find an antiderivative F(x) with F'(x)=f(x) and F(0)=1
f(x)=x^3+x
Int(x^3 + x)dx =
(x^4)/4 + (x^2)/2 + c
With x = 0, Int = 1 therefore c = 1
The complete antiderivative is:
(x^4)/4 + (x^2)/2 + 1
QED
ty!
To find an antiderivative F(x) of the given function f(x), we need to reverse the process of taking derivatives.
Step 1: Find the antiderivative of each term in f(x).
The antiderivative of x^n is (1/(n+1)) * x^(n+1) for any real number n (except -1). Applying this rule to each term in f(x), we get:
∫(x^3 + x) dx = (1/4) * x^4 + (1/2) * x^2
Step 2: Add the constant of integration.
Since an antiderivative is not unique, we need to include a constant of integration (+ C) when finding F(x). In this case, we know that F(0) = 1, so we can determine the value of the constant.
F(x) = (1/4) * x^4 + (1/2) * x^2 + C
To find the value of C, substitute the given value F(0) = 1:
1 = (1/4) * 0^4 + (1/2) * 0^2 + C
1 = C
Therefore, the antiderivative F(x) with F'(x) = f(x) and F(0) = 1 is:
F(x) = (1/4) * x^4 + (1/2) * x^2 + 1