What is #F(x) = int e^(x-2) - 2x^2 dx# if #F(0) = 1 #?

1 Answer
Ken C.
Mar 12, 2016

#F(x)=e^(x-2)-2/3x^3+1-e^-2#

Explanation:

Step 1: Break it Up
Always, always look for ways to simplify a problem before you start solving it. Using the properties of integrals, we can break this big integral up into two smaller ones:
#inte^(x-2)-2x^2dx=inte^(x-2)dx-int2x^2dx#

Step 2: Solve the Integrals
The first integral is very easy - if you know your exponent rules. #e^(x-2)# can be rewritten as #e^xe^-2#, using the sum rule for exponents (#e^(a+b)=e^ae^b#). That means our new integral is:
#inte^xe^-2dx#
Because #e^-2# is a constant, we can pull it out:
#inte^(x-2)=e^-2inte^xdx#
#=e^-2(e^x+C)=e^(x-2)+C#

The second integral is also simple - just some reverse power rule:
#int2x^2dx=2/3x^3+C#

Step 3: Constant of Integration
Our solution is #F(x)=e^(x-2)-2/3x^3+C#. We are told #F(0)=1#; that is to say:
#1=e^(0-2)-2/3(0)^3+C#
Solving for #C# gives:
#1=e^-2+C#
#C=1-e^-2#

Therefore, #F(x)=e^(x-2)-2/3x^3+1-e^-2#, or #F(x)=e^(x-2)-2/3x^3+0.865#.

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