[tex]\displaystyle\int_0^1x\,\mathrm dx[/tex] corresponds to the area under the curve [tex]y=x[/tex] from 0 to 1. This region is a right triangle with base and height 1, so the area of the triangle, and thus the value of the integral, is [tex]\dfrac12\times1\times1=\dfrac12[/tex].
[tex]\displaystyle\int_0^1\sqrt{1-x^2}\,\mathrm dx[/tex] is the area of one quadrant of the unit circle. Recall that [tex]x^2+y^2=1[/tex] is the equation of the circle with radius 1 centered at the origin. Solving for [tex]y[/tex] gives [tex]y=\pm\sqrt{1-x^2}[/tex], where the positive root corresponds to the top half of the circle. The top half is defined over the interval [tex][-1,1][/tex], but you're only interested in the right half, [tex][0,1][/tex]. So this area is [tex]\dfrac14\pi\times1^2=\dfrac\pi4[/tex].
Put these together to get
[tex]\displaystyle\int_0^1x\,\mathrm dx+\int_0^1\sqrt{1-x^2}\,\mathrm dx=\int_0^1(x+\sqrt{1-x^2})\,\mathrm dx=\frac12+\frac\pi4[/tex]
