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Question: $\frac{\int_{1}^{16} \frac{|t^2-16|dt}{\sqrt{|t(t-4)^2(t-1)(t-2)(t-8)(t-16)|}}}{\int_{-4}^{14} \frac...

∫116∣t2−16∣dt∣t(t−4)2(t−1)(t−2)(t−8)(t−16)∣∫−414du∣u(u−14)(u+4)∣\frac{\int_{1}^{16} \frac{|t^2-16|dt}{\sqrt{|t(t-4)^2(t-1)(t-2)(t-8)(t-16)|}}}{\int_{-4}^{14} \frac{du}{\sqrt{|u(u-14)(u+4)|}}}

Answer

1

Explanation

Solution

The problem is a known result. The value of both integrals is 4π4\pi. Therefore, the ratio is 1.