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

The integral from-2 to 2 of (4-x^2)^2 is 512/15

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integral from-2 to 2 of (4-x^2)^2

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

\frac{512}{15}

Decimal Notation

34.13333 \dots

Solution steps

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

Expand

(4 - x^{2})^{2} : 16 - 8 x^{2} + x^{4}

= \int_{- 2}^{2} 16 - 8 x^{2} + x^{4} 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}^{2} 16 d x - \int_{- 2}^{2} 8 x^{2} d x + \int_{- 2}^{2} x^{4} d x

\int_{- 2}^{2} 16 d x = 64

\int_{- 2}^{2} 8 x^{2} d x = \frac{128}{3}

\int_{- 2}^{2} x^{4} d x = \frac{64}{5}

= 64 - \frac{128}{3} + \frac{64}{5}

Simplify

64 - \frac{128}{3} + \frac{64}{5} : \frac{512}{15}

= \frac{512}{15}

\int_{- 1}^{1} \frac{1}{x + 2} d x

\int_{1}^{2} \frac{x + 1}{x} d x

\frac{d ^{4} y}{d x ^{4}} - 23 \frac{d ^{2} y}{d x ^{2}} - 50 y = 0

5 x + \frac{4}{x}

x \to 0 lim (- (\frac{1}{x} - \frac{1}{∣ x ∣}))

What is the integral from-2 to 2 of (4-x^2)^2 ?
The integral from-2 to 2 of (4-x^2)^2 is 512/15