QuestionReportMathematics Question on integralIf the value of ∫−0.150.15∣100x2−1∣dx=k3000∫_{-0.15}^{0.15} |100x^{2}-1|dx = \frac{k}{3000}∫−0.150.15∣100x2−1∣dx=3000k, then the value of k is?AnswerThe correct answer is 575.
