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

Annette Sabin 2022-01-05 Answered
Evaluate the following definite integrals.
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}\)

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deginasiba
Answered 2022-01-06 Author has 1976 answers
Step 1
The given integral is \(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}\).
Let \(\displaystyle{x}^{{{2}}}={9}+{p}^{{{2}}}\).
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
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}={\int_{{{3}}}^{{{5}}}}{\frac{{{x}{\left.{d}{x}\right.}}}{{{x}}}}\)
\(\displaystyle={\int_{{{3}}}^{{{5}}}}{\left.{d}{x}\right.}\)
=5-3
=2
Thus, \(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}={2}\).
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xandir307dc
Answered 2022-01-07 Author has 556 answers
Step 1
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}\)
Step 2
Let \(\displaystyle{u}={9}+{p}^{{{2}}}\Rightarrow{d}{u}={2}{p}{d}{p}\), and
at \(\displaystyle{p}={0}\Rightarrow{u}={9}\),
at \(\displaystyle{p}={4}\Rightarrow{u}={25}\)
Step 3
Then
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}={\frac{{{1}}}{{{2}}}}{\int_{{{9}}}^{{{25}}}}{\frac{{{d}{u}}}{{\sqrt{{{u}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}\sqrt{{{u}}}{{\mid}_{{{9}}}^{{{25}}}}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left(\sqrt{{{25}}}-\sqrt{{{9}}}\right)}\)
=2
Result:
2
0
karton
Answered 2022-01-11 Author has 8659 answers

\(\int_{0}^{4}\frac{p}{\sqrt{9+p^{2}}}dp\)
For the integrand \(\frac{p}{\sqrt{9+p^{2}}}\), we substitute:
\(u=9+p^{2}, du=2pdp\Rightarrow dp=\frac{1}{2p}du\)
The new limits of integration are:
\(u(0)=9+0^{2}=9\ \text{and}\ u(4)=9+4^{2}=25\)
Thus:
\(I=\int_{9}^{25}\frac{1}{2\sqrt{u}}du \\=\frac{1}{2}\int_{9}^{25}u^{-\frac{1}{2}}du \\=\frac{1}{2}(2u^{\frac{1}{2}})|_{9}^{25} \\=\frac{1}{2}(2*\sqrt{25}-2*\sqrt{9})\)
=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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