What is the integral from-2 to 2 of (-x^4)-(x^4-32) ?

The integral from-2 to 2 of (-x^4)-(x^4-32) is 512/5

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integral from-2 to 2 of (-x^4)-(x^4-32)

\int_{- 2}^{2} (- x^{4}) - (x^{4} - 32) d x

\frac{512}{5}

Decimal Notation

102.4

Solution steps

\int_{- 2}^{2} - x^{4} - (x^{4} - 32) d x

Simplify

= \int_{- 2}^{2} - 2 x^{4} + 32 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} 2 x^{4} d x + \int_{- 2}^{2} 32 d x

\int_{- 2}^{2} 2 x^{4} d x = \frac{128}{5}

\int_{- 2}^{2} 32 d x = 128

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

Simplify

- \frac{128}{5} + 128 : \frac{512}{5}

= \frac{512}{5}

\frac{1}{1 - x ^{2}}

f (x) = \frac{12}{x}

(4 + x) y^{^{'}} = 5 y

\frac{d}{d x} (1 - x^{3})

What is the integral from-2 to 2 of (-x^4)-(x^4-32) ?
The integral from-2 to 2 of (-x^4)-(x^4-32) is 512/5