What is the integral from-3 to 3 of (16-x^2)-(x^2-2) ?

The integral from-3 to 3 of (16-x^2)-(x^2-2) is 72

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integral from-3 to 3 of (16-x^2)-(x^2-2)

\int_{- 3}^{3} (16 - x^{2}) - (x^{2} - 2) d x

72

Solution steps

\int_{- 3}^{3} 16 - x^{2} - (x^{2} - 2) d x

Simplify

= \int_{- 3}^{3} - 2 x^{2} + 18 d x

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}^{3} 2 x^{2} d x + \int_{- 3}^{3} 18 d x

\int_{- 3}^{3} 2 x^{2} d x = 36

\int_{- 3}^{3} 18 d x = 108

= - 36 + 108

Simplify

= 72

y = 2 sec (x) - csc (x)

2 x^{2} y^{^{'}} = y^{^{'}} + 6 x e - y

\frac{d}{d x} (- x^{5} + 3 x^{4} - 2 x^{3} - 2 x^{2} + 3 x - 1)

t^{5} e^{- t}

x \to 0 - lim (x)

What is the integral from-3 to 3 of (16-x^2)-(x^2-2) ?
The integral from-3 to 3 of (16-x^2)-(x^2-2) is 72