What is the integral from-3 to 4 of [y-(y^2-12)] ?

The integral from-3 to 4 of [y-(y^2-12)] is 343/6

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integral from-3 to 4 of [y-(y^2-12)]

\int_{- 3}^{4} [y - (y^{2} - 12)] d y

\frac{343}{6}

Decimal Notation

57.16666 \dots

Solution steps

\int_{- 3}^{4} y - (y^{2} - 12) d y

Simplify

= \int_{- 3}^{4} y - y^{2} + 12 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_{- 3}^{4} y d y - \int_{- 3}^{4} y^{2} d y + \int_{- 3}^{4} 12 d y

\int_{- 3}^{4} y d y = \frac{7}{2}

\int_{- 3}^{4} y^{2} d y = \frac{91}{3}

\int_{- 3}^{4} 12 d y = 84

= \frac{7}{2} - \frac{91}{3} + 84

Simplify

\frac{7}{2} - \frac{91}{3} + 84 : \frac{343}{6}

= \frac{343}{6}

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What is the integral from-3 to 4 of [y-(y^2-12)] ?
The integral from-3 to 4 of [y-(y^2-12)] is 343/6