Question

consider the function f(x) = 10/x^2 - 8/x^7, x > 0. let f(x) be the antiderivative of f(x) with f(1) = 0. then f(x) = video example: solving a similar problem

Explanation:

Step1: Recall antiderivative rules

The antiderivative of \(x^n\) is \(\frac{x^{n + 1}}{n+1}+C\) for \(n
eq - 1\). For \(f(x)=\frac{10}{x^{2}}-\frac{8}{x^{7}}\), rewrite it as \(f(x)=10x^{-2}-8x^{-7}\).

Step2: Find the antiderivative of each term

The antiderivative of \(10x^{-2}\) is \(10\times\frac{x^{-2 + 1}}{-2+1}=- \frac{10}{x}\), and the antiderivative of \(-8x^{-7}\) is \(-8\times\frac{x^{-7 + 1}}{-7 + 1}=\frac{4}{3x^{6}}\). So \(F(x)=-\frac{10}{x}+\frac{4}{3x^{6}}+C\).

Step3: Use the initial - condition

Given \(F(1) = 0\), substitute \(x = 1\) into \(F(x)\): \(0=-10+\frac{4}{3}+C\). Solve for \(C\): \(C=10-\frac{4}{3}=\frac{30 - 4}{3}=\frac{26}{3}\).

Step4: Write the final antiderivative

\(F(x)=-\frac{10}{x}+\frac{4}{3x^{6}}+\frac{26}{3}\).

Answer:

\(F(x)=-\frac{10}{x}+\frac{4}{3x^{6}}+\frac{26}{3}\)