Bài 22. 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) Đặt \(u = 1 - x \Rightarrow du = - dx\)
Advertisements (Quảng cáo)
\(\int\limits_0^1 {f\left( x \right)} dx = \int\limits_1^0 {f\left( {1 - u} \right)} \left( { - du} \right) = \int\limits_0^1 {f\left( {1 - u} \right)} du = \int\limits_0^1 {f\left( {1 - x} \right)} dx\)
b) \(\int\limits_{ - 1}^1 {f\left( x \right)} dx = \int\limits_{-1}^0 {f\left( x \right)} dx + \int\limits_0^1 {f\left( x \right)} dx\) với \(\int\limits_{ - 1}^0 {f\left( x \right)} dx\)
Đặt \(u = - x \Rightarrow du = - dx\)
Khi đó \(\int\limits_{ - 1}^0 {f\left( x \right)dx = \int\limits_1^0 {f\left( { - u} \right)} } \left( { - du} \right) = \int\limits_0^1 {f\left( { - u} \right)} du = \int\limits_0^1 {f\left( { - x} \right)} dx\)
Do đó \(\int\limits_{ - 1}^1 {f\left( x \right)} dx = \int\limits_0^1 {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} dx\)