Question: "Evaluate the following definite integrals. ∫04p9+p2dp"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Annette Sabin

Answered question

2022-01-05

Evaluate the following definite integrals.

\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p

Answer & Explanation

deginasiba

Beginner2022-01-06Added 31 answers

Step 1
The given integral is

\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p

.
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}}} d p = \int_{3}^{5} \frac{x d x}{x}

= \int_{3}^{5} d x

=5-3
=2
Thus,

\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p = 2

xandir307dc

Beginner2022-01-07Added 35 answers

Step 1

\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p

Step 2
Let

u = 9 + p^{2} \Rightarrow d u = 2 p d p

, 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}}} d p = \frac{1}{2} \int_{9}^{25} \frac{d u}{\sqrt{u}}

= \frac{1}{4} \sqrt{u} ∣_{9}^{25}

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

=2
Result:
2

karton

Expert2022-01-11Added 613 answers

$\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p$
For the integrand $\frac{p}{\sqrt{9 + p^{2}}}$ , we substitute:
$u = 9 + p^{2}, d u = 2 p d p \Rightarrow d p = \frac{1}{2 p} d u$
The new limits of integration are:
$u (0) = 9 + 0^{2} = 9 and u (4) = 9 + 4^{2} = 25$
Thus:
$I = \int_{9}^{25} \frac{1}{2 \sqrt{u}} d u = \frac{1}{2} \int_{9}^{25} u^{- \frac{1}{2}} d u = \frac{1}{2} (2 u^{\frac{1}{2}}) |_{9}^{25} = \frac{1}{2} (2 * \sqrt{25} - 2 * \sqrt{9})$
=5-3
=2
$\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p = 2$
Result:
$\int_{0}^{4} \frac{p}{\sqrt{9 + p^{2}}} d p = 2$

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