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

The integral from 0 to 3-x of 3-x-y is x^2-6x-((-x+3)^2)/2+9

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integral from 0 to 3-x of 3-x-y

\int_{0}^{3 - x} 3 - x - y d y

x^{2} - 6 x - \frac{( - x + 3 ) ^{2}}{2} + 9

Solution steps

\int_{0}^{3 - x} 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}^{3 - x} 3 d y - \int_{0}^{3 - x} x d y - \int_{0}^{3 - x} y d y

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

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

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

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

Simplify

3 (- x + 3) - x (- x + 3) - \frac{( - x + 3 ) ^{2}}{2} : x^{2} - 6 x - \frac{( - x + 3 ) ^{2}}{2} + 9

= x^{2} - 6 x - \frac{( - x + 3 ) ^{2}}{2} + 9

\frac{d}{d x} (x tan (\frac{1}{x}))

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

cos^{3} (12 x)

y = 3 x^{2} + 2 x, (2, 4)

What is the integral from 0 to 3-x of 3-x-y ?
The integral from 0 to 3-x of 3-x-y is x^2-6x-((-x+3)^2)/2+9