Cho hàm số \(y = f\left( x \right) = \left\{ \begin{array}{l} {x^2} + 3{x^2};x \ge 1\\ 5 - x\,;x < 1 \end{array} \right.\). Tính \(I = 2\int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\cos x{\rm{d}}x + 3\int\limits_0^1 {f\left( {3 - 2x} \right){\rm{d}}x} } \).
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Lời giải:
Báo saiXét tích phân \({{I}_{1}}=\int\limits_{0}^{\frac{\pi }{2}}{f\left( \sin x \right)\cos x\text{d}x}\). Đặt \(t=sinx\Rightarrow \text{d}t=\cos x\text{d}x\)
Đổi cận
Ta có \({{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right)\text{d}t=}\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( 5-x \right)\text{d}x=}\left. \left( 5x-\frac{{{x}^{2}}}{2} \right) \right|_{0}^{1}=\frac{9}{2}\)
Xét tích phân \({{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}\). Đặt \(t=3-2x\Rightarrow \text{d}t=-2\text{d}x\Rightarrow \text{d}x=\frac{-\text{d}t}{2}\)
Đổi cận
Ta có \({{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}=\frac{1}{2}\int\limits_{1}^{3}{f\left( t \right)\text{d}t=}\frac{1}{2}\int\limits_{1}^{3}{f\left( x \right)\text{d}x=}\frac{1}{2}\int\limits_{1}^{3}{\left( {{x}^{2}}+3 \right)\text{d}x=}\frac{1}{2}\left. \left( \frac{{{x}^{3}}}{3}+3x \right) \right|_{1}^{3}=\frac{1}{2}\left( 18-\frac{10}{3} \right)=\frac{22}{3}\)
Vậy \(I=2\int\limits_{0}^{\frac{\pi }{2}}{f\left( \sin x \right)\cos x\text{d}x+3\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}}=9+22=31\).