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
- A \(1\)
- B \(\frac{3}{2}\)
- C \(2\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\( \int_1^2|2 x-[3 x]| d x=\int_1^{\frac{4}{3}}|2 x-3| d x+\int_{\frac{4}{3}}^{\frac{5}{3}}|2 x-4| d x+\) \(\int_{\frac{5}{3}}^2|2 x-5| d x \)
\( =\int_1^{\frac{4}{3}}(3-2 x) d x+\int_{\frac{4}{3}}^{\frac{5}{3}}(4-2 x) d x+\int_{\frac{5}{3}}^2(5-2 x) d x \)
\( =\left[3 x-x^2\right]_1^{4 / 3}+\left[4 x-x^2\right]_{4 / 3}^{5 / 3}+\left[5 x-x^2\right]_{5 / 3}^2 \)
\( =3[x]_1^{4 / 3}+[4 x]_{4 / 3}^{5 / 3}+5\left[x^2\right]_{5 / 3}^2-\left[x^2\right]_1^2 \)
\( =(3+4+5) \times \frac{1}{3}-\left(2^2-1^2\right) \)
\( =4-3=1\)
\( =\int_1^{\frac{4}{3}}(3-2 x) d x+\int_{\frac{4}{3}}^{\frac{5}{3}}(4-2 x) d x+\int_{\frac{5}{3}}^2(5-2 x) d x \)
\( =\left[3 x-x^2\right]_1^{4 / 3}+\left[4 x-x^2\right]_{4 / 3}^{5 / 3}+\left[5 x-x^2\right]_{5 / 3}^2 \)
\( =3[x]_1^{4 / 3}+[4 x]_{4 / 3}^{5 / 3}+5\left[x^2\right]_{5 / 3}^2-\left[x^2\right]_1^2 \)
\( =(3+4+5) \times \frac{1}{3}-\left(2^2-1^2\right) \)
\( =4-3=1\)
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