Cho hàm số \(f(x) = \left\{ \begin{array}{l} 2x - 1\,\,\,{\rm{khi}}\,x \ge 1\\ {x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,x < 1 \end{array} \right.\). Tính tích phân \(\int\limits_{1}^{13}{f\left( \sqrt{x+3}-2 \right)}\text{d}x\).
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Lời giải:
Báo saiXét \(I=\int\limits_{1}^{13}{f\left( \sqrt{x+3}-2 \right)\text{d}x}\)
Đặt \(\sqrt{x+3}-2=t\Rightarrow \sqrt{x+3}=t+2\Rightarrow x+3={{(t+2)}^{2}}\Rightarrow \text{d}x=2(t+2)\text{d}t\)
Với \(x=1\)\(\Rightarrow \)\(t=0\)
\(x=13\)\(\Rightarrow \)\(t=2\)
\(\Rightarrow I=2\int\limits_{0}^{2}{(t+2)f\left( t \right)\text{d}t}=2\int\limits_{0}^{2}{(x+2)f\left( x \right)\text{d}x}=2\int\limits_{0}^{1}{(x+2)f\left( x \right)\text{d}x}+2\int\limits_{1}^{2}{(x+2)f\left( x \right)\text{d}x}\)
\(=2\int\limits_{0}^{1}{(x+2){{x}^{2}}\text{d}x}+2\int\limits_{1}^{2}{(2x-1)(x+2)\text{d}x}=\frac{97}{6}.\)
