Q90 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $[\cdot]$ denotes the greatest integer function, then $\int_1^4 x[x] dx$ is equal to
- A$34$
- B$-17$
- C$-34$
- Correct answerD$17$
Solution
The given integral can be split at integer values since $[x]$ changes its value at integers.
$I = \int_1^4 x[x] dx$
$I = \int_1^2 x[x] dx + \int_2^3 x[x] dx + \int_3^4 x[x] dx$
Substituting the values of $[x]$ in the respective intervals:
$I = \int_1^2 x(1) dx + \int_2^3 x(2) dx + \int_3^4 x(3) dx$
$I = \left[ \dfrac{x^2}{2} \right]_1^2 + \left[ x^2 \right]_2^3 + \left[ \dfrac{3x^2}{2} \right]_3^4$
$I = \left( \dfrac{4}{2} - \dfrac{1}{2} \right) + (9 - 4) + \left( \dfrac{48}{2} - \dfrac{27}{2} \right)$
$I = \dfrac{3}{2} + 5 + \dfrac{21}{2}$
$I = \dfrac{24}{2} + 5$
$I = 12 + 5 = 17$
Answer: $17$
$I = \int_1^4 x[x] dx$
$I = \int_1^2 x[x] dx + \int_2^3 x[x] dx + \int_3^4 x[x] dx$
Substituting the values of $[x]$ in the respective intervals:
$I = \int_1^2 x(1) dx + \int_2^3 x(2) dx + \int_3^4 x(3) dx$
$I = \left[ \dfrac{x^2}{2} \right]_1^2 + \left[ x^2 \right]_2^3 + \left[ \dfrac{3x^2}{2} \right]_3^4$
$I = \left( \dfrac{4}{2} - \dfrac{1}{2} \right) + (9 - 4) + \left( \dfrac{48}{2} - \dfrac{27}{2} \right)$
$I = \dfrac{3}{2} + 5 + \dfrac{21}{2}$
$I = \dfrac{24}{2} + 5$
$I = 12 + 5 = 17$
Answer: $17$
Hence, the correct answer is option: 'D'.