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This is small part of a much larger question. My trouble is calculating with the modulus function on. Since I don’t really know which function is greater than the other at, I can’t tell how it will open up. Assuming that there is no graphing calculator (or a calculator of any sort) at my disposal, how should I calculate it?

Gert
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Aditya
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    Of the two functions in the absolute values, one is increasing and the other decreasing. It helps to find when the two functions are equal, then you know the sign of their difference. – player3236 Nov 09 '20 at 14:30
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    Show that $\frac 32 e^{1-x} -\frac 12 e^{x-1} \ge 0$ on the interval. As @player3236 said, it is easy to solve $\frac 32 e^{1-x} -\frac 12 e^{x-1} = 0$ – GEdgar Nov 09 '20 at 14:33
  • To me a substituion of u +1 = x also seems useful. – Imago Nov 09 '20 at 15:22

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hint

$$F(x)=\frac 32e^{1-x}-\frac 12e^{1+x}=$$

$$\frac 12e^{1-x}\Bigl(3-e^{2x}\Bigr)$$

$$e^{2x}<3\iff x<\ln(\sqrt{3})$$

So

$$\int_0^{1+\ln(\sqrt{3})}|F|=\int_0^{\ln(\sqrt{3})}F+\int_{\ln(\sqrt{3})}^{1+\ln(\sqrt{3})}(-F)$$

hamam_Abdallah
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