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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 \(u = 1 – x \Rightarrow du = – dx\)
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\(\int\limits_0^1 {f\left( x \right)} dx = \int\limits_1^0 {f\left( {1 – u} \right)} \left( { – du} \right) = \int\limits_0^1 {f\left( {1 – u} \right)} du = \int\limits_0^1 {f\left( {1 – x} \right)} dx\)
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 \(\int\limits_{ – 1}^1 {f\left( x \right)} dx = \int\limits_0^1 {\left[ {f\left( x \right) + f\left( { – x} \right)} \right]} dx\)