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

The integral from-1 to 2 of (y+2)-(y^2) is 9/2

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

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

\frac{9}{2}

Solution steps

\int_{- 1}^{2} y + 2 - 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_{- 1}^{2} y d y + \int_{- 1}^{2} 2 d y - \int_{- 1}^{2} y^{2} d y

\int_{- 1}^{2} y d y = \frac{3}{2}

\int_{- 1}^{2} 2 d y = 6

\int_{- 1}^{2} y^{2} d y = 3

= \frac{3}{2} + 6 - 3

Simplify

\frac{3}{2} + 6 - 3 : \frac{9}{2}

= \frac{9}{2}

\int tan (x) x d x

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 7 )} d x

\int \frac{8}{θ ^{2}} cos (\frac{8}{θ}) d θ

y = ln (8 - 4 x)

f (x) = 21 x

What is the integral from-1 to 2 of (y+2)-(y^2) ?
The integral from-1 to 2 of (y+2)-(y^2) is 9/2