How do you find the indefinite integral of #int (7u^(3/2)+2u^(1/2))du#?

Question

Cameron Alexander · Accepted Answer

#14/5u^(5/2)+4/3u^(3/2)+c#

We can integrate each term using the #color(blue)"power rule"#

That is #color(red)(bar(ul(|color(white)(a/a)color(black)(int(ax^n)dx=a/(n+1)x^(n+1))color(white)(a/a)|)))#

#rArrint(7u^(3/2)+2u^(1/2))du#

#=7/(5/2)u^(3/2+1)+2/(3/2)u^(1/2+1)+c#

#=14/5u^(5/2)+4/3u^(3/2)+c#

where c is the constant of integration.

Elijah Abram · Answer

undefined To find the indefinite integral of $\int (7u^{3/2} + 2u^{1/2}) \, du$, use the power rule for integration. The power rule states that for $ \int u^n \, du$, where $n \neq -1$, the result is $\frac{u^{n+1}}{n+1} + C$, where $C$ is the constant of integration.

Applying the power rule:

$$
\begin{align*}
\int (7u^{3/2} + 2u^{1/2}) \, du &= \frac{7}{\frac{3}{2}+1}u^{\frac{3}{2}+1} + \frac{2}{\frac{1}{2}+1}u^{\frac{1}{2}+1} + C \
&= \frac{7}{\frac{5}{2}}u^{\frac{5}{2}} + \frac{2}{\frac{3}{2}}u^{\frac{3}{2}} + C \
&= \frac{14}{5}u^{\frac{5}{2}} + \frac{4}{3}u^{\frac{3}{2}} + C
\end{align*}
$$

So, the indefinite integral of $\int (7u^{3/2} + 2u^{1/2}) \, du$ is $\frac{14}{5}u^{\frac{5}{2}} + \frac{4}{3}u^{\frac{3}{2}} + C$, where $C$ is the constant of integration.