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

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

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

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

1

Solution steps

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

Take the constant out:

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

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

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

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

= \frac{3}{4} (2 - \frac{2}{3})

Simplify

\frac{3}{4} (2 - \frac{2}{3}) : 1

= 1

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