What is the integral from-4 to 4 of 1/2 (16^2-x^4) ?

The integral from-4 to 4 of 1/2 (16^2-x^4) is 4096/5

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integral from-4 to 4 of 1/2 (16^2-x^4)

\int_{- 4}^{4} \frac{1}{2} (1 6^{2} - x^{4}) d x

\frac{4096}{5}

Decimal Notation

819.2

Solution steps

\int_{- 4}^{4} \frac{1}{2} (1 6^{2} - x^{4}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- 4}^{4} 1 6^{2} - x^{4} d x

Simplify

= \frac{1}{2} \cdot \int_{- 4}^{4} 256 - 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

= \frac{1}{2} (\int_{- 4}^{4} 256 d x - \int_{- 4}^{4} x^{4} d x)

\int_{- 4}^{4} 256 d x = 2048

\int_{- 4}^{4} x^{4} d x = \frac{2048}{5}

= \frac{1}{2} (2048 - \frac{2048}{5})

Simplify

\frac{1}{2} (2048 - \frac{2048}{5}) : \frac{4096}{5}

= \frac{4096}{5}

\int cos (3 x + 1) d x

x \to 0 lim (4 sin (x) - 1)

\int_{0}^{4} x d x

x \to 0 - lim (x - 2)

f (x) = 2 x^{2} - 3, (2, 5)

What is the integral from-4 to 4 of 1/2 (16^2-x^4) ?
The integral from-4 to 4 of 1/2 (16^2-x^4) is 4096/5