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Popular Calculus >

integral from 0 to 2 of (4-x-y)

Solution

\int_{0}^{2} (4 - x - y) d y

Solution

- 2 x + 6

Solution steps

\int_{0}^{2} 4 - 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} 4 d y - \int_{0}^{2} x d y - \int_{0}^{2} y d y

\int_{0}^{2} 4 d y = 8

\int_{0}^{2} x d y = 2 x

\int_{0}^{2} y d y = 2

= 8 - 2 x - 2

Simplify

= - 2 x + 6

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Frequently Asked Questions (FAQ)

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