Analytical Solution for the Convolution of Signal with a Box Filter
DO NOT PASS GO; DO NOT Collect $200; DO NOT Take Fourier transforms, or worse yet, FFTs


You do the convolution exactly the way you would do any other convolution: start with the basic convolution integral (not what you wrote) and apply the properties of the signals that you are using to come up with an easier calculation.  Begin with
$$\int_{-\infty}^\infty y(x^\prime)h(x^\prime-x)\mathrm dx^\prime\tag{1}$$
and use the result that $h(\cdot)$ is nonzero only when its argument ($x^\prime-x$ in this instance) lies in the interval $\left[-\frac{\Delta}2, +\frac{\Delta}2\right]$. So, the integrand of the convolution integral is $0$ whenever $x^\prime$ is such that
$$x^\prime-x &gt; \frac{\Delta}2 \implies  x^\prime &gt; x + \frac{\Delta}2$$
or that
$$x^\prime-x &lt; -\frac{\Delta}2 \implies  x^\prime &lt; x - \frac{\Delta}2.$$
This allows us to simplify the convolution integral $(1)$ into
$$\int_{x - \frac{\Delta}2}^{x + \frac{\Delta}2} y(x^\prime)\frac{1}{\Delta}\mathrm dx^\prime\tag{2}$$
where I have substituted the nonzero value of $h$ in the integral to save a step.  Can you take it from here?