consider the function f(x)=4/x^2 -2/x^6. let f(x) be the antiderivative of f(x) with f(1)=0. Then f(5) equals
"let f(x) be the antiderivative of f(x) "
this makes no sense unless f(x) = e^x
so is f(x) the antiderivative or .... ?
suppose f ' (x) = f(x)=4/x^2 -2/x^6
= 4x^-2 - 2x^-6
then
f(x) = (-4x^-1) + (2/5)x^-5 + c
if f(1) = 0
-4 + 2/5 + c = 0, c = 18/5
f(x) = -4/x + (2/5) /x^5 + 18/5
f(5) = -4/5 + 2/15625 + 18/5 = 2.8
To find the value of f(5), we need to evaluate the antiderivative of f(x) at x = 5. Given that f(x) = 4/x^2 - 2/x^6, we can find its antiderivative by integrating term by term.
The integral of 4/x^2 is 4 * integral(1/x^2) = 4 * (-1/x) = -4/x.
The integral of 2/x^6 is 2 * integral(1/x^6) = 2 * (1/5 * x^-5) = 2/5x^5.
Therefore, the antiderivative of f(x) is F(x) = -4/x + 2/5x^5 + C, where C is the constant of integration.
We are given that f(1) = 0. Substituting x = 1 into the antiderivative function, we get:
0 = -4/1 + 2/5(1^5) + C
0 = -4 + 2/5 + C
4 = 2/5 + C
C = 4 - 2/5
C = 18/5
So, the specific antiderivative function f(x) with f(1) = 0 is F(x) = -4/x + 2/5x^5 + 18/5.
To find f(5), we substitute x = 5 into the antiderivative function:
f(5) = -4/5 + 2/5(5^5) + 18/5
f(5) = -4/5 + 2/5(3125) + 18/5
f(5) = -4/5 + 625/5 + 18/5
f(5) = (625 - 4 + 18)/5
f(5) = 639/5
Therefore, f(5) equals 639/5.