Calculate the area between y=(e^x+1)/e^x and y=1 and to the right of x=0.
It is clear that:lim(x→∞)(f(x)-g(x))=lim((e^x+1)/e^x -1)=lim(x→∞)〖1/e^x 〗=0Therefore:A=∫_0^∞((e^x+1)/e^x -1) dx=lim(t→∞)∫_0^t1/e^x dx=lim(t→∞)∫_0^te^(-x) dx=lim(t→∞)〖[-e^(-x) ]_0^t 〗=lim(t→∞)[-e^(-t)+1]=1 square unit
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It is clear that:
need an explanation for this answer? contact us directly to get an explanation for this answerlim(x→∞)(f(x)-g(x))=lim((e^x+1)/e^x -1)=lim(x→∞)〖1/e^x 〗=0
Therefore:
A=∫_0^∞((e^x+1)/e^x -1) dx=lim(t→∞)∫_0^t1/e^x dx=lim(t→∞)∫_0^te^(-x) dx
=lim(t→∞)〖[-e^(-x) ]_0^t 〗=lim(t→∞)[-e^(-t)+1]=1 square unit