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$\int_0^4 x[x] d x=($ where $[x]$ denotes greatest integer function not greater than $\mathrm{x}$ )
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Verified Answer
The correct answer is:
17
Let $I=\int_0^4 x[x] d x$
$$
\begin{aligned}
& \therefore I=\int_0^1(0) d x+\int_1^2 x d x+\int_2^3 2 x d x+\int_3^4 3 x d x \\
& =\left[\frac{x^2}{2}\right]_1^2+\left[\frac{2 x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}\right]_3^4 \\
& =\frac{1}{2}(4-1)+(9-4)+\left(\frac{3}{2}\right)(16-9)=\frac{3}{2}+5+\frac{21}{2}=17
\end{aligned}
$$
$$
\begin{aligned}
& \therefore I=\int_0^1(0) d x+\int_1^2 x d x+\int_2^3 2 x d x+\int_3^4 3 x d x \\
& =\left[\frac{x^2}{2}\right]_1^2+\left[\frac{2 x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}\right]_3^4 \\
& =\frac{1}{2}(4-1)+(9-4)+\left(\frac{3}{2}\right)(16-9)=\frac{3}{2}+5+\frac{21}{2}=17
\end{aligned}
$$
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