Q93 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $[x]$ is the greatest integer function not greater than $x$, then the value of $\displaystyle\int_0^2 x[x^2]\,dx$ is...
- A$\dfrac{3}{2}$
- B$\dfrac{5}{2}$
- Correct answerC$3$
- D$5$
Solution
Let $I = \displaystyle\int_0^2 x[x^2]\,dx$
Substitute $x^2 = t \Rightarrow 2x\,dx = dt \Rightarrow x\,dx = \dfrac{dt}{2}$
When $x = 0$, $t = 0$ and when $x = 2$, $t = 4$
$I = \dfrac{1}{2} \displaystyle\int_0^4 [t]\,dt$
$I = \dfrac{1}{2} \left( \displaystyle\int_0^1 0\,dt + \displaystyle\int_1^2 1\,dt + \displaystyle\int_2^3 2\,dt + \displaystyle\int_3^4 3\,dt \right)$
$I = \dfrac{1}{2} (0 + 1(2-1) + 2(3-2) + 3(4-3))$
$I = \dfrac{1}{2} (1 + 2 + 3) = \dfrac{6}{2} = 3$
Answer: $3$
Substitute $x^2 = t \Rightarrow 2x\,dx = dt \Rightarrow x\,dx = \dfrac{dt}{2}$
When $x = 0$, $t = 0$ and when $x = 2$, $t = 4$
$I = \dfrac{1}{2} \displaystyle\int_0^4 [t]\,dt$
$I = \dfrac{1}{2} \left( \displaystyle\int_0^1 0\,dt + \displaystyle\int_1^2 1\,dt + \displaystyle\int_2^3 2\,dt + \displaystyle\int_3^4 3\,dt \right)$
$I = \dfrac{1}{2} (0 + 1(2-1) + 2(3-2) + 3(4-3))$
$I = \dfrac{1}{2} (1 + 2 + 3) = \dfrac{6}{2} = 3$
Answer: $3$
Hence, the correct answer is option: 'C'.