Question
Evaluate: $\int_0^2(x-[x]) d x$
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Answer
$\text { (c) : Let } I=\int_0^2(x-[x]) d x=\int_0^2 x d x-\int_0^2[x] d x$
$=\left[\frac{x^2}{2}\right]_0^2-\int_0^1[x] d x-\int_1^2[x] d x=\frac{4}{2}-\int_0^1 0 d x-\int_1^2 1 d x$
$=2-0-[x]_1^2=2-[2-1]=2-1=1$
$=\left[\frac{x^2}{2}\right]_0^2-\int_0^1[x] d x-\int_1^2[x] d x=\frac{4}{2}-\int_0^1 0 d x-\int_1^2 1 d x$
$=2-0-[x]_1^2=2-[2-1]=2-1=1$
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