Question

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

Answer 1

`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`.