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

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

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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

= 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}})

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

d er i v a t i v e f (x) = ln ∣ cot (3 \cdot x) ∣

((3 x^{2} + 3 x + 4)^{^{'}} (3 x^{2} + 6 x)^{^{'}})^{^{'}}

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