What is the integral from 0 to x of (4-x^2) ?

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

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

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

4 x - \frac{x ^{3}}{3}

Solution steps

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

\int_{0}^{x} 4 d x = 4 x

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

= 4 x - \frac{x ^{3}}{3}

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