Question

assignment 10: problem 5
(1 point)
calculate the following antiderivatives:
(a) \\( \int 9t - 2t^4 + 1 \\, dt = \square + c \\).
(b) \\( \int \frac{1}{u^{1/4}} + 3\sqrt{u} \\, du = \square + c \\).
(c) \\( \int \frac{1}{6x^3} \\, dx = \square + c \\).
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Explanation:

Part (a)

Step 1: 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 \(\int9t\ dt\), using the power rule with \(n = 1\), we have \(9\times\frac{t^{1 + 1}}{1+1}=\frac{9t^{2}}{2}\)
For \(\int- 2t^{4}\ dt\), using the power rule with \(n=4\), we have \(-2\times\frac{t^{4 + 1}}{4 + 1}=-\frac{2t^{5}}{5}\)
For \(\int1\ dt\), using the rule for constant integration, we have \(t\)

Step 2: Combine the results

Adding the three results together, we get \(\frac{9t^{2}}{2}-\frac{2t^{5}}{5}+t\)

Part (b)

Step 1: Rewrite the integrand

First, rewrite \(\frac{1}{u^{1/4}}\) as \(u^{-1/4}\) and \(3\sqrt{u}\) as \(3u^{1/2}\)

Step 2: Integrate term by term

For \(\int u^{-1/4}\ du\), using the power rule with \(n=-\frac{1}{4}\), we have \(\frac{u^{-\frac{1}{4}+1}}{-\frac{1}{4}+1}=\frac{u^{\frac{3}{4}}}{\frac{3}{4}}=\frac{4}{3}u^{\frac{3}{4}}\)
For \(\int3u^{1/2}\ du\), using the power rule with \(n = \frac{1}{2}\), we have \(3\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}=3\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}} = 2u^{\frac{3}{2}}\)

Step 3: Combine the results

Adding the two results together, we get \(\frac{4}{3}u^{\frac{3}{4}}+2u^{\frac{3}{2}}\)

Part (c)

Step 1: Rewrite the integrand

Rewrite \(\frac{1}{6x^{3}}\) as \(\frac{1}{6}x^{-3}\)

Step 2: Integrate using the power rule

Using the power rule with \(n=-3\), we have \(\frac{1}{6}\times\frac{x^{-3 + 1}}{-3+1}=\frac{1}{6}\times\frac{x^{-2}}{-2}=-\frac{1}{12x^{2}}\)

Answer:

s:
(a) \(\frac{9t^{2}}{2}-\frac{2t^{5}}{5}+t\)

(b) \(\frac{4}{3}u^{\frac{3}{4}}+2u^{\frac{3}{2}}\)

(c) \(-\frac{1}{12x^{2}}\)