15 ∫ 0 [ X 2 ] D X - Mathematics

Sum

$\int\limits_0^{15} \left[ x^2 \right] dx$

Solution

We have,

$I = \int\limits_0^{1 . 5} \left[ x^2 \right] dx$

$= \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx + \int\limits_\sqrt{2}^{1 . 5} \left[ x^2 \right] dx$

$= \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^\sqrt{2} \left( 1 \right) dx + \int\limits_\sqrt{2}^{1 . 5} \left( 2 \right) dx ..............\left(\because \left[ x^2 \right] = \begin{cases}0 &where,& 0 < x < 1 \\ 1 &where,& 1 < x < \sqrt{2}\\2 &where,& \sqrt{2} < x < 1.5 \end{cases}\right)$

$= 0 + \left[ x \right]_1^\sqrt{2} + \left[ 2x \right]_\sqrt{2}^{1 . 5}$

$= \left[ x \right]_1^\sqrt{2} + 2 \left[ x \right]_\sqrt{2}^{1 . 5}$

$= \left( \sqrt{2} - 1 \right) + 2\left( 1 . 5 - \sqrt{2} \right)$

$= \sqrt{2} - 1 + 3 - 2\sqrt{2}$

$= 2 - \sqrt{2}$

Concept: Definite Integrals Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 20 Definite Integrals
Revision Exercise | Q 45 | Page 122