Asked by Pratibha
Find the first quadrant area under y = e -2x
Answers
Answered by
Reiny
I assume you mean y = e^(-2x)
area = ∫e^(-2x) dx from 0 to ∞
= [ (-1/2)e^(-2x) ] from 0 to ∞
let's look at (-1/2) e(^(-2x)
or -1/(2e^(2x) as x ---> ∞
so as the denominator becomes larger and larger, the quotient becomes smaller and smaller
that is ,
limit -1/(2e^(2x) as x---> ∞ = 0
so area = 0 - (-1/e^0
= 1
let's test our answer, let the area be from 0 to 100
area = -1/e^200 - ( -1/e^0)
= 1.38x10^-87 + 1
= pretty close to 1
area = ∫e^(-2x) dx from 0 to ∞
= [ (-1/2)e^(-2x) ] from 0 to ∞
let's look at (-1/2) e(^(-2x)
or -1/(2e^(2x) as x ---> ∞
so as the denominator becomes larger and larger, the quotient becomes smaller and smaller
that is ,
limit -1/(2e^(2x) as x---> ∞ = 0
so area = 0 - (-1/e^0
= 1
let's test our answer, let the area be from 0 to 100
area = -1/e^200 - ( -1/e^0)
= 1.38x10^-87 + 1
= pretty close to 1
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