If _-0.15^0.15|100x^2-1|dx=(k)/(3000), then k is equal to ___________.

Question

If $\int _{-0.15}^{0.15}|100{x}^{2}-1|dx=\frac{k}{3000}$, then $k$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 575

Accepted Answer

$\int _{-0.15}^{0.15}|100{x}^{2}-1|dx=2\int _{0}^{0.15}|100{x}^{2}-1|dx$

$100{x}^{2}-1=0\Rightarrow {x}^{2}=\frac{1}{100}\Rightarrow x=0.1$

$I=2[{\int }_{0.1}^{0}(1-100{x}^{2})dx+{\int }_{0.15}^{0.1}(100{x}^{2}-1)dx]$

$\begin{gathered} I =2{[x-\frac{100}{3}{x}^{3}]}_{0.1}^{0}+2{[\frac{100{x}^{3}}{3}-x]}_{0.15}^{0.1} \ \ =2[0.1-\frac{0.1}{3}]+2[\frac{0.3375}{3}-0.15-\frac{0.1}{3}+0.1] \ \ =2[0.2-\frac{0.2}{3}+0.1125-0.15] \ \ =2[\frac{5}{100}-\frac{2}{30}+\frac{1125}{10000}]=2(\frac{1500-2000+3375}{30000}) \ \ =\frac{575}{3000}\Rightarrow k=575 \end{gathered}$

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