What is the integral from 0 to 4 of x^3-8x^2+16x ?

The integral from 0 to 4 of x^3-8x^2+16x is 64/3

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integral from 0 to 4 of x^3-8x^2+16x

\int_{0}^{4} x^{3} - 8 x^{2} + 16 x d x

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

\int_{0}^{4} x^{3} - 8 x^{2} + 16 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^{3} d x - \int_{0}^{4} 8 x^{2} d x + \int_{0}^{4} 16 x d x

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

\int_{0}^{4} 8 x^{2} d x = \frac{512}{3}

\int_{0}^{4} 16 x d x = 128

= 64 - \frac{512}{3} + 128

Simplify

64 - \frac{512}{3} + 128 : \frac{64}{3}

= \frac{64}{3}

\int_{1}^{3} t^{2} + 3 t - 18 d t

\int_{0}^{x^{3}} t^{2} d t

\int_{- 1}^{1} cos (2 π x) d x

\int_{0}^{5} \frac{1}{96} y^{3} e^{- \frac{y}{2}} d y

\int_{0}^{5} x e^{- 0.4 x} d x

What is the integral from 0 to 4 of x^3-8x^2+16x ?
The integral from 0 to 4 of x^3-8x^2+16x is 64/3