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

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

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

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

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

Solution steps

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

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

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

= \frac{3}{4} (x + 1 - \frac{x ^{3} + 1}{3})

Simplify

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

= \frac{1}{4} (- x^{3} + 3 x + 2)

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