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Evaluate the integral. 14₀∫¹x³√1-x²

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Hello!

Sure, I can help you with that. The integral you've given is:

∫(0 to 1) 14x³√(1 - x²) dx

Let's perform the integration step by step:

1. Use substitution: Let u = 1 - x². Then, du/dx = -2x, and dx = du / (-2x).

2. Substitute the limits of integration as well: When x = 0, u = 1; when x = 1, u = 0.

Now the integral becomes:

∫(1 to 0) 14x³√u * (du / -2x)

Cancel out the x term:

∫(1 to 0) -7x²√u du

Integrate with respect to u:

[-7/3 * u^(3/2)] from 1 to 0

Substitute back u = 1 - x²:

-7/3 * (1 - x²)^(3/2) - (-7/3 * 1^(3/2))

Simplify:

-7/3 * (1 - x²)^(3/2) + 7/3

So, the value of the integral is:

-7/3 * (1 - x²)^(3/2) + 7/3

Hope this help you!

If you have more questions, feel free to ask!

User Nosov Pavel
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