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

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

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

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

\frac{4}{3}

Decimal Notation

2.30940 \dots

Solution steps

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

Take the constant out:

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

= 3 \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

= 3 (\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}

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

Simplify

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

= \frac{4}{3}

\int \frac{1}{h} d h

\int sec^{4} (x t a) n^{4} x d x

sin^{9} (x)

x \to \infty lim (\frac{3 ^{x + 1}}{3 ^{x}})

(- 3, - 3), (- 5, - 2)

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