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

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

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

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

\frac{32}{3}

Solution steps

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

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

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

= - \frac{64}{3} - (- 32)

Simplify

- \frac{64}{3} - (- 32) : \frac{32}{3}

= \frac{32}{3}

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

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

\int_{0}^{2 + y} 1 - 3 z^{2} d z

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

\int_{- a}^{a} x^{2} a^{2} - x^{2} d x

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