Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)

q(s)=8e^5+3.7

Answer 1

The most general antiderivative of the function q(s) = 8
`e^5` + 3.7 is Q(s) = 8
`e^s` + 3.7s + C.

Step-by-step explanation:

To find the antiderivative of q(s), recall that the antiderivative of
`e^x` is (
`e^x`) itself, and the antiderivative of a constant is that constant times s. Applying this, the antiderivative of 8
`e^5` becomes 8
`e^s` and the antiderivative of 3.7 is 3.7s. Hence, the antiderivative of q(s) is 8
`e^s` + 3.7s.

However, when finding the antiderivative, it's essential to include the constant of integration, denoted as C, as any indefinite integral will have an arbitrary constant attached to it. Therefore, the complete antiderivative is Q(s) = 8
`e^s` + 3.7s + C, where C accounts for any potential initial conditions or factors that might affect the specific function.

By differentiating Q(s) using the rules of differentiation, the derivative of (8
`e^s`) remains (8
`e^s`) and the derivative of 3.7s is 3.7. As the derivative of a constant is zero, the derivative of C would be zero. Thus, the derivative of (Q(s)) returns to the original function (q(s) = 8
`e^5` + 3.7), confirming the correctness of the antiderivative.