What is the integral from 0 to y of (1-xy) ?

The integral from 0 to y of (1-xy) is y-(y^3)/2

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integral from 0 to y of (1-xy)

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

y - \frac{y ^{3}}{2}

Solution steps

\int_{0}^{y} 1 - x y d x

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}^{y} 1 d x - \int_{0}^{y} y x d x

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

\int_{0}^{y} y x d x = \frac{y ^{3}}{2}

= y - \frac{y ^{3}}{2}

\int_{1}^{2} (0.5 - 0.125 x) d x

\int_{1}^{3} \frac{1}{( 1 + x ) x} d x

\int_{0}^{2} (2 x + 1) (4 x^{2} + 1) d x

\int_{1}^{4} ∣ 4 x - 12 ∣ d x

\int_{b}^{a} \frac{1}{x ^{2}} d x

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