Annette Sabin

2022-01-05

Evaluate the following definite integrals.

${\int}_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$

deginasiba

Beginner2022-01-06Added 31 answers

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,${\int}_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=2$ .

The given integral is

Let

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

=5-3

=2

Thus,

xandir307dc

Beginner2022-01-07Added 35 answers

Step 1

${\int}_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp$

Step 2

Let$u=9+{p}^{2}\Rightarrow du=2pdp$ , and

at$p=0\Rightarrow u=9$ ,

at$p=4\Rightarrow u=25$

Step 3

Then

$\int}_{0}^{4}\frac{p}{\sqrt{9+{p}^{2}}}dp=\frac{1}{2}{\int}_{9}^{25}\frac{du}{\sqrt{u}$

$=\frac{1}{4}\sqrt{u}{\mid}_{9}^{25}$

$=\frac{1}{4}(\sqrt{25}-\sqrt{9})$

=2

Result:

2

Step 2

Let

at

at

Step 3

Then

=2

Result:

2

karton

Expert2022-01-11Added 613 answers

For the integrand

The new limits of integration are:

Thus:

=5-3

=2

Result:

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