What is the integral from-2 to 2 of 3/32 (4-x^2) ?

The integral from-2 to 2 of 3/32 (4-x^2) is 1

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integral from-2 to 2 of 3/32 (4-x^2)

\int_{- 2}^{2} \frac{3}{32} (4 - x^{2}) d x

1

Solution steps

\int_{- 2}^{2} \frac{3}{32} (4 - x^{2}) d x

Take the constant out:

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

= \frac{3}{32} \cdot \int_{- 2}^{2} 4 - x^{2} 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{3}{32} (\int_{- 2}^{2} 4 d x - \int_{- 2}^{2} x^{2} d x)

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

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

= \frac{3}{32} (16 - \frac{16}{3})

Simplify

\frac{3}{32} (16 - \frac{16}{3}) : 1

= 1

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What is the integral from-2 to 2 of 3/32 (4-x^2) ?
The integral from-2 to 2 of 3/32 (4-x^2) is 1