Answer:
[tex]\frac{4}{\pi}+1[/tex]
Step-by-step explanation:
Find intersection points:
[tex]x^2-x=2\\x^2-x-2=0\\(x-2)(x+1)=0[/tex]
x = 2 and x = -1 are intersection points.
[tex]x^2-x=sin(\pi x)[/tex]
x = 0 and x = 1 are intersection points.
So,
[tex]\int\limits^1_0 (sin(\pi x)-(x^2-x))dx + \int\limits^2_1 ((x^2-x)-sin(\pi x))dx=\\\\=(-\frac{1}{\pi}cox(\pi x)-\frac{1}{3}x^3 +\frac{1}{2}x^2)|\limits^1_0+(\frac{1}{3}x^3-\frac{1}{2}x^2)+\frac{1}{\pi}cox(\pi x)|\limits^2_1=\\\\=\frac{4}{\pi}+1[/tex]
