How do you find the antiderivative of the $\sqrt{}$ $sqrt$ 4-x^2#?

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How do you find the antiderivative of the $\sqrt{}$ $sqrt$ 4-x^2#?

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Answer 1 · 2015-09-18T22:53:41

$\int (4 - x^{2}) d x = 4 x - \frac{x^{3}}{3} + c$ $int(4-x^2)dx = 4x - x^3/3 + c$

Explanation:

Okay, the antiderivative is nothing more, nothing less than the integral. So:

$\int (4 - x^{2}) d x$ $int(4-x^2)dx$

The integral of any constant, is itself times the variable whereas the integral of $x^{2}$ $x^2$ is, naturally $\frac{x^{3}}{3}$ $x^3/3$ because that's the function we derive to get $x^{2}$ $x^2$ , or, because of the formula $\frac{x^{n + 1}}{n + 1}$ $x^(n+1)/(n+1)$ , whichever you prefer.

So, in the end we have:
$\int (4 - x^{2}) d x = 4 x - \frac{x^{3}}{3} + c$ $int(4-x^2)dx = 4x - x^3/3 + c$

Whereas $c$ $c$ is the integrating constant because for any value of $c$ $c$ that function would still derive to $4 - x^{2}$ $4-x^2$

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How do you find the antiderivative of the $\sqrt{}$ $sqrt$ 4-x^2#?

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