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(B) We split the integral based on the definition of the greatest integer function $[x]$: $\int_{-0.5}^{1.5} x^2[x] d x = \int_{-0.5}^{0} x^2[x] d x +

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(B) We split the integral based on the definition of the greatest integer function $[x]$: $\int_{-0.5}^{1.5} x^2[x] d x = \int_{-0.5}^{0} x^2[x] d x + \int_{0}^{1} x^2[x] d x + \int_{1}^{1.5} x^2[x] d x$ For $-0.5 \le x For $0 \le x For $1 \le x Substituting these values: $\int_{-0.5}^{0} x^2(-1) d x + \int_{0}^{1} x^2(0) d x + \int_{1}^{1.5} x^2(1) d x$ $= -\int_{-0.5}^{0} x^2 d x + 0 + \int_{1}^{1.5} x^2 d x$ $= -\left[ \frac{x^3}{3} \right]_{-0.5}^{0} + \left[ \frac{x^3}{3} \right]_{1}^{1.5}$ $= -\left( 0 - \frac{(-0.5)^3}{3} \right) + \left( \frac{(1.5)^3}{3} - \frac{1^3}{3} \right)$ $= -\left( 0 - \frac{-0.125}{3} \right) + \left( \frac{3.375}{3} - \frac{1}{3} \right)$ $= -\frac{0.125}{3} + \frac{2.375}{3} = \frac{2.25}{3} = 0.75 = \frac{3}{4}$

If $[x]$ is the greatest integer not exceeding $x$,then $\int_{-0.5}^{1.5} x^2[x] d x=$

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