What is the integral from 1 to 4 of (u-8)/(sqrt(u)) ?

The integral from 1 to 4 of (u-8)/(sqrt(u)) is -34/3

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integral from 1 to 4 of (u-8)/(sqrt(u))

\int_{1}^{4} \frac{u - 8}{u} d u

- \frac{34}{3}

Decimal Notation

- 11.33333 \dots

Solution steps

\int_{1}^{4} \frac{u - 8}{u} d u

Expand

\frac{u - 8}{u} : u - \frac{8}{u}

= \int_{1}^{4} u - \frac{8}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{4} u d u - \int_{1}^{4} \frac{8}{u} d u

\int_{1}^{4} u d u = \frac{14}{3}

\int_{1}^{4} \frac{8}{u} d u = 16

= \frac{14}{3} - 16

Simplify

\frac{14}{3} - 16 : - \frac{34}{3}

= - \frac{34}{3}

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\int_{0}^{2} (3 x + 1) e^{2 x} d x

What is the integral from 1 to 4 of (u-8)/(sqrt(u)) ?
The integral from 1 to 4 of (u-8)/(sqrt(u)) is -34/3