# What is # int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx #?

# int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx = 11/405 sqrt(3) #

We seek:

We can perform a substitution of the form:

And we manipulate the integral as follows:

So the last answer is correct.

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To solve the integral ∫(1/3)^(2/3) x^3√(4-9x^2) dx, we can use trigonometric substitution. Let's let x = (2/3)sin(θ). Then dx = (2/3)cos(θ)dθ.

Now, substitute x and dx into the integral: ∫(1/3)^(2/3) x^3√(4-9x^2) dx = ∫sin^3(θ)√(4-9(2/3)^2sin^2(θ))(2/3)cos(θ)dθ.

This simplifies to: (2/3)∫sin^3(θ)√(4-4sin^2(θ))cos(θ)dθ.

Using the identity sin^2(θ) + cos^2(θ) = 1, we have sin^2(θ) = 1 - cos^2(θ). Substitute this into the integral: (2/3)∫sin^3(θ)√(4-4(1-cos^2(θ)))cos(θ)dθ = (2/3)∫sin^3(θ)√(4cos^2(θ))cos(θ)dθ.

Now, simplify inside the square root: (2/3)∫sin^3(θ)√(4cos^2(θ))cos(θ)dθ = (2/3)∫sin^3(θ)(2cos(θ))dθ = (4/3)∫sin^3(θ)cos(θ)dθ.

Now, we can use the trigonometric identity ∫sin^m(θ)cos^n(θ)dθ = -(cos^(n-1)(θ)sin^(m+1)(θ))/(m+1) + (n-1)/(m+1)∫sin^m(θ)cos^(n-2)(θ)dθ when m is odd.

Applying this identity with m = 3 and n = 1: (4/3)∫sin^3(θ)cos(θ)dθ = (4/3) * [-(cos^0(θ)sin^4(θ))/4 + 3/4∫sin^3(θ)cos(θ)dθ].

Now, we have a simpler integral: ∫sin^3(θ)cos(θ)dθ. Using the same trigonometric identity, we can solve this integral. After solving for θ, we'll need to go back to x using the substitution x = (2/3)sin(θ) to obtain the final answer in terms of x.

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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.

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.

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.

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