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

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

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

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

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

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

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