What is the integral from sqrt(x) to 1+x of 2xy ?

The integral from sqrt(x) to 1+x of 2xy is x(x^2+x+1)

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integral from sqrt(x) to 1+x of 2xy

\int_{x}^{1 + x} 2 x y d y

x (x^{2} + x + 1)

Solution steps

\int_{x}^{1 + x} 2 x y d y

Take the constant out:

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

= 2 x \cdot \int_{x}^{1 + x} y d y

Apply the Power Rule

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

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

Compute the boundaries:

\frac{( 1 + x ) ^{2}}{2} - \frac{x}{2}

= 2 x (\frac{( 1 + x ) ^{2}}{2} - \frac{x}{2})

Simplify

= x (x^{2} + x + 1)

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What is the integral from sqrt(x) to 1+x of 2xy ?
The integral from sqrt(x) to 1+x of 2xy is x(x^2+x+1)