∫01.5 x2dx is equal to - Infinity Learn by Sri Chaitanya

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

Solution:

We have,

$\begin{array}{l} \int_{0}^{1.5} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{1.5} [x^{2}] d x \\ \Rightarrow \int_{0}^{1.5} [x^{2}] d x = \int_{0}^{1} 0 d x + \int_{1}^{1.5} 1 \cdot d x + \int_{\sqrt{2}}^{1} 2 d x \end{array}$

$\Rightarrow \int_{0}^{1.5} [x^{2}] d x = 0 + (\sqrt{2} - 1) + 2 (1.5 - \sqrt{2}) = 2 - \sqrt{2}$

$\int_{0}^{1.5} [x^{2}] d x$ is equal to

Solution:

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