Home > Calculus > Evaluating the Constant of Integration

What is #F(x) = int x-e^(-x) dx# if #F(0) = 2 #?

Answer 1

Emily Ammerman

#F(x)=x^2/2+e^(-x)+1#

#F(x)=intx-e^(-x)dx# #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#

#(0)^2/2+e^(-0)+C=2#

#0+1+C=2# #C=2-1# #C=1#

Therefore, #F(x)=x^2/2+e^(-x)+1#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To find ( F(x) ) given ( F(0) = 2 ), integrate ( x - e^{-x} ) with respect to ( x ):

[ F(x) = \int (x - e^{-x}) , dx ]

[ F(x) = \frac{x^2}{2} + e^{-x} + C ]

Given that ( F(0) = 2 ), substitute ( x = 0 ) into ( F(x) ) and solve for ( C ):

[ 2 = \frac{0^2}{2} + e^{0} + C ] [ 2 = 0 + 1 + C ] [ C = 1 ]

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

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.