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

The integral from 0 to 4 of (4-x^2)-(4-4x) is 32/3

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

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

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

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

Simplify

= \int_{0}^{4} - x^{2} + 4 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

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

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

= - \frac{64}{3} + 32

Simplify

- \frac{64}{3} + 32 : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 0 to 4 of (4-x^2)-(4-4x) ?
The integral from 0 to 4 of (4-x^2)-(4-4x) is 32/3