Let X be random variable with the probability density function
$f(x)= \begin{cases} \ ax^2 & { if \ \ \ 0 < x<1}\\ \ bx^{-4}& \text { if $\ \ \ 1 \le x<\infty$} \\ \end{cases} $
Where $a$ and $b$ are positive real numbers .If $E(X)=1$, then $E(X^2)$ equals___
$\int_{-\infty}^{\infty}f(x)=\int_{0}^{1} ax^2dx+\int_{1}^{\infty}bx^{-4}dx=\frac{ax^3}{3}\bigg|_{0}^{1}+\frac{bx^{-3}}{-3}\bigg|_{1}^{\infty}\implies a+b=3---(1)$
$E(X)=\int_{-\infty}^{\infty}xf(x)=\int_{0}^{1} ax^3dx+\int_{1}^{\infty}bx^{-3}dx=\frac{ax^4}{4}\bigg|_{0}^{1}+\frac{bx^{-2}}{-2}\bigg|_{1}^{\infty}\implies \frac{a}{4}+\frac{b}{2}\implies a+ 2b=4---(1)$
Solving $(1)$and $(2)$ we get $a=2$ and $b=1$
$E(X^2)=\int_{-\infty}^{\infty}x^2f(x)=\int_{0}^{1} ax^4dx+\int_{1}^{\infty}bx^{-2}dx=\frac{ax^5}{5}\bigg|_{0}^{1}+\frac{bx^{-1}}{-1}\bigg|_{1}^{\infty}\implies \frac{a}{5}+{b}\implies \frac{2}{5}+1=\frac{7}{5}=1.4$
Can anyone point out my mistake did I do everything right? My answer key says $3.5$.