Question

Find the antiderivative, F(x), of the function f(x)=2e^(2x)+x-2 that satisfies F(0)=5.

Answer 1

The antiderivative of the function f(x)=2e^(2x)+x-2 is F(x) = e^(2x) + (1/2)x^2 - 2x + 4. To find the constant of integration, we can substitute x=0 into the formula and solve for the constant. Given F(0) = 5, the constant of integration is 4.

Step-by-step explanation:

To find the antiderivative of the function f(x)=2e^(2x)+x-2, we can use the power rule for integration and the rule for integrating exponential functions.

1. For the term 2e^(2x), we can integrate it using the rule ∫ e^u du = e^u, where u = 2x. So the integral of 2e^(2x) with respect to x is 2e^(2x)/2 = e^(2x).
2. For the term x, the integral is ∫ x dx = (1/2)x^2.
3. For the term -2, the integral is -2x.

Putting it all together, the antiderivative F(x) of f(x) is F(x) = e^(2x) + (1/2)x^2 - 2x.

Given that F(0) = 5, we can substitute x=0 into the formula for F(x) to find the constant of integration. F(0) = e^(2*0) + (1/2)*0^2 - 2*0 = 1 + 0 - 0 = 1. So the constant of integration is 5 - 1 = 4.

Therefore, the antiderivative F(x) of f(x) that satisfies F(0) = 5 is F(x) = e^(2x) + (1/2)x^2 - 2x + 4.