What is the integral from 0 to 1 of 4/3 (1-xy) ?

The integral from 0 to 1 of 4/3 (1-xy) is 4/3 (1-x/2)

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integral from 0 to 1 of 4/3 (1-xy)

\int_{0}^{1} \frac{4}{3} (1 - x y) d y

\frac{4}{3} (1 - \frac{x}{2})

Solution steps

\int_{0}^{1} \frac{4}{3} (1 - x y) d y

Take the constant out:

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

= \frac{4}{3} \cdot \int_{0}^{1} 1 - 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{4}{3} (\int_{0}^{1} 1 d y - \int_{0}^{1} x y d y)

\int_{0}^{1} 1 d y = 1

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

= \frac{4}{3} (1 - \frac{x}{2})

\int_{- 2}^{0} ∣ x + 1 ∣ d x

\int_{- 7}^{- 3} (- x^{2} - 3 x) d x

\int_{0}^{x} \frac{1}{4} d x

\int_{- x}^{x^{2}} (8 x - 10 y + 4) d x

\int_{- 7}^{- 3} - x^{2} - 3 x d x

What is the integral from 0 to 1 of 4/3 (1-xy) ?
The integral from 0 to 1 of 4/3 (1-xy) is 4/3 (1-x/2)