What is the integral from 9 to 10 of-f(u) ?

The integral from 9 to 10 of-f(u) is -(19f)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 9 to 10 of-f(u)

\int_{9}^{10} - f (u) d u

- \frac{19 f}{2}

Solution steps

\int_{9}^{10} - f u d u

Take the constant out:

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

= - f \cdot \int_{9}^{10} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{9}^{10}

= - f [\frac{u ^{2}}{2}]_{9}^{10}

Compute the boundaries:

\frac{19}{2}

= - f \frac{19}{2}

Simplify

= - \frac{19 f}{2}

\int_{1}^{0} \frac{1}{x ^{2}} d x

\int_{0}^{- \infty} e^{x} d x

\int_{1}^{3} (4 x - 1) d x

\int_{0}^{\infty} y^{5} e^{- y} d y

\int_{0}^{π^{2}} cos (x) d x

What is the integral from 9 to 10 of-f(u) ?
The integral from 9 to 10 of-f(u) is -(19f)/2