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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.)

  1. A 3
  2. B 4
  3. C 1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(D) 2

Step-by-step Solution

Detailed explanation

\( \int_0^2[x] d x+\int_0^2[x-1] d x \)
\( =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2\) \((x-1) d x \)
\( =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^2 \)
\( =0+1+\frac{1}{2}+\frac{1}{2}=2\)