What is the integral from 0 to 6 of p/(sqrt(64+p^2)) ?

The integral from 0 to 6 of p/(sqrt(64+p^2)) is 2

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integral from 0 to 6 of p/(sqrt(64+p^2))

\int_{0}^{6} \frac{p}{64 + p ^{2}} d p

2

Solution steps

\int_{0}^{6} \frac{p}{64 + p ^{2}} d p

Apply u-substitution

: \int_{64}^{100} \frac{1}{2 u} d u

= \int_{64}^{100} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{64}^{100} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{64}^{100}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{64}^{100}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{64}^{100}

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

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What is the integral from 0 to 6 of p/(sqrt(64+p^2)) ?
The integral from 0 to 6 of p/(sqrt(64+p^2)) is 2