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MHT CET · Maths · Definite Integration

Let \([t]\) denot the greatest integer less than or equal to \(t\). Then the value of \(\int_1^2|2 x-[3 x]| d x\) is

  1. A \(1\)
  2. B \(\frac{3}{2}\)
  3. C \(2\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

\( \int_1^2|2 x-[3 x]| d x=\int_1^{\frac{4}{3}}|2 x-3| d x+\int_{\frac{4}{3}}^{\frac{5}{3}}|2 x-4| d x+\) \(\int_{\frac{5}{3}}^2|2 x-5| d x \)
\( =\int_1^{\frac{4}{3}}(3-2 x) d x+\int_{\frac{4}{3}}^{\frac{5}{3}}(4-2 x) d x+\int_{\frac{5}{3}}^2(5-2 x) d x \)
\( =\left[3 x-x^2\right]_1^{4 / 3}+\left[4 x-x^2\right]_{4 / 3}^{5 / 3}+\left[5 x-x^2\right]_{5 / 3}^2 \)
\( =3[x]_1^{4 / 3}+[4 x]_{4 / 3}^{5 / 3}+5\left[x^2\right]_{5 / 3}^2-\left[x^2\right]_1^2 \)
\( =(3+4+5) \times \frac{1}{3}-\left(2^2-1^2\right) \)
\( =4-3=1\)