What is the integral from-1/3 to 0 of 3/4 (1-x^2) ?

The integral from-1/3 to 0 of 3/4 (1-x^2) is 13/54

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integral from-1/3 to 0 of 3/4 (1-x^2)

\int_{- \frac{1}{3}}^{0} \frac{3}{4} (1 - x^{2}) d x

\frac{13}{54}

Decimal Notation

0.24074 \dots

Solution steps

\int_{- \frac{1}{3}}^{0} \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_{- \frac{1}{3}}^{0} 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_{- \frac{1}{3}}^{0} 1 d x - \int_{- \frac{1}{3}}^{0} x^{2} d x)

\int_{- \frac{1}{3}}^{0} 1 d x = \frac{1}{3}

\int_{- \frac{1}{3}}^{0} x^{2} d x = \frac{1}{81}

= \frac{3}{4} (\frac{1}{3} - \frac{1}{81})

Simplify

\frac{3}{4} (\frac{1}{3} - \frac{1}{81}) : \frac{13}{54}

= \frac{13}{54}

x + x^{2} + ln (x)

x \to + 2 π lim (f (x))

ln (x^{3} + 3 x)

\frac{d x}{d y} = 5 x + y

\frac{d y}{d x} = y^{2}, y (1) = 1

What is the integral from-1/3 to 0 of 3/4 (1-x^2) ?
The integral from-1/3 to 0 of 3/4 (1-x^2) is 13/54