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

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

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deginasiba
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, $$\displaystyle{\int_{{{0}}}^{{{4}}}}{\frac{{{p}}}{{\sqrt{{{9}+{p}^{{{2}}}}}}}}{d}{p}={2}$$.
###### Not exactly what youâ€™re looking for?
xandir307dc
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
karton

$$\int_{0}^{4}\frac{p}{\sqrt{9+p^{2}}}dp$$
For the integrand $$\frac{p}{\sqrt{9+p^{2}}}$$, we substitute:
$$u=9+p^{2}, du=2pdp\Rightarrow dp=\frac{1}{2p}du$$
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:
$$\int_{0}^{4}\frac{p}{\sqrt{9+p^{2}}}dp=2$$