Cho hàm số \(f(x) = \left\{ \begin{array}{l} 2{x^2} + 1\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi }}x \ge 0\\ 2{x^2} - x + 1\,\,\,\,\,{\rm{khi }}x < 0 \end{array} \right.\). Tính tích phân \(\int\limits_{0}^{\frac{\pi }{3}}{f\left( 3\cos x-2 \right)}\sin x\text{d}x\).
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Lời giải:
Báo saiXét \(I=\int\limits_{0}^{\frac{\pi }{3}}{f\left( 3\cos x-2 \right)\sin x\text{d}x}\)
Đặt \(3\cos x-2=t\)\(\Rightarrow -3\sin x\text{d}x=\text{d}t\Rightarrow \sin x\text{d}x=-\frac{1}{3}\text{d}t\)
Với \(x=0\)\(\Rightarrow \)\(t=1\)
\(x=\frac{\pi }{3}\)\(\Rightarrow \)\(t=-\frac{1}{2}\)
\(\Rightarrow \)\(I=\frac{1}{3}\int\limits_{-\frac{1}{2}}^{1}{f\left( t \right)\text{d}t}=\frac{1}{3}\int\limits_{-\frac{1}{2}}^{1}{f\left( x \right)\text{d}x}=\frac{1}{3}\int\limits_{-\frac{1}{2}}^{0}{f(x)\text{d}x}+\frac{1}{3}\int\limits_{0}^{1}{f(x)\text{d}x}\)
\(=\frac{1}{3}\int\limits_{-\frac{1}{2}}^{0}{\left( 2{{x}^{2}}-x+1 \right)\text{d}x}+\frac{1}{3}\int\limits_{0}^{1}{\left( 2{{x}^{2}}+1 \right)\text{d}x}=\frac{19}{24}.\)
