If ∫ - 0 . 15 0 . 15 100 x 2 - 1 d x = k 3000 , then k is equal to _____.

Given,

${\int }_{-0.15}^{0.15}\left|100x^2-1\right|dx=\frac{k}{3000}$

Now we know that, ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx \text{if} f\left(x\right) \text{is even}$

So, ${\int }_{-0.15}^{0.15}\left|100x^2-1\right|dx=2{\int }_0^{0.15}\left|100x^2-1\right|dx$

Now solving, $2{\int }_0^{0.15}\left|100x^2-1\right|dx=2\left({\int }_0^{0.1}-\left(100x^2-1\right)dx+{\int }_{0.1}^{0.15}100x^2-1dx\right)$

$\Rightarrow 2{\int }_0^{0.15}\left|100x^2-1\right|dx=2\left({\left[x-\frac{100x^3}{3}\right]}_0^{0.1}+{\left[\frac{100x^3}{3}-x\right]}_{0.1}^{0.15}\right)$

$\Rightarrow 2{\int }_0^{0.15}\left|100x^2-1\right|dx=2\left(\left(\frac{1}{10}-\frac{1}{30}\right)+\left(\frac{100\times {15}^3}{3\times {100}^3}-\frac{15}{100}\right)-\left(\frac{1}{30}-\frac{1}{10}\right)\right)$

$\Rightarrow 2{\int }_0^{0.15}\left|100x^2-1\right|dx=2\left(\frac{2}{30}+\left(\frac{225}{2000}-\frac{15}{100}\right)+\frac{2}{30}\right)$

$\Rightarrow 2{\int }_0^{0.15}\left|100x^2-1\right|dx=2\left(\frac{4}{30}+\left(\frac{-3}{80}\right)\right)=\left(\frac{8}{30}-\frac{3}{40}\right)=\frac{32-9}{120}=\frac{23}{120}=\frac{575}{3000}$

$\Rightarrow 2{\int }_0^{0.15}\left|100x^2-1\right|dx=\frac{575}{3000}$

Now comparing with ${\int }_{-0.15}^{0.15}\left|100x^2-1\right|dx=\frac{k}{3000}$, we get $k=575$