What is the integral from 0 to 4 of (y-2)(2y+1) ?

The integral from 0 to 4 of (y-2)(2y+1) is 32/3

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integral from 0 to 4 of (y-2)(2y+1)

\int_{0}^{4} (y - 2) (2 y + 1) d y

\frac{32}{3}

Solution steps

\int_{0}^{4} (y - 2) (2 y + 1) d y

Expand

(y - 2) (2 y + 1) : 2 y^{2} - 3 y - 2

= \int_{0}^{4} 2 y^{2} - 3 y - 2 d y

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} 2 y^{2} d y - \int_{0}^{4} 3 y d y - \int_{0}^{4} 2 d y

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

\int_{0}^{4} 3 y d y = 24

\int_{0}^{4} 2 d y = 8

= \frac{128}{3} - 24 - 8

Simplify

\frac{128}{3} - 24 - 8 : \frac{32}{3}

= \frac{32}{3}

\int_{0}^{3} (3 x - 8) d x

\int_{3}^{4} x^{4} d x

\int_{1}^{2} (2 x + \frac{1}{x}) d x

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

\int_{0}^{3} 2 t 4 t^{2} + 4 d t

What is the integral from 0 to 4 of (y-2)(2y+1) ?
The integral from 0 to 4 of (y-2)(2y+1) is 32/3