What is #F(x) = int 3x^2-e^(2-x) dx# if #F(0) = 1 #?

Question

What is #F(x) = int 3x^2-e^(2-x)    dx# if #F(0) = 1   #?

Cameron Alexander · Accepted Answer

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

#F(x)=x^3+e^2(e^-x -1)+1.# #F(x)=int3x^2-e^(2-x)dx.# #=3intx^2dx-inte^2*e^-xdx,# #=3(x^3/3)-e^2inte^-xdx# #F(x)=x^3-e^2*(e^-x/-1)+C=x^3+e^(2-x)+C.# But, #F(0)=1 rArr 0^3+e^(2-0)+C=1.# # rArr C=1-e^2.# Therefore, #F(x)=x^3+e^(2-x)+1-e^2, or, # #F(x)=x^3+e^2(e^-x -1)+1.#

Evelyn Agard · Answer

undefined To find $ F(x) $, integrate the given function $ 3x^2 - e^{2-x} $ with respect to $ x $ and then use the initial condition $ F(0) = 1 $.

$$
F(x) = \int (3x^2 - e^{2-x}) \, dx
$$

Integrating $ 3x^2 $ and $ e^{2-x} $ separately:

$$
\int 3x^2 \, dx = x^3 + C_1
$$

$$
\int e^{2-x} \, dx = -e^{2-x} + C_2
$$

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

Using the initial condition $ F(0) = 1 $:

$$
1 = 0^3 - e^{2-0} + C
$$

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

$$
C = 1 + e^2
$$

Therefore, the solution is:

$$
F(x) = x^3 - e^{2-x} + (1 + e^2)
$$

Christopher Allers · Answer

undefined To find the function $ F(x) $, we need to evaluate the integral of $ 3x^2 - e^{2-x} $ with respect to $ x $. Then, we can determine the constant of integration using the given initial condition $ F(0) = 1 $.

After integrating, the function $ F(x) $ is:

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

Given that $ F(0) = 1 $, we substitute $ x = 0 $ into the function:

$$ 1 = 0^3 + e^{2-0} + C $$
$$ 1 = 1 + e^2 + C $$

Solving for $ C $, we find:

$$ C = 1 - e^2 $$

So, the function $ F(x) $ is:

$$ F(x) = x^3 + e^{2-x} + (1 - e^2) $$