Q93 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
If $[x]$ denotes the greatest integer less than or equal to $x$, then the value of the integral $\int_0^2 x^2[x]\, dx$ is equal to
- A$\dfrac{8}{3}$
- B$\dfrac{3}{8}$
- Correct answerC$\dfrac{7}{3}$
- D$0$
Solution
The given integral is $I = \int_0^2 x^2[x]\, dx$.
Using the property of the greatest integer function, we split the integral at integer points within the limits of integration:
$I = \int_0^1 x^2[x]\, dx + \int_1^2 x^2[x]\, dx$
For $x \in [0, 1)$, $[x] = 0$.
For $x \in [1, 2)$, $[x] = 1$.
Substituting these values into the integrals:
$I = \int_0^1 x^2(0)\, dx + \int_1^2 x^2(1)\, dx$
$I = 0 + \left[ \dfrac{x^3}{3} \right]_1^2$
$I = \dfrac{2^3}{3} - \dfrac{1^3}{3}$
$I = \dfrac{8}{3} - \dfrac{1}{3} = \dfrac{7}{3}$
Answer: $\dfrac{7}{3}$
Using the property of the greatest integer function, we split the integral at integer points within the limits of integration:
$I = \int_0^1 x^2[x]\, dx + \int_1^2 x^2[x]\, dx$
For $x \in [0, 1)$, $[x] = 0$.
For $x \in [1, 2)$, $[x] = 1$.
Substituting these values into the integrals:
$I = \int_0^1 x^2(0)\, dx + \int_1^2 x^2(1)\, dx$
$I = 0 + \left[ \dfrac{x^3}{3} \right]_1^2$
$I = \dfrac{2^3}{3} - \dfrac{1^3}{3}$
$I = \dfrac{8}{3} - \dfrac{1}{3} = \dfrac{7}{3}$
Answer: $\dfrac{7}{3}$
Hence, the correct answer is option: 'C'.