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

The integral from 0 to 2 of (y-2)(2y+1) is -14/3

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

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

- \frac{14}{3}

Decimal Notation

- 4.66666 \dots

Solution steps

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

Expand

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

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

\int_{0}^{2} 2 y^{2} d y = \frac{16}{3}

\int_{0}^{2} 3 y d y = 6

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

= \frac{16}{3} - 6 - 4

Simplify

\frac{16}{3} - 6 - 4 : - \frac{14}{3}

= - \frac{14}{3}

\int 1 + cos (π x) d x

\frac{\partial}{\partial y} (cos (x) sin (y))

x \to + 0 lim (\frac{sin ( 40 )}{x})

f (x) = cos (cos (x))

\frac{d ^{4}}{d x ^{4}} (\frac{1}{x})

What is the integral from 0 to 2 of (y-2)(2y+1) ?
The integral from 0 to 2 of (y-2)(2y+1) is -14/3