From density function to cumulative distribution function? Given $$f(y)=\theta/(\delta^{\theta}y^{\theta+1})\mathbb{1(y>1/\delta)}$$ where the last factor is the indicator function, and I am asked to compute the Cumulative Distribution Function of y:
$$\int\theta/\delta^{\theta}y^{\theta+1}dy$$
and I get a result different from what the solution says, which would be $$1-(1/\delta^{\theta}y^{\theta})$$.
The $1-$ part makes me think I should consider the complement because we have that in the indicator function $y>1/\delta$
I would really appreciate an intuitive explanation behind this.

When computing a CDF from a density (PDF), you may safely multiply the density by any positive constant you like.  The reason is that any CDF must rise from a value of $0$ for extremely negative arguments to a value of $1$ for extremely positive arguments.  The value your modified CDF attains in the latter case tells you how to rescale it to be correct.
Exploit this flexibility by choosing a constant that simplifies your calculations.  In this case, because obviously $\theta \gt 0,$ observe that $f$ is proportional to
$$f(y) \ \propto \ \theta y^{-\theta -1}, \quad y \gt 1/\delta.$$
I chose to write it like this because $-\theta y^{-\theta-1}$ fits the pattern of differentiating a power: it is the derivative of $-y^{-\theta}.$  No further effort will be required for integration.  Already this tells us the CDF must be in the form
$$F(x) = A - B x^{-\theta}$$
for $x \gt 1/\delta.$
$A$ is a constant of integration (which we must find) and $B$ is the constant of proportionality we introduced into $f,$ which we also must find.
At this juncture you could just "guess and check" since
$$f(x) = \frac{\mathrm d}{\mathrm d x} F(x) = B\theta x^{-\theta-1}$$
(which, when compared to the original expression for $f(x),$ easily yields the value of $B$) and
$$1 = \lim_{x\to\infty} F(x) = \lim_{x\to\infty} A - B x^{-\theta} = A.$$
This often works well.  But if somehow it doesn't produce an answer, let's pull out the machinery of integral Calculus and proceed.
The CDF $F$ is the integral of the density beginning at $-\infty,$ whence
$$F(x)\ \propto\ \int_{-\infty}^x f(y)\,\mathrm{d}y = \int_{-\infty}^{1/\delta} f(y)\,\mathrm{d}y + \int_{1/\delta}^x f(y)\,\mathrm{d}y = \int_{1/\delta}^x f(y)\,\mathrm{d}y$$
because on the interval $(-\infty, x]$ the integrand is $0.$  Thus, for $x\ge 1/\delta,$
$$F(x) \ \propto \ \int_{1/\delta}^x \theta y^{-\theta - 1}\,\mathrm{d}\theta = -y^{-\theta}\bigg|_{1/\delta}^x = \delta^\theta - x^{-\theta}.$$
The limit of this as $x\to\infty$ is $\delta^\theta.$  Dividing $F$ by this limiting value yields the CDF,
$$F(x) = \frac{\delta^\theta - x^{-\theta}}{\delta^\theta} \mathbb{I}\left(x \gt \frac{1}{\delta}\right) = \left(1 - \frac{1}{\left(\delta x\right)^\theta}\right)\mathbb{I}\left(x \gt \frac{1}{\delta}\right).$$
