What is the integral from-2 to 2 of (x+7)sqrt(4-x^2) ?

The integral from-2 to 2 of (x+7)sqrt(4-x^2) is 14pi

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integral from-2 to 2 of (x+7)sqrt(4-x^2)

\int_{- 2}^{2} (x + 7) 4 - x^{2} d x

14 π

Decimal Notation

43.98229 \dots

Solution steps

\int_{- 2}^{2} (x + 7) 4 - x^{2} d x

Expand

(x + 7) 4 - x^{2} : x 4 - x^{2} + 7 4 - x^{2}

= \int_{- 2}^{2} x 4 - x^{2} + 7 4 - 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

= \int_{- 2}^{2} x 4 - x^{2} d x + \int_{- 2}^{2} 7 4 - x^{2} d x

\int_{- 2}^{2} x 4 - x^{2} d x = 0

\int_{- 2}^{2} 7 4 - x^{2} d x = 14 π

= 0 + 14 π

Simplify

= 14 π

x \to 0 lim (\frac{csc ( x )}{3 x})

\frac{\partial}{\partial y} (\frac{x}{y} + \frac{y}{z})

\frac{\partial}{\partial z} (x^{2} y + y z^{3} + z^{5} x)

y \to 0 lim (\frac{y ^{2}}{y ^{2}})

\int x^{3} 25 + x^{2} d x

What is the integral from-2 to 2 of (x+7)sqrt(4-x^2) ?
The integral from-2 to 2 of (x+7)sqrt(4-x^2) is 14pi