7) Consider the function f(x)=(10/x^2)−(2/x^6). Let F(x) be the antiderivative of f(x) with F(1)=0.
Then F(x)= ?
F=INt 10/x^2 dx -INT 2/x^6 dx
= -10/x +2/5 * x^-5+c
F(1)=0=-10+2/5 _+ C solve for c.
F' = 10 x^-2 dx - 2 x^-6 dx
F = -10x^-1 + (2/5) x^-5 + c
when x = 1 , F = 0
0 = -10 + 2/5 + c
c = 48/5
F = -10 /x + 2/(5x^5) +48/5
To find the antiderivative F(x) for the function f(x)=(10/x^2)−(2/x^6), we will apply the power rule for antiderivatives.
The power rule states that for a term of the form x^n, the antiderivative is (1/(n+1)) * x^(n+1), except when n = -1. In that case, the antiderivative is natural logarithm of the absolute value of x.
Applying the power rule to each term in f(x), we have:
∫(10/x^2)dx = 10∫x^(-2)dx = 10 * (1/(-1+1)) * x^(-1+1)
Similarly,
∫(2/x^6)dx = 2∫x^(-6)dx = 2 * (1/(-6+1)) * x^(-6+1)
Now, we can simplify and combine these terms:
F(x) = 10 * (1/(-1+1)) * x^(-1+1) + 2 * (1/(-6+1)) * x^(-6+1)
= 10 * (1/0) * x^0 + 2 * (1/-5) * x^-5
= 10 * 0 + (-2/5) * x^-5
= (-2/5) * x^-5
Lastly, we know that F(1) = 0, so we can substitute x = 1 in the equation:
F(1) = (-2/5) * (1)^-5
= (-2/5) * 1
= -2/5
Therefore, the antiderivative of f(x) with F(1) = 0 is F(x) = (-2/5) * x^-5