"\\text{As shown in the figure},\\\\\n\\text{ the area is divided to parts}\\\\\n\nA=\\int\\limits_{-7}^1(-x+1-0)dx+\\int\\limits_1^{11}(x-1-0)dx\\\\\n\n=\\int\\limits_{-7}^1(-x+1)dx+\\int\\limits_1^{11}(x-1)dx\\\\\n\n=(-x^2\/2+x)|_{-7}^1+(x^2\/2-x)|_1^{11}\\\\\n\n=-0.5+1+49\/2+7\\\\\n+121\/2-11-0.5+1\\\\\n\n=82"
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Since the function y=|x-1| has different signs on x1, area as the definite integrals should be considered for these two cases separately. The bounds a=-7, b=1 were chosen because x1 was considered and the curve is bounded by the line x=11.
HI. Since we know that we are given the value a=-7 and b=11, do we just assume that since were not given a b for both integrals, we will then evaluate by 1, b=-7 and a=1, 11
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