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

The integral from 1 to 2 of 3/32*(4-x^2) is 5/32

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

\int_{1}^{2} \frac{3}{32} \cdot (4 - x^{2}) d x

\frac{5}{32}

Decimal Notation

0.15625

Solution steps

\int_{1}^{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_{1}^{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_{1}^{2} 4 d x - \int_{1}^{2} x^{2} d x)

\int_{1}^{2} 4 d x = 4

\int_{1}^{2} x^{2} d x = \frac{7}{3}

= \frac{3}{32} (4 - \frac{7}{3})

Simplify

\frac{3}{32} (4 - \frac{7}{3}) : \frac{5}{32}

= \frac{5}{32}

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What is the integral from 1 to 2 of 3/32*(4-x^2) ?
The integral from 1 to 2 of 3/32*(4-x^2) is 5/32