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

Question

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

Cameron Alexander · Accepted Answer

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

Samantha Adkison · Answer

undefined To find $ F(x) = \int e^{x-2} - 2x^2 \, dx $ given $ F(0) = 1 $, first integrate $ e^{x-2} - 2x^2 $ with respect to $ x $ to get $ F(x) $, then solve for the constant of integration using the given initial condition $ F(0) = 1 $.

$ F(x) = \int e^{x-2} - 2x^2 \, dx $

$ F(x) = \int e^{x-2} \, dx - \int 2x^2 \, dx $

$ F(x) = e^{x-2} - \frac{2}{3}x^3 + C $

Now, use the given initial condition $ F(0) = 1 $ to solve for $ C $:

$ F(0) = e^{-2} - \frac{2}{3}(0)^3 + C = 1 $

$ e^{-2} + C = 1 $

$ C = 1 - e^{-2} $

Thus, $ F(x) = e^{x-2} - \frac{2}{3}x^3 + (1 - e^{-2}) $.