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How do you integrate #intsqrt(9-x^2)(-2x)dx#?

Answer 1

Christopher Allers

Let #u = 9 - x^2#. Then, by the power rule, #du = -2xdx# and #dx = (du)/(-2x)#.

#=intsqrt(u)(-2x) * (du)/(-2x)#

#=intsqrt(u)du#

#=u^(3/2) + C#

#=2/3(9 - x^2)^(3/2) + C#

Hopefully this helps!

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Answer 2

Ezekiel Alexander

To integrate (\int \sqrt{9-x^2}(-2x) , dx), we can use the substitution method. Let (u = 9 - x^2), then (du = -2x , dx).

Substituting: [u = 9 - x^2] [du = -2x , dx]

This gives: [\int \sqrt{9-x^2}(-2x) , dx = -\int \sqrt{u} , du]

Now, integrate (-\int \sqrt{u} , du), which is: [-\frac{2}{3} u^\frac{3}{2} + C]

Substitute (u = 9 - x^2) back in: [-\frac{2}{3} (9 - x^2)^\frac{3}{2} + C]

So, (\int \sqrt{9-x^2}(-2x) , dx = -\frac{2}{3} (9 - x^2)^\frac{3}{2} + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.