Giải bài 22 trang 162 SGK giải tích nâng cao 12
Chứng minh rằng
a) \(\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx};\) b) \(\int\limits_{-1}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left[ f\left( x \right)+f\left( -x \right) \right]dx}\)
a) \(\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}\)
Đặt \(u=1-x\Rightarrow du=-dx\)
| x | 0 | 1 |
| u | 1 | 0 |
\(\Rightarrow \int\limits_{0}^{1}{f\left( 1-x \right)dx}=\int\limits_{1}^{0}{f\left( u \right)\left( -du \right)}=\int\limits_{0}^{1}{f\left( u \right)du}=\int\limits_{0}^{1}{f\left( x \right)dx} \\ \)
b) \(\int\limits_{-1}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left[ f\left( x \right)+f\left( -x \right) \right]dx}\)
Ta có \(\int\limits_{-1}^{1}{f\left( x \right)dx}=\int\limits_{-1}^{0}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( x \right)dx}\)
Xét \(\int\limits_{-1}^{0}{f\left( x \right)dx}\)
Đặt \(u=-x\Rightarrow du=-dx\)
| x | \(-1\) | 0 |
| u | 1 | 0 |
\(\Rightarrow \int\limits_{-1}^{0}{f\left( x \right)dx}=\int\limits_{1}^{0}{f\left( -u \right)\left( -du \right)}=\int\limits_{0}^{1}{f\left( -u \right)du}=\int\limits_{0}^{1}{f\left( -x \right)dx} \\ \)
Vậy \( \int\limits_{-1}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( -x \right)dx}=\int\limits_{0}^{1}{\left[ f\left( x \right)+f\left( -x \right) \right]dx} \)