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The Integral given:

$$\int_{z+2}^{z+4}e^{-{|x|}}dx$$

The problem is that if we divide the integral on two parts where $(|x|\ge0$ and $|x|<0)$ we will get this equation: $$\int_{z+2}^0e^x dx+\int_0^{z+4}e^{-x}dx = 2-e^{z+2}-\frac{1}{e^{z+4}}$$ which is a solution for the interval $[-4,2]$ only. How to solve it for $(z<-4$ and $z>-2)$ ?

Никита
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    If $z \gt -2$ then you have $\int\limits_{z+2}^{z+4}e^{-{x}}dx$; while if $z \lt -4$ then you have $\int\limits_{z+2}^{z+4}e^{{x}}dx$ – Henry Mar 31 '20 at 14:05

1 Answers1

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For $z<-4$, both the limits of the integral are negative, hence it becomes $$\int_{z+2}^{z+4}e^xdx = e^{z+2}(e^2-1)$$ For $z>-2$, both the limits are positive, so it becomes $$\int_{z+2}^{z+4}e^{-x}dx = -e^{-x}\bigg|_{z+2}^{z+4}\\ =e^{-z-2}-e^{-z-4}\\ =e^{-z-4}(e^2-1)$$

Martund
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