Question:

021xdx\int\limits^2_0 |1 -x| dx is equal to

Updated On: Jan 30, 2025
  • 0
  • 1
  • 32\frac{3}{2}
  • 12\frac{1}{2}
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The Correct Option is B

Solution and Explanation

021xdx=01(1x)dx+12(x1)dx\int\limits_{0}^{2}|1-x| d x=\int\limits_{0}^{1}(1-x) d x+\int\limits_{1}^{2}(x-1) d x
=[xx22]01+[x22x]12=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{2}
=112+[22(121)]=1-\frac{1}{2}+\left[2-2-\left(\frac{1}{2}-1\right)\right]
=12+12=1=\frac{1}{2}+\frac{1}{2}=1
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