What is the integral from 0 to 4 of x^2(-0.5x+2) ?

The integral from 0 to 4 of x^2(-0.5x+2) is 32/3

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integral from 0 to 4 of x^2(-0.5x+2)

\int_{0}^{4} x^{2} (- 0.5 x + 2) d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{0}^{4} x^{2} (- 0.5 x + 2) d x

Expand

x^{2} (- 0.5 x + 2) : - 0.5 x^{3} + 2 x^{2}

= \int_{0}^{4} - 0.5 x^{3} + 2 x^{2} 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} 0.5 x^{3} d x + \int_{0}^{4} 2 x^{2} d x

\int_{0}^{4} 0.5 x^{3} d x = 32

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

= - 32 + \frac{128}{3}

Simplify

- 32 + \frac{128}{3} : \frac{32}{3}

= \frac{32}{3}

\int_{0}^{2} 10 d x

\int_{- 1}^{1} (y^{2} - 4) - (e^{y}) d y

\int_{- 1}^{3} (x^{2} - 2) d x

\int_{0}^{9} x^{2} - 4 d x

\int_{- 2}^{3} (x + 2) - (x^{2} - 4) d x

What is the integral from 0 to 4 of x^2(-0.5x+2) ?
The integral from 0 to 4 of x^2(-0.5x+2) is 32/3