If int_0^1 f(x) dx = 1,int_0^1 x f(x) dx = a, | vedclass

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(D) We are given the integrals $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$. Expanding the integrand $(x-a)^2 f(x)

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(D) We are given the integrals $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$. Expanding the integrand $(x-a)^2 f(x) = (x^2 - 2ax + a^2) f(x) = x^2 f(x) - 2ax f(x) + a^2 f(x)$. Now,integrate term by term: $\int_0^1 (x-a)^2 f(x) dx = \int_0^1 x^2 f(x) dx - 2a \int_0^1 x f(x) dx + a^2 \int_0^1 f(x) dx$. Substituting the given values: $= a^2 - 2a(a) + a^2(1) = a^2 - 2a^2 + a^2 = 0$.

If $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$,then $\int_0^1 (x-a)^2 f(x) dx$ is equal to

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