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}

\int_{1}^{4} - 2 x + 6 d x

\int_{- 1}^{2} \frac{1}{x ^{2} - 4 x + 13} d x

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

\int_{1}^{2} π (4 - \frac{1}{2} x)^{2} d x

\int_{0}^{1} x^{3} + x e^{- x} d x

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