int_0^1 f(1 - x) , dx has the same value as w | vedclass

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Question

(A) Let $I = \int_0^1 f(1 - x) \, dx$. Substitute $1 - x = t$,which implies $x = 1 - t$ and $dx = -dt$. When $x = 0$,$t = 1$. When $x = 1$,$t = 0$. Su

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$\int_0^1 f(x) \, dx$

Vedclass · Answer

(A) Let $I = \int_0^1 f(1 - x) \, dx$. Substitute $1 - x = t$,which implies $x = 1 - t$ and $dx = -dt$. When $x = 0$,$t = 1$. When $x = 1$,$t = 0$. Substituting these into the integral: $I = \int_1^0 f(t) (-dt) = -\int_1^0 f(t) \, dt = \int_0^1 f(t) \, dt$. Since the variable of integration is a dummy variable,$\int_0^1 f(t) \, dt = \int_0^1 f(x) \, dx$. Thus,$\int_0^1 f(1 - x) \, dx = \int_0^1 f(x) \, dx$.

$\int_0^1 f(1 - x) \, dx$ has the same value as which of the following integrals?

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