Bài tập 22 trang 162 SGK Toán 12 NC
Bài tập 22 trang 162 SGK Toán 12 NC
Chứng minh rằng
\(\begin{array}{l}
a)\int \limits_0^1 f\left( x \right)dx = \int \limits_0^1 f\left( {1 - x} \right)dx\\
b)\int \limits_{ - 1}^1 f\left( x \right)dx = \int \limits_0^1 \left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx
\end{array}\)
Đặt u = 1 − x ⇒ du = −dx
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\(\begin{array}{l}
\int\limits_0^1 f \left( x \right)dx = \int\limits_1^0 f \left( {1 - u} \right)\left( { - du} \right)\\
= \int\limits_0^1 f \left( {1 - u} \right)dx = \int\limits_0^1 f \left( {1 - x} \right)dx
\end{array}\)
b)
\(\int \limits_{ - 1}^1 f\left( x \right)dx = \int \limits_{ - 1}^0 f\left( x \right)dx + \int \limits_0^1 f\left( x \right)dx\)
Tính \(\int \limits_{ - 1}^0 f\left( x \right)dx\)
Đặt u = −x ⇒ du = − dx
Khi đó:
\(\begin{array}{l}
\int\limits_{ - 1}^0 f \left( x \right)dx = \int\limits_1^0 f \left( { - u} \right)\left( { - du} \right)\\
= \int\limits_0^1 f \left( { - u} \right)du = \int\limits_0^1 f \left( { - x} \right)dx
\end{array}\)
Do đó:
\(\int \limits_{ - 1}^1 f\left( x \right)dx = \int \limits_0^1 \left[ {f\left( x \right) + f\left( { - x} \right)} \right]dx\)
-- Mod Toán 12
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