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

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Answer 1 · 2016-09-15T17:38:44

$F (x) = \frac{x^{2}}{2} + e^{- x} + 1$ $F(x)=x^2/2+e^(-x)+1$

$F (x) = \int x - e^{- x} d x$ $F(x)=intx-e^(-x)dx$
$F (x) = \frac{x^{2}}{2} + e^{- x} + C$ $F(x)=x^2/2+e^(-x)+C$ where C is a constant number

Let's find C given $F (0) = 2$ $F(0)=2$

$F (0) = 2$ $F(0)=2$

$\frac{{(0)}^{2}}{2} + e^{- 0} + C = 2$ $(0)^2/2+e^(-0)+C=2$

$0 + 1 + C = 2$ $0+1+C=2$
$C = 2 - 1$ $C=2-1$
$C = 1$ $C=1$

Therefore, $F (x) = \frac{x^{2}}{2} + e^{- x} + 1$ $F(x)=x^2/2+e^(-x)+1$

What is F(x) = int x-e^(-x) dxF(x)=∫x−e−xdxF(x) = int x-e^(-x) dx if F(0) = 2 F(0)=2F(0) = 2 ?