What is $f (x) = \int e^{x} d x$ $f(x) = int e^x dx$ if $f (0) = - 2$ $f(0)=-2$ ?

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What is $f (x) = \int e^{x} d x$ $f(x) = int e^x dx$ if $f (0) = - 2$ $f(0)=-2$ ?

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Answer 1 · 2018-03-28T06:39:04

$e^{x} - 3$ $e^x -3$

Explanation:

We have, $f (x) = \int e^{x} d x$ $f(x)=inte^xdx$ which comes out to be $e^{x} + C$ $e^x +C$ where $C$ $C$ is the constant of integration.
We also have $f (0) = - 2$ $f(0)=-2$ .Plugging in this value in the expression we derived earlier, we get $e^{0} + C$ $e^0 +C$ =-2 or $C$ $C$ =-2-1 which is -3. Thus,the expression comes out to be $e^{x} - 3$ $e^x -3$

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What is $f (x) = \int e^{x} d x$ $f(x) = int e^x dx$ if $f (0) = - 2$ $f(0)=-2$ ?

1 Answer

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What is $f (x) = \int e^{x} d x$ $f(x) = int e^x dx$ if $f (0) = - 2$ $f(0)=-2$ ?