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

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

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

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

\frac{3}{4} x - \frac{x ^{3}}{4} + C

Solution steps

\int \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^{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 d x - \int x^{2} d x)

\int 1 d x = x

\int x^{2} d x = \frac{x ^{3}}{3}

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

Simplify

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

= \frac{3}{4} x - \frac{x ^{3}}{4}

Add a constant to the solution

= \frac{3}{4} x - \frac{x ^{3}}{4} + C

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