What is the integral from y to 1 of 8xy ?

The integral from y to 1 of 8xy is 4y(-y^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from y to 1 of 8xy

\int_{y}^{1} 8 x y d x

4 y (- y^{2} + 1)

Solution steps

\int_{y}^{1} 8 x y d x

Take the constant out:

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

= 8 y \cdot \int_{y}^{1} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{y}^{1}

= 8 y [\frac{x ^{2}}{2}]_{y}^{1}

Compute the boundaries:

\frac{1}{2} - \frac{y ^{2}}{2}

= 8 y (\frac{1}{2} - \frac{y ^{2}}{2})

Simplify

= 4 y (- y^{2} + 1)

\frac{d}{d x} (\frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}})

\frac{d}{d x} (\frac{1}{5 x ^{8}})

\int \frac{9}{3 sin ( x ) - 4 cos ( x )} d x

n \to \infty lim (n^{- 1})

y^{^{''}} - y^{^{'}} - 3 y = 0

What is the integral from y to 1 of 8xy ?
The integral from y to 1 of 8xy is 4y(-y^2+1)