Question

find the antiderivative for each function when c equals 0. check your answers by differentiation.
(a) ( h(x)=\frac{4}{9} x^{-\frac{1}{9}} ) (b) ( g(x)=\frac{1}{9} x^{-\frac{4}{9}} ) (c) ( k(x)=-\frac{1}{9} x^{-\frac{10}{9}} )
(a) ( h(x)=\frac{1}{2} x^{\frac{8}{9}} )
(b) ( g(x)=\frac{1}{5} x^{\frac{5}{9}} )
(c) ( k(x)=square )

Explanation:

Step1: Recall the power - rule for antiderivatives

The power - rule for antiderivatives is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\), where \(n
eq - 1\). For the function \(k(x)=-\frac{1}{9}x^{-\frac{10}{9}}\), we have \(n=-\frac{10}{9}\).

Step2: Apply the power - rule

Using the power - rule \(\int k(x)dx=\int-\frac{1}{9}x^{-\frac{10}{9}}dx\).
First, factor out the constant: \(-\frac{1}{9}\int x^{-\frac{10}{9}}dx\).
Then, apply the formula \(\int x^{n}dx=\frac{x^{n + 1}}{n + 1}\) with \(n=-\frac{10}{9}\). So \(n+1=-\frac{10}{9}+1=-\frac{10}{9}+\frac{9}{9}=-\frac{1}{9}\).
\(-\frac{1}{9}\times\frac{x^{-\frac{10}{9}+1}}{-\frac{10}{9}+1}=-\frac{1}{9}\times\frac{x^{-\frac{1}{9}}}{-\frac{1}{9}}\).
The \(-\frac{1}{9}\) terms cancel out, and we get \(x^{-\frac{1}{9}}\).

Step3: Check by differentiation

Differentiate \(K(x)=x^{-\frac{1}{9}}\) using the power - rule \((x^{n})^\prime=nx^{n - 1}\).
\(K^\prime(x)=-\frac{1}{9}x^{-\frac{1}{9}-1}=-\frac{1}{9}x^{-\frac{10}{9}}\), which is equal to \(k(x)\).

Answer:

\(K(x)=x^{-\frac{1}{9}}\)