If $\int_0^1 f(x) d x=1, \int_0^1 x f(x) d x=a$ and $\int_0^1 x^2 f(x) d x=a^2$, then $\int_0^1(x-a)^2 f(x)…
If $\int_0^1 f(x) d x=1, \int_0^1 x f(x) d x=a$ and $\int_0^1 x^2 f(x) d x=a^2$, then $\int_0^1(x-a)^2 f(x) d x$ is equal to
$a^2$
$a^2+1$
$a^2-1$
0
Solution
$
\text { } \begin{aligned}
& \int_0^1(x-a)^2 f(x)=\int_0^1\left(x^2 f(x)+a^2 f(x)-2 a x f(x)\right) d x \\
& =\int_0^1 x^2 f(x) d x+a^2 \int_0^1 f(x) d x-2 a \int_0^1 x f(x) d x \\
& =a^2+a^2(1)-2 a(a) \\
& =a^2+a^2-2 a^2=0
\end{aligned}
$