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

Answer:

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

Explanation:

Step1: Simplify the function

[ \begin{align*} g(t)&=\frac{7 + t + t^{2}}{\sqrt{t}}\ &=7t^{-\frac{1}{2}}+t^{\frac{1}{2}}+t^{\frac{3}{2}} \end{align*} ]

Step2: Integrate term - by - term

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