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

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

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

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

\frac{343}{6}

Decimal

57.16666 \dots

Solution steps

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

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

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

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

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

Simplify

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

= \frac{343}{6}

y^{^{''}} + 7 y^{^{'}} + 6 y = 0

f (x) = 2 (x)^{3} ln (x)

x \to - 2 lim (\frac{x ^{2} + 4}{x + 2})

y = - x^{2} + 10, y = - 1.5 x + 7

\frac{d}{d x} (\frac{x ^{2} - x ^{4}}{8})

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