What is the integral from 0 to 8 of x/(sqrt(36+x^2)) ?

The integral from 0 to 8 of x/(sqrt(36+x^2)) is 4

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integral from 0 to 8 of x/(sqrt(36+x^2))

\int_{0}^{8} \frac{x}{36 + x ^{2}} d x

4

Solution steps

\int_{0}^{8} \frac{x}{36 + x ^{2}} d x

Apply u-substitution

: \int_{36}^{100} \frac{1}{2 u} d u

= \int_{36}^{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_{36}^{100} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{36}^{100}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{36}^{100}

Apply radical rule:

= \frac{1}{2} [2 u]_{36}^{100}

Compute the boundaries:

8

= \frac{1}{2} \cdot 8

Simplify

= 4

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What is the integral from 0 to 8 of x/(sqrt(36+x^2)) ?
The integral from 0 to 8 of x/(sqrt(36+x^2)) is 4