If $\int_\limits{-0.15}^{0.15}\left|100 x^{2}-1\right| d x=\frac{k}{3000}$, then $k$ is equal to ___________.

$$ \int\limits_{-0.15}^{0.15}\left|100 x^2-1\right| d x=2 \int\limits_0^{0.15}\left|100 x^2-1\right| \mathrm{dx} $$ <br/><br/>$\text { Now } 100 x^2-1=0 \Rightarrow x^2=\frac{1}{100} \Rightarrow x=0.1$ <br/><br/>$$ I=2\left[\int_0^{0.1}\left(1-100 x^2\right) d x+\int_{0.1}^{0.15}\left(100 x^2-1\right) d x\right] $$ <br/><br/>$$ \begin{aligned} I & =2\left[x-\frac{100}{3} x^3\right]_0^{0.1}+2\left[\frac{100 x^3}{3}-x\right]_{0.1}^{0.15} \\\\ & =2\left[0.1-\frac{0.1}{3}\right]+2\left[\frac{0.3375}{3}-0.15-\frac{0.1}{3}+0.1\right] \\\\ & =2\left[0.2-\frac{0.2}{3}+0.1125-0.15\right] \\\\ & =2\left[\frac{5}{100}-\frac{2}{30}+\frac{1125}{10000}\right]=2\left(\frac{1500-2000+3375}{30000}\right) \\\\ & =\frac{575}{3000} \Rightarrow \mathrm{k}=575 \end{aligned} $$