Sure, let's go through the steps together to find the antiderivative [tex]\( F(x) \)[/tex] of the function [tex]\( f(x) = \frac{6}{x^3} - \frac{6}{x^7} \)[/tex], given the initial condition [tex]\( F(1) = 0 \)[/tex].
1. Express the function in a more integrable form:
First, rewrite [tex]\( f(x) \)[/tex] with negative exponents:
[tex]\[
f(x) = 6x^{-3} - 6x^{-7}
\][/tex]
2. Find the antiderivative:
To find the antiderivative [tex]\( F(x) \)[/tex], we integrate [tex]\( f(x) \)[/tex] term by term:
[tex]\[
F(x) = \int (6x^{-3} - 6x^{-7}) \, dx
\][/tex]
Using the power rule for integration [tex]\( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)[/tex], we get:
[tex]\[
\int 6x^{-3} \, dx = 6 \int x^{-3} \, dx = 6 \left( \frac{x^{-3+1}}{-3+1} \right) = 6 \left( \frac{x^{-2}}{-2} \right) = -3x^{-2}
\][/tex]
[tex]\[
\int 6x^{-7} \, dx = 6 \int x^{-7} \, dx = 6 \left( \frac{x^{-7+1}}{-7+1} \right) = 6 \left( \frac{x^{-6}}{-6} \right) = -x^{-6}
\][/tex]
So, the general antiderivative is:
[tex]\[
F(x) = -3x^{-2} - x^{-6} + C
\][/tex]
3. Apply the initial condition:
We are given that [tex]\( F(1) = 0 \)[/tex]. Substituting [tex]\( x = 1 \)[/tex] and [tex]\( F(1) = 0 \)[/tex] into the antiderivative, we get:
[tex]\[
0 = -3(1)^{-2} - (1)^{-6} + C
\][/tex]
Simplifying, we find:
[tex]\[
0 = -3(1) - 1 + C \implies 0 = -3 - 1 + C \implies 0 = -4 + C \implies C = 4
\][/tex]
4. Write the final solution:
Substituting the value of [tex]\( C \)[/tex] back into the antiderivative:
[tex]\[
F(x) = -3x^{-2} - x^{-6} + 4
\][/tex]
We can rewrite [tex]\( x^{-2} \)[/tex] and [tex]\( x^{-6} \)[/tex] back using positive exponents:
[tex]\[
F(x) = -\frac{3}{x^2} - \frac{1}{x^6} + 4
\][/tex]
So, the antiderivative of the function [tex]\( f(x) = \frac{6}{x^3} - \frac{6}{x^7} \)[/tex] with the initial condition [tex]\( F(1) = 0 \)[/tex] is:
[tex]\[
F(x) = -\frac{3}{x^2} - \frac{1}{x^6} + 4 \quad \text{or equivalently,} \quad F(x) = 4 - \frac{3}{x^2} - \frac{1}{x^6}
\][/tex]
The function [tex]\( F(x) = 2 + \frac{1 - 3x^4}{x^6} \)[/tex] satisfies all these conditions.
