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

The integral from-1 to 1 of (2sqrt(1-x^2))^2 is 16/3

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Simplify

(2 1 - x^{2})^{2} : 4 (1 - x^{2})

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

Take the constant out:

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

= 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

= 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}

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

Simplify

4 (2 - \frac{2}{3}) : \frac{16}{3}

= \frac{16}{3}

\int (\frac{2}{3 x}) d x

x \to \infty lim (- 2 x^{3})

d er i v a t i v e x a + b x

\int \frac{ln ( 6 x )}{x ^{4}} d x

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

What is the integral from-1 to 1 of (2sqrt(1-x^2))^2 ?
The integral from-1 to 1 of (2sqrt(1-x^2))^2 is 16/3