Mathematics

Find the value of $$\displaystyle \int _{0}^{2}|1-x|\ dx$$.


SOLUTION
Now,
$$\displaystyle \int _{0}^{2}|1-x|\ dx$$
$$=\displaystyle \int _{0}^{1}(x-1)\ dx$$$$+\displaystyle \int _{1}^{2}(1-x)\ dx$$
$$=\left[\dfrac{x^2}{2}-x\right]_0^1$$$$+\left[-\dfrac{x^2}{2}+x\right]_1^2$$
$$=\dfrac{1}{2}-1-\dfrac{3}{2}+1$$
$$=-1$$.
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Subjective Medium Published on 17th 09, 2020
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