Graphic integral

$\begin{aligned} I & = \int_0^{2020 \pi} \big|\sin x + \cos x\big| \ dx \\ & = \int_0^{2020 \pi} \left|\sqrt 2 \sin \left(x + \frac \pi 4\right) \right| \ dx & \small \blue{\text{Since the integrand has a period of }\pi,} \\ & = 2020\sqrt 2 \int_0^\pi \left|\sin \left(x + \frac \pi 4\right) \right| \ dx \\ & = 2020\sqrt 2 \left(\int_0^{\frac 34 \pi} \sin \left(x + \frac \pi 4\right) dx - \int_{\frac 34 \pi}^\pi \sin \left(x + \frac \pi 4\right) dx \right) & \small \blue{\text{Let }\theta = x + \frac \pi 4 \implies d\theta = d x} \\ & = 2020\sqrt 2 \left(\int_{\frac 14 \pi}^\pi \sin \theta \ d\theta - \int_\pi^{\frac 54 \pi} \sin \theta \ d\theta \right) \\ & = 2020\sqrt 2 \left(\int_{\frac 14 \pi}^\pi \sin \theta \ d\theta + \int_0^{\frac 14 \pi} \sin \theta \ d\theta \right) \\ & = 2020\sqrt 2 \int_0^\pi \sin \theta \ d\theta \\ & = 2020\sqrt 2 \bigg[-\cos \theta\bigg]_0^\pi \\ & = 4040 \sqrt 2 \end{aligned}$

Therefore $a+b = 4040 + 2 = \boxed{4042}$ .

$|\sin x+\cos x|=√2|\sin {(x+\dfrac{π}{4})}|$ . Therefore the value of the integral is $√2\times 2\times {2020}=4040√2$ . So $a=4040, b=2$ and $a+b=\boxed {4042}$