Finding $Var(X)$ without integrating:
$Var(X) = E(X^2) - E(X)^2$ by definition.
If $X\sim Norm(60, 8^2)$ where $Norm(\mu = 60, \sigma^2 = 8^2)$. Find $Var(X)$. Given that $E(X) = \mu$
$$Var(Z) = E(Z^2) + 0 = \int_{-\infty}^{\infty}z^2 \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\,dx = 1$$
Find $Var(X)$ without integrating.
My attempt:
$$Z = \frac{X-u}{\sigma}\sim Norm(0, 1) \to Z = \frac{X-60}{8}\sim Norm(0, 1)$$
$$1 = Var(Z) = Var\left[\frac{X}{8} - \frac{60}{8} \right]$$
I believe this property could be useful from my textbook $Var(a + bx) = b^2Var(X), \forall a, b, \in \mathbb R$
$1 = Var(Z) = Var\left[\frac{X}{8} - \frac{60}{8} \right] = \frac{1}{8^2}Var(X)$. Therefore $Var(X) = 64$
Not sure if this is possible, and why its wrong. Below is the actual answer of $3664$
$$E(X^2) = \int_{-\infty}^{\infty} x^2\frac{1}{8\sqrt{2\pi}}e^{-1/128(x-60)^2} \, dx = 3664$$
You have the correct $\mathsf E(X^2)$, however, that is not the variance.
$\mathsf{Var}(X)$ $=\mathsf E(X^2)-\mathsf E(X)^2 \\= 3664-(60)^2\\= 64\\=8^2$
Now, can you verify that $\mathsf E(X)=60$?