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

The integral from-2 to 2 of (2sqrt(4-x^2))^2 is 128/3

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integral from-2 to 2 of (2sqrt(4-x^2))^2

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

\frac{128}{3}

Decimal Notation

42.66666 \dots

Solution steps

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

Apply Integration By Parts

: [4 x (4 - x^{2}) - \int - 8 x^{2} d x]_{- 2}^{2}

= [4 x (4 - x^{2}) - \int - 8 x^{2} d x]_{- 2}^{2}

\int - 8 x^{2} d x = - \frac{8 x ^{3}}{3}

= [4 x (4 - x^{2}) - (- \frac{8 x ^{3}}{3})]_{- 2}^{2}

Simplify

= [4 x (4 - x^{2}) + \frac{8 x ^{3}}{3}]_{- 2}^{2}

Compute the boundaries:

\frac{128}{3}

= \frac{128}{3}

\frac{d}{d x} (\frac{( x ^{2} - 6 x )}{x + 1})

\int x x^{2} + 9 d x

\int (e^{x} + 2)^{2} d x

\frac{\partial}{\partial x} (- 5 y^{3} sin (5 x))

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8} x

What is the integral from-2 to 2 of (2sqrt(4-x^2))^2 ?
The integral from-2 to 2 of (2sqrt(4-x^2))^2 is 128/3