Paul's Online Notes
Home / Calculus I / Integrals / Computing Indefinite Integrals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5.2 : Computing Indefinite Integrals

23. Determine $$g\left( z \right)$$ given that $$g'\left( z \right) = 3{z^3} + \frac{7}{{2\sqrt z }} - {{\bf{e}}^z}$$ and $$g\left( 1 \right) = 15 - {\bf{e}}$$.

Show All Steps Hide All Steps

Hint : We know that integration is simply asking what function we differentiated to get the integrand and so we should be able to use this idea to arrive at a general formula for the function.
Start Solution

Recall from the notes in this section that we saw,

$g\left( z \right) = \int{{g'\left( z \right)\,\,dz}}$

and so to arrive at a general formula for $$g\left( z \right)$$ all we need to do is integrate the derivative that we’ve been given in the problem statement.

$g\left( z \right) = \int{{3{z^3} + \frac{7}{2}{z^{ - \,\frac{1}{2}}} - {{\bf{e}}^z}\,dz}} = \frac{3}{4}{z^4} + 7{z^{\,\frac{1}{2}}} - {{\bf{e}}^z} + c$

Don’t forget the “+c”!

Hint : To determine the value of the constant of integration, $$c$$, we have the value of the function at $$z = 1$$.
Show Step 2

Because we have the condition that $$g\left( 1 \right) = 15 - {\bf{e}}$$ we can just plug $$z = 1$$ into our answer from the previous step, set the result equal to $$15 – \bf{e}$$ and solve the resulting equation for $$c$$.

Doing this gives,

$15 - {\bf{e}} = g\left( 1 \right) = \frac{{31}}{4} - {\bf{e}} + c\hspace{0.25in} \Rightarrow \hspace{0.5in}c = \frac{{29}}{4}$

The function is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{g\left( z \right) = \frac{3}{4}{z^4} + 7{z^{\,\frac{1}{2}}} - {{\bf{e}}^z} + \frac{{29}}{4}}}$