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

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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 Notation

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}

3 sec (3 x)

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

cos (t - \frac{π}{2}) H (t - \frac{π}{2})

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