Question : Integrate $\displaystyle \int_{-1}^{1} |1- x| dx$
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Originally Posted by zorro Question : Integrate $\displaystyle \int_{-1}^{1} |1- x| dx$ Note that $\displaystyle |1 - x| = 1 - x$ when $\displaystyle x < 1$ and |$\displaystyle 1 - x| = -(1 - x) = x - 1$ when $\displaystyle x \geq 1$.
or note that the integrand is non-negative for $\displaystyle -1\le x\le1,$ so the absolute value is irrelevant there and you can simply evaluate $\displaystyle \int_{-1}^1(1-x)\,dx.$
Originally Posted by zorro Question : Integrate $\displaystyle \int_{-1}^{1} |1- x| dx$ $\displaystyle \forall x\in [-1,1],1-x\geq 0$
Originally Posted by Raoh $\displaystyle \forall x\in [-1,1],1-x\geq 0$ $\displaystyle \int_{-1}^{1} |1-x|dx \ $= $\displaystyle \int_{-1}^{1} (1-x)dx$ = ........ = $\displaystyle 2$ Is this correct???
Originally Posted by zorro $\displaystyle \int_{-1}^{1} |1-x|dx \ $= $\displaystyle \int_{-1}^{1} (1-x)dx$ = ........ = $\displaystyle 2$ Is this correct??? i believe you're right
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