Find the antiderivative, F(x), with F′(x) = f(x) and F(0) = 1.
f(x) = −7 x
if F' (x) = -7x
then F(x) = -(7/2)x^2 + c
but F(0) = 1
1 = 0(-7/2) + c
c = 1
F(x) = (-7/2)x^2 + 1
To find the antiderivative F(x) of f(x) = -7x, we can apply the power rule for integration.
The power rule states that if f(x) = x^n, then F(x) = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In this case, we have f(x) = -7x, which can be expressed as f(x) = -7 * x^1.
Using the power rule, we can integrate f(x) as follows:
F(x) = (1/(1+1)) * x^(1+1) + C
= (1/2) * x^2 + C
Given that F(0) = 1, we can substitute x = 0 into the equation to solve for C:
1 = (1/2) * 0^2 + C
C = 1
Therefore, the antiderivative F(x) with F'(x) = f(x) and F(0) = 1 is:
F(x) = (1/2) * x^2 + 1.
To find the antiderivative of f(x) = -7x, we can use the power rule of integration.
The power rule states that if f(x) = x^n, then the antiderivative F(x) = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In this case, f(x) = -7x, which can be written as f(x) = -7 * x^1.
Applying the power rule, we get F(x) = (1/(1+1)) * x^(1+1) + C = (1/2) * x^2 + C.
To find the constant of integration (C), we can use the condition F(0) = 1. Plugging in x = 0 and F(0) = 1 into the equation, we have:
1 = (1/2) * 0^2 + C
1 = 0 + C
C = 1
Therefore, the antiderivative F(x) = (1/2) * x^2 + 1.