Question

assignment 10: problem 2
(1 point)
find the most general antiderivative of ( f(u) = \frac{9u^4 - 4sqrt{u}}{u^2} ).
note: any arbitrary constants used must be an upper-case \c\.
( f(u) = square )

Explanation:

Step1: Simplify the function

First, we simplify the function \( f(u)=\frac{9u^{4}-4\sqrt{u}}{u^{2}} \) by dividing each term in the numerator by \( u^{2} \). Recall that \( \sqrt{u}=u^{\frac{1}{2}} \), so we have:
\( f(u)=\frac{9u^{4}}{u^{2}}-\frac{4u^{\frac{1}{2}}}{u^{2}} \)
Using the exponent rule \( \frac{a^{m}}{a^{n}} = a^{m - n} \), we get:
\( f(u)=9u^{4 - 2}-4u^{\frac{1}{2}-2}=9u^{2}-4u^{-\frac{3}{2}} \)

Step2: Find the antiderivative

To find the antiderivative \( F(u) \), we use the power rule for integration, which states that \( \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C \) (where \( n
eq - 1 \)).
For the first term \( 9u^{2} \):
\( \int9u^{2}du = 9\times\frac{u^{2 + 1}}{2+1}=9\times\frac{u^{3}}{3}=3u^{3} \)
For the second term \( - 4u^{-\frac{3}{2}} \):
\( \int-4u^{-\frac{3}{2}}du=-4\times\frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}=-4\times\frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \)
Simplify the coefficient: \( - 4\times\frac{1}{-\frac{1}{2}}=-4\times(-2) = 8 \)
So the integral of \( - 4u^{-\frac{3}{2}} \) is \( 8u^{-\frac{1}{2}} \) (we can also write \( u^{-\frac{1}{2}}=\frac{1}{\sqrt{u}} \), so it's \( \frac{8}{\sqrt{u}} \))
Putting it together with the constant of integration \( C \), we get:
\( F(u)=3u^{3}+8u^{-\frac{1}{2}}+C \) or \( F(u)=3u^{3}+\frac{8}{\sqrt{u}}+C \)

Answer:

\( 3u^{3}+\frac{8}{\sqrt{u}}+C \) (or \( 3u^{3}+8u^{-\frac{1}{2}}+C \))