What is the integral from-2 to 3 of (x^2+4)-(-x^2-1) ?

The integral from-2 to 3 of (x^2+4)-(-x^2-1) is 145/3

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integral from-2 to 3 of (x^2+4)-(-x^2-1)

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

\frac{145}{3}

Decimal

48.33333 \dots

Solution steps

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

Simplify

= \int_{- 2}^{3} 2 x^{2} + 5 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_{- 2}^{3} 2 x^{2} d x + \int_{- 2}^{3} 5 d x

\int_{- 2}^{3} 2 x^{2} d x = \frac{70}{3}

\int_{- 2}^{3} 5 d x = 25

= \frac{70}{3} + 25

Simplify

\frac{70}{3} + 25 : \frac{145}{3}

= \frac{145}{3}

x \to \infty lim (8^{- x})

p (x) = 2 x^{4} + 3 x^{3} - 6 x^{2} + 9

\frac{d}{d x} (4 cos (3 x) - 3 sin (4 x))

\frac{d}{d x} (sin (x^{3} + 3))

\frac{d}{d x} (e^{a x} + a e^{a x} x)

What is the integral from-2 to 3 of (x^2+4)-(-x^2-1) ?
The integral from-2 to 3 of (x^2+4)-(-x^2-1) is 145/3