What is the integral from-1 to x of 1 (x^2-x) ?

The integral from-1 to x of 1 (x^2-x) is d((x^3+1)/3-(x^2-1)/2)

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integral from-1 to x of 1 (x^2-x)

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

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

Solution steps

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

Take the constant out:

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

= d \cdot \int_{- 1}^{x} x^{2} - x 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

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

\int_{- 1}^{x} x^{2} d x = \frac{x ^{3} + 1}{3}

\int_{- 1}^{x} x d x = \frac{x ^{2} - 1}{2}

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

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What is the integral from-1 to x of 1 (x^2-x) ?
The integral from-1 to x of 1 (x^2-x) is d((x^3+1)/3-(x^2-1)/2)