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Find the y-intercept (the value of y when x=0) of the curve y=f(x) that goes though the...
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You need to find the function `f(x)` , such that:
`(dy)/(dx) = 6x*e^(-x^2)`
Separating the variables, yields:
`dy = 6x*e^(-x^2) dx`
Integrating both sides, yields:
int dy = int 6x*e^(-x^2) dx
`y = 6 int x*e^(-x^2) dx`
You should come up with the following substitution, such that:
`-x^2 = t => -2xdx = dt => xdx = -(dt)/2`
Changing the variable, yields:
`y = -6/2 int e^t dt => y = -3 e^t + c`
Substituting back `-x^2` for `t` yields:
`y = -3 e^(-x^2) + c`
Considering x = 1 and y = 0 yields:
`0 = -3*e^-1 + c => 1 = -3/e + c => c = 1.1
`Hence, evaluating the function `f(x) = y` yields `f(x) = -3 e^(-x^2) + 1.1` and the curve f(x) intercepts y axis at `y = 1.1` .
Posted by sciencesolve on March 2, 2013 at 6:43 PM (Answer #1)
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