∫_0^1.5 [x^2 ] d x is equal to

Correct answer: (B) 2-√2. ∫_0^1.5 [x^2 ] d x =∫_0^1 [x^2 ] d x+∫_1^√2 [x^2 ] d x+∫_√2^3 / 2 [x^2 ] d x =0+1(√2-1)+2(1.5-√2) =2-√2

Q70 / 155 · WBJEE 2025 · Mathematics

$\int_0^{1.5}\left[x^2\right] d x$ is equal to

A
$2$
Correct answer
B
$2-\sqrt{2}$
C
$2+\sqrt{2}$
D
$\sqrt{2}$

Solution

$\begin{aligned} & \int_0^{1.5}\left[x^2\right] d x \\ & =\int_0^1\left[x^2\right] d x+\int_1^{\sqrt{2}}\left[x^2\right] d x+\int_{\sqrt{2}}^{3 / 2}\left[x^2\right] d x \\ & =0+1(\sqrt{2}-1)+2(1.5-\sqrt{2}) \\ & =2-\sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'B'.

$\int_0^{1.5}\left[x^2\right] d x$ is equal to

Solution

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