What is the integral from x-1 to sqrt(2x+6 of)xy ?

The integral from x-1 to sqrt(2x+6 of)xy is x(3+x-((-1+x)^2)/2)

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integral from x-1 to sqrt(2x+6 of)xy

\int_{x - 1}^{2 x + 6} x y d y

x (3 + x - \frac{( - 1 + x ) ^{2}}{2})

Solution steps

\int_{x - 1}^{2 x + 6} x y d y

Take the constant out:

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

= x \cdot \int_{- 1 + x}^{6 + 2 x} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{- 1 + x}^{6 + 2 x}

= x [\frac{y ^{2}}{2}]_{- 1 + x}^{6 + 2 x}

Compute the boundaries:

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

= x (3 + x - \frac{( - 1 + x ) ^{2}}{2})

\int_{1}^{3} (t^{2} - t - 6) d t

\int_{4}^{6} \frac{1}{x x ^{2} - 4} d x

\int_{4}^{7} (2 x^{2} - x) d x

\int_{0}^{6} 80 x e^{\frac{- x}{10}} d x

\int_{- 7}^{0} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

What is the integral from x-1 to sqrt(2x+6 of)xy ?
The integral from x-1 to sqrt(2x+6 of)xy is x(3+x-((-1+x)^2)/2)