What is the integral from 1 to 4 of (u-2)/(sqrt(u)) ?

The integral from 1 to 4 of (u-2)/(sqrt(u)) is 2/3

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integral from 1 to 4 of (u-2)/(sqrt(u))

\int_{1}^{4} \frac{u - 2}{u} d u

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{1}^{4} \frac{u - 2}{u} d u

Expand

\frac{u - 2}{u} : u - \frac{2}{u}

= \int_{1}^{4} u - \frac{2}{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{2}{u} d u

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

\int_{1}^{4} \frac{2}{u} d u = 4

= \frac{14}{3} - 4

Simplify

\frac{14}{3} - 4 : \frac{2}{3}

= \frac{2}{3}

\frac{\partial}{\partial y} ((x^{2} y - y^{3})^{5})

\frac{d y}{d x} = \frac{cos ( 9 x )}{e ^{9 y}}

\frac{\partial}{\partial x} (z - x^{2} + y^{2})

\int \frac{4 x ^{2} - 2}{x ^{3} - x} d x

\int x^{\frac{1}{3}} + 3 d x

What is the integral from 1 to 4 of (u-2)/(sqrt(u)) ?
The integral from 1 to 4 of (u-2)/(sqrt(u)) is 2/3