f\left( x \right)\text{d}x}=\frac{1}{7}\int\limits_{-5}^{0}{f(x)\text{d}x}+\frac{1}{7}\int\limits_{0}^{2}{f(x)\text{d}x}+\frac{1}{7}\int\limits_{2}^{9}{f(x)\text{d}x}\)
\(=\frac{1}{7}\int\limits_{-5}^{0}{\left( 2{{x}^{2}}-1 \right)\text{d}x}+\frac{1}{7}\int\limits_{0}^{2}{\left( x-1 \right)\text{d}x}+\frac{1}{7}\int\limits_{2}^{9}{\left( 5-2x \right)\text{d}x}=\frac{109}{21}.\)
