# The rate of growth of the profit (in millions of dollars) from a new technology is approximated by

P'(x)=xe^(-x^2)
Where x represents time measured in years. The total profit in the thrid year that the new technology is in operation is \$10,000.

A. Find the total profit function

B. What happens to the total amount of profit in the long run?

## To find the total profit function, we need to integrate the given rate of growth equation.

A. Find the total profit function:

The rate of growth equation is P'(x) = xe^(-x^2). To find the total profit function, we need to integrate P'(x) with respect to x.

∫P'(x) dx = ∫xe^(-x^2) dx

To integrate xe^(-x^2), we can use the substitution method:

Let u = -x^2 (differentiating, we get du = -2x dx)
dx = -du / (2x)

Substituting the values, the integral becomes:

∫xe^(u) (-du / (2x)) = -1/2 ∫e^u du

Integrating e^u yields:

-1/2 ∫e^u du = -1/2 e^u + C = -1/2 e^(-x^2) + C

Therefore, the total profit function is:

P(x) = -1/2 e^(-x^2) + C

To find the value of C, we can use the given information about the total profit in the third year.

When x = 3 (third year), P(x) = \$10,000.

10,000 = -1/2 e^(-3^2) + C
10,000 = -1/2 e^(-9) + C

Solving for C:

10,000 + 1/2 e^(-9) = C

Therefore, the total profit function is:

P(x) = -1/2 e^(-x^2) + 10,000 + 1/2 e^(-9)

B. What happens to the total amount of profit in the long run?

In the long run, as x approaches positive or negative infinity, the value of e^(-x^2) becomes extremely small, approaching zero. This means that the total profit function approaches a constant value:

As x approaches infinity, P(x) approaches -1/2 e^(-∞) + 10,000 + 1/2 e^(-9) = 10,000 + 1/2 e^(-9)

As x approaches negative infinity, P(x) approaches -1/2 e^(-∞) + 10,000 + 1/2 e^(-9) = 10,000 + 1/2 e^(-9)

Therefore, in the long run, the total amount of profit approaches the constant value of 10,000 + 1/2 e^(-9) dollars.