Question
If $[x]$ denotes the greatest integer function, then find $\int_0^{3 / 2}\left[x^2\right] d x$
✓
Answer
$\int_0^{3 / 2}\left[x^2\right] d x$
For $\quad 0 \leq x<1,0 \leq x^2<1$, hence $\left[x^2\right]=0$
For $1 \leq x<\sqrt{2}, 1 \leq x^2<2$, hence $\left[x^2\right]=1$
For $\quad \sqrt{2} \leq x<\frac{3}{2}, 2 \leq x^2<\frac{9}{4}$, hence $\left[x^2\right]=2$
$\begin{aligned} \int_0^{3 / 2}\left[x^2\right] d x & =\int_0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{3 / 2} 2 d x \\ & =0[x]_0^1+1[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{3 / 2} \\ & =0+\sqrt{2}-1+3-2 \sqrt{2} \\ & =2-\sqrt{2}\end{aligned}$
For $\quad 0 \leq x<1,0 \leq x^2<1$, hence $\left[x^2\right]=0$
For $1 \leq x<\sqrt{2}, 1 \leq x^2<2$, hence $\left[x^2\right]=1$
For $\quad \sqrt{2} \leq x<\frac{3}{2}, 2 \leq x^2<\frac{9}{4}$, hence $\left[x^2\right]=2$
$\begin{aligned} \int_0^{3 / 2}\left[x^2\right] d x & =\int_0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{3 / 2} 2 d x \\ & =0[x]_0^1+1[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{3 / 2} \\ & =0+\sqrt{2}-1+3-2 \sqrt{2} \\ & =2-\sqrt{2}\end{aligned}$
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