What is the integral from 0 to 2-2x of 1+3x+y ?

The integral from 0 to 2-2x of 1+3x+y is -6x^2+4x+((-2x+2)^2)/2+2

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integral from 0 to 2-2x of 1+3x+y

\int_{0}^{2 - 2 x} 1 + 3 x + y d y

- 6 x^{2} + 4 x + \frac{( - 2 x + 2 ) ^{2}}{2} + 2

Solution steps

\int_{0}^{2 - 2 x} 1 + 3 x + y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2 - 2 x} 1 d y + \int_{0}^{2 - 2 x} y d y + \int_{0}^{2 - 2 x} 3 x d y

\int_{0}^{2 - 2 x} 1 d y = - 2 x + 2

\int_{0}^{2 - 2 x} y d y = \frac{( - 2 x + 2 ) ^{2}}{2}

\int_{0}^{2 - 2 x} 3 x d y = 3 x (- 2 x + 2)

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

Simplify

- 2 x + 2 + \frac{( - 2 x + 2 ) ^{2}}{2} + 3 x (- 2 x + 2) : - 6 x^{2} + 4 x + \frac{( - 2 x + 2 ) ^{2}}{2} + 2

= - 6 x^{2} + 4 x + \frac{( - 2 x + 2 ) ^{2}}{2} + 2

\int (5 x^{2} + 1) (5 x^{3} + 3 x - 8)^{6} d x

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

4 y^{2} + 7 x^{2}

\frac{\partial}{\partial x} (2 x y^{6} + x^{4} y^{8})

f (x) = x^{2} - \frac{1}{x}, at x = 1

What is the integral from 0 to 2-2x of 1+3x+y ?
The integral from 0 to 2-2x of 1+3x+y is -6x^2+4x+((-2x+2)^2)/2+2