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The area of the region bounded by the curves y=1+∣x−1∣,x=−1,x=2,y=0y = 1+|x-1|, x=-1, x=2, y=0 is

Solution

✅ Correct Option: 1

For x<1x<1, y=1+(1−x)=2−xy=1+(1-x)=2-x; for x≥1x \geq 1, y=xy=x. So the area is ∫−11(2−x)dx+∫12x dx=4+32=112\int_{-1}^{1}(2-x)dx+\int_{1}^{2}x\,dx=4+\frac{3}{2}=\frac{11}{2} square units.

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