Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\) và \(\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=4\), \(\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=6\). Tính \(I=\int\limits_{-1}^{1}{f\left( \left| 2x+1 \right| \right)\text{d}x}\)
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Lời giải:
Báo saiĐặt \(u=2x+1\)\(\Rightarrow \operatorname{d}x=\frac{1}{2}\operatorname{d}u\). Khi \(x=-1\) thì \(u=-1\). Khi \(x=1\) thì \(u=3\).
Nên \(I=\frac{1}{2}\int\limits_{-1}^{3}{f\left( \left| u \right| \right)\operatorname{d}u} =\frac{1}{2}\left( \int\limits_{-1}^{0}{f\left( \left| u \right| \right)\operatorname{d}u}+\int\limits_{0}^{3}{f\left( \left| u \right| \right)\operatorname{d}u} \right)\)
\(=\frac{1}{2}\left( \int\limits_{-1}^{0}{f\left( -u \right)\operatorname{d}u}+\int\limits_{0}^{3}{f\left( u \right)\operatorname{d}u} \right)\).
Xét \(\int\limits_{0}^{1}{f\left( x \right)\operatorname{d}x}=4\). Đặt \(x=-u\)\(\Rightarrow \operatorname{d}x=-\operatorname{d}u\).
Khi \(x=0\) thì \(u=0\). Khi \(x=1\) thì \(u=-1\).
Nên \(4=\int\limits_{0}^{1}{f\left( x \right)\operatorname{d}x}= -\int\limits_{0}^{-1}{f\left( -u \right)\operatorname{d}u} =\int\limits_{-1}^{0}{f\left( -u \right)\operatorname{d}u}\).
Ta có \(\int\limits_{0}^{3}{f\left( x \right)\operatorname{d}x}=6\)\(\Rightarrow \int\limits_{0}^{3}{f\left( u \right)\operatorname{d}u}=6\).
Nên \(I=\frac{1}{2}\left( \int\limits_{-1}^{0}{f\left( -u \right)\operatorname{d}u}+\int\limits_{0}^{3}{f\left( u \right)\operatorname{d}u} \right) =\frac{1}{2}\left( 4+6 \right)=5\).
