MHT CET · Maths · Definite Integration
\(\int_0^2[x] d x+\int_0^2|x-1| d x=\)
(where \([x]\) denotes the greatest integer function.)
- A 3
- B 4
- C 1
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
\( \int_0^2[x] d x+\int_0^2[x-1] d x \)
\( =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2\) \((x-1) d x \)
\( =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^2 \)
\( =0+1+\frac{1}{2}+\frac{1}{2}=2\)
\( =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2\) \((x-1) d x \)
\( =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^2 \)
\( =0+1+\frac{1}{2}+\frac{1}{2}=2\)
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