If ∫ _ - 0 . 15 ^ 0 . 15 |100 x ^ 2 - 1|dx = (k)/(3000), then k is...

Given,

${∫}_{- 0 . 15}^{0 . 15}\left|100{x}^{2} - 1\right| d x = \dfrac{k}{3000}$

Now we know that, ${∫}_{- a}^{a}f\left(x\right) d x = 2{∫}_{0}^{a}f\left(x\right) d x\text{if}f\left(x\right)\text{is even}$

So, ${∫}_{- 0 . 15}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x$

Now solving, $2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2\left({∫}_{0}^{0 . 1} - \left(100{x}^{2} - 1\right) d x + {∫}_{0 . 1}^{0 . 15}100{x}^{2} - 1 d x\right)$

$⇒ 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2\left({\left[x - \dfrac{100{x}^{3}}{3}\right]}_{0}^{0 . 1} + {\left[\dfrac{100{x}^{3}}{3} - x\right]}_{0 . 1}^{0 . 15}\right)$

$⇒ 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2\left(\left(\dfrac{1}{10} - \dfrac{1}{30}\right) + \left(\dfrac{100 \times {15}^{3}}{3 \times {100}^{3}} - \dfrac{15}{100}\right) - \left(\dfrac{1}{30} - \dfrac{1}{10}\right)\right)$

$⇒ 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2\left(\dfrac{2}{30} + \left(\dfrac{225}{2000} - \dfrac{15}{100}\right) + \dfrac{2}{30}\right)$

$⇒ 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = 2\left(\dfrac{4}{30} + \left(\dfrac{- 3}{80}\right)\right) = \left(\dfrac{8}{30} - \dfrac{3}{40}\right) = \dfrac{32 - 9}{120} = \dfrac{23}{120} = \dfrac{575}{3000}$

$⇒ 2{∫}_{0}^{0 . 15}\left|100{x}^{2} - 1\right| d x = \dfrac{575}{3000}$

Now comparing with ${∫}_{- 0 . 15}^{0 . 15}\left|100{x}^{2} - 1\right| d x = \dfrac{k}{3000}$, we get $k = 575$