Integral from normal variable.

Supposoe we have normal variable $ N(mu_t, sigma_t^2 ) $.

Let’s define X = $int_0^T N(mu_t, sigma_t^2 ) dt$.

Is X also a normal variable?

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It’s simple to check that expectation of X is simply $int_0^T mu_t dt$ as
$mathbb{E} (int) = int (mathbb{E})$.

BUT it is not the case for variance as

$mathbb{V} (X) = mathbb{V} (int_0^T N(mu_t, sigma_t^2 ) dt)
approx mathbb{V} (sum_i N(mu_i, sigma_i^2 ) triangle) =
> mathbb{V} (sum_i N(mu_i, triangle^2sigma_i^2 ) )$

Entering the sum into normal variable:

$mathbb{V} (N( sum_i mu_i, sum_i triangle^2sigma_i^2 ) ) =
sum_i triangle^2sigma_i^2$

Now I would like to get back to integrals and do something like this:

$sum_i triangle^2sigma_i^2 approx int_0^T triangle sigma_t^2
dt, triangle to 0$

But it seems weird…

Could someone help me to find the good variance and figure out where I
was wrong?