I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 \(g\left( 1 \right) = 15 - {\bf{e}}\).
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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”!
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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{g\left( z \right) = \frac{3}{4}{z^4} + 7{z^{\,\frac{1}{2}}} - {{\bf{e}}^z} + \frac{{29}}{4}}}\]