Question

homework9: problem 1
(1 point)
calculate the following antiderivatives:
(a) $\int 5 t-5 t^{9}-10 d t=\square+c$
(b) $\int \frac{1}{u^{3 / 4}}+2 \sqrt{u} d u=\square+c$
(c) $\int \frac{1}{2 x^{3}} d x=\square+c$
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Explanation:

Part (a)

Step1: Integrate term by term

We use the power rule for integration, \(\int t^n dt=\frac{t^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) and \(\int kdt=kt + C\) (where \(k\) is a constant).
For \(\int(5t-5t^{9}-10)dt\), we integrate each term:

• \(\int5t\ dt = 5\times\frac{t^{1 + 1}}{1+1}=\frac{5t^{2}}{2}\)
• \(\int - 5t^{9}\ dt=-5\times\frac{t^{9 + 1}}{9+1}=-\frac{5t^{10}}{10}=-\frac{t^{10}}{2}\)
• \(\int-10\ dt=-10t\)

Step2: Combine the results

Adding the three integrated terms together, we get \(\frac{5t^{2}}{2}-\frac{t^{10}}{2}-10t\)

Step1: Simplify the integrand

First, rewrite \(\frac{1}{u^{3/4}}\) as \(u^{-3/4}\) and \(2\sqrt{u}\) as \(2u^{1/2}\). So the integrand becomes \(u^{-3/4}+2u^{1/2}\)

Step2: Integrate term by term

Using the power rule \(\int u^n du=\frac{u^{n + 1}}{n + 1}+C\) (\(n
eq - 1\)):

• \(\int u^{-3/4}\ du=\frac{u^{-3/4+1}}{-3/4 + 1}=\frac{u^{1/4}}{1/4}=4u^{1/4}\)
• \(\int2u^{1/2}\ du=2\times\frac{u^{1/2+1}}{1/2 + 1}=2\times\frac{u^{3/2}}{3/2}=\frac{4}{3}u^{3/2}\)

Step3: Combine the results

Adding the two integrated terms together, we get \(4u^{1/4}+\frac{4}{3}u^{3/2}\)

Step1: Simplify the integrand

Rewrite \(\frac{1}{2x^{3}}\) as \(\frac{1}{2}x^{-3}\)

Step2: Integrate using power rule

Using the power rule \(\int x^n dx=\frac{x^{n+1}}{n + 1}+C\) (\(n
eq - 1\)):
\(\int\frac{1}{2}x^{-3}\ dx=\frac{1}{2}\times\frac{x^{-3+1}}{-3 + 1}=\frac{1}{2}\times\frac{x^{-2}}{-2}=-\frac{1}{4x^{2}}\)

Answer:

\(\frac{5t^{2}}{2}-\frac{t^{10}}{2}-10t\)

Part (b)