Annette Sabin

2022-01-05

Evaluate the following definite integrals.
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$

deginasiba

Step 1
The given integral is ${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$.
Let ${x}^{2}=9+{p}^{2}$.
This gives
2xdx=2pdp.
xdx=pdp
when p=0, x=3.
when p=4, x=5.
Substitute these values in the above integral as follows.
Step 2
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp={\int }_{3}^{5}\frac{xdx}{x}$
$={\int }_{3}^{5}dx$
=5-3
=2
Thus, ${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=2$.

xandir307dc

Step 1
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$
Step 2
Let $u=9+{p}^{2}⇒du=2pdp$, and
at $p=0⇒u=9$,
at $p=4⇒u=25$
Step 3
Then
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=\frac{1}{2}{\int }_{9}^{25}\frac{du}{\sqrt{u}}$
$=\frac{1}{4}\sqrt{u}{\mid }_{9}^{25}$
$=\frac{1}{4}\left(\sqrt{25}-\sqrt{9}\right)$
=2
Result:
2

karton

${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$
For the integrand $\frac{p}{\sqrt{9+{p}^{2}}}$, we substitute:
$u=9+{p}^{2},du=2pdp⇒dp=\frac{1}{2p}du$
The new limits of integration are:

Thus:
$I={\int }_{9}^{25}\frac{1}{2\sqrt{u}}du\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\int }_{9}^{25}{u}^{-\frac{1}{2}}du\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(2{u}^{\frac{1}{2}}\right){|}_{9}^{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(2\ast \sqrt{25}-2\ast \sqrt{9}\right)$
=5-3
=2
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=2$
Result:
${\int }_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=2$

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