What is the integral from 0 to 2 of 1/8 (x+y) ?

The integral from 0 to 2 of 1/8 (x+y) is (x+1)/4

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integral from 0 to 2 of 1/8 (x+y)

\int_{0}^{2} \frac{1}{8} (x + y) d y

\frac{x + 1}{4}

Solution steps

\int_{0}^{2} \frac{1}{8} (x + y) d y

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{2} 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

= \frac{1}{8} (\int_{0}^{2} x d y + \int_{0}^{2} y d y)

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

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

= \frac{1}{8} (2 x + 2)

Simplify

\frac{1}{8} (2 x + 2) : \frac{x + 1}{4}

= \frac{x + 1}{4}

f (x) = e^{2 x} - 1

\int 5 cos^{5} (x) sin^{4} (x) d x

y = x^{3} - 2 x^{2} + 5, at x = 2

ln (π)

x \to - 0.1 lim (\frac{sin ( x )}{x})

What is the integral from 0 to 2 of 1/8 (x+y) ?
The integral from 0 to 2 of 1/8 (x+y) is (x+1)/4