Question

find the most general antiderivative of the function. (check your answer by diffe
g(t)=\frac{7+t+t^{2}}{sqrt{t}}
g(t)=

Explanation:

Step1: Simplify the function

$$
LATEXBLOCK0
$$

Step2: Integrate term - by - term

Recall the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\))

• For the first term \(\int7t^{-\frac{1}{2}}dt\):

Using the power rule with \(n=-\frac{1}{2}\), we have \(7\times\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=7\times\frac{t^{\frac{1}{2}}}{\frac{1}{2}} = 14t^{\frac{1}{2}}\)

• For the second term \(\int t^{\frac{1}{2}}dt\):

Using the power rule with \(n = \frac{1}{2}\), we get \(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3}t^{\frac{3}{2}}\)

• For the third term \(\int t^{\frac{3}{2}}dt\):

Using the power rule with \(n=\frac{3}{2}\), we obtain \(\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}=\frac{2}{5}t^{\frac{5}{2}}\)

Step3: Combine the results

The antiderivative \(G(t)=\int g(t)dt=\int(7t^{-\frac{1}{2}}+t^{\frac{1}{2}}+t^{\frac{3}{2}})dt = 14t^{\frac{1}{2}}+\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+C\)

Step4: Check the answer by differentiation

Differentiate \(G(t)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\)

• \((14t^{\frac{1}{2}})^\prime=14\times\frac{1}{2}t^{\frac{1}{2}-1}=7t^{-\frac{1}{2}}\)
• \((\frac{2}{3}t^{\frac{3}{2}})^\prime=\frac{2}{3}\times\frac{3}{2}t^{\frac{3}{2}-1}=t^{\frac{1}{2}}\)
• \((\frac{2}{5}t^{\frac{5}{2}})^\prime=\frac{2}{5}\times\frac{5}{2}t^{\frac{5}{2}-1}=t^{\frac{3}{2}}\)

And \((C)^\prime = 0\). So \(G^\prime(t)=7t^{-\frac{1}{2}}+t^{\frac{1}{2}}+t^{\frac{3}{2}}=g(t)\)

Answer:

\(14t^{\frac{1}{2}}+\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+C\)