Chapter 4, Problem 52RE

To determine

Tofind:the most general anti derivative of the function.

Answer to Problem 52RE

Themost general anti derivative of the function is ${\displaystyle \int \text{g}\left(\text{t}\right){\text{dt=2t}}^{\frac{\text{1}}{\text{2}}}}\text{+}\frac{2}{3}{\text{t}}^{\frac{\text{3}}{\text{2}}}\text{+C}\text{.}$ .

Explanation of Solution

Given:

  $\text{g}\left(\text{t}\right)\text{=}\frac{\left(\text{1+t}\right)}{\sqrt{\text{t}}}$ .

Concept used:

Antiderivative=integration

Integration formula:

  ${\displaystyle \int {\text{x}}^{\text{n}}}\text{dx=}\frac{{\text{x}}^{\text{n+1}}}{\text{n+1}}$ .

Calculation:

  $\text{g}\left(\text{t}\right)\text{=}\frac{\left(\text{1+t}\right)}{\sqrt{\text{t}}}$ .

  ${\displaystyle \int \text{g}\left(\text{t}\right)\text{dt=}{\displaystyle \int \frac{\left(\text{1+t}\right)}{\sqrt{\text{t}}}}}\text{dt}$

  ${\displaystyle \int \text{g}\left(\text{t}\right)\text{dt=}{\displaystyle \int \frac{\text{1}}{\sqrt{\text{t}}}\text{dt}}}\text{+}{\displaystyle \int \sqrt{\text{t}}\text{dt}}$

  ${\displaystyle \int \text{g}\left(\text{t}\right)\text{dt=}{\displaystyle \int {\text{t}}^{\frac{\text{-1}}{\text{2}}}\text{dt}}}\text{+}{\displaystyle \int {\text{t}}^{\frac{\text{1}}{\text{2}}}}\text{dt}$ .

  ${\displaystyle \int \text{g}\left(\text{t}\right)\text{dt=}\frac{{\text{t}}^{\frac{\text{1}}{\text{2}}}}{\frac{\text{1}}{\text{2}}}}\text{+}\frac{{\text{t}}^{\frac{\text{3}}{\text{2}}}}{\frac{\text{3}}{\text{2}}}\text{+C}$ .

  ${\displaystyle \int \text{g}\left(\text{t}\right){\text{dt=2t}}^{\frac{\text{1}}{\text{2}}}}\text{+}\frac{2}{3}{\text{t}}^{\frac{\text{3}}{\text{2}}}\text{+C}\text{.}$

Hence the most general anti derivative of the function is ${\displaystyle \int \text{g}\left(\text{t}\right){\text{dt=2t}}^{\frac{\text{1}}{\text{2}}}}\text{+}\frac{2}{3}{\text{t}}^{\frac{\text{3}}{\text{2}}}\text{+C}\text{.}$ .