Let's represent the required area on the coordinate plane:
We can see that it consists of two isosceles right triangles with sides 8 and 10, so we can find the required area as a sum of two areas of triangles.
"A=\\frac{1}{2}*8*8+\\frac{1}{2}*10*10=32+50=82."
Another way to find the area is using integrals. As our area consists of two parts, we have to split integration area in two areas, separated by point x=1.
"y=|x-1|=\\begin{cases} 1-x, & x < 1, \\\\ x-1, & x\\ge 1. \\end{cases}"
"A=\\int_{-7}^{11} |x-1|dx=\\int_{-7}^{1} (1-x)dx+\\int_{1}^{11} (x-1)dx="
"=(x-\\frac{x^2}{2}) |_{-7}^{1} +(\\frac{x^2}{2}-x) |_{1}^{11} ="
"=(1-\\frac{1^2}{2})-(-7-\\frac{(-7)^2}{2})+(\\frac{11^2}{2}-11)-(\\frac{1^2}{2}-1)="
"=1-\\frac{1}{2}+7+\\frac{49}{2}+\\frac{121}{2}-11-\\frac{1}{2}+1=(1+7-11+1)+(-\\frac{1}{2}+\\frac{49}{2}+\\frac{121}{2}-\\frac{1}{2})="
"=-2+\\frac{168}{2}=-2+84=82."
Answer: 82.
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