# How do you find the definite integral of #(x^3)/sqrt(16 - x^2)# in the interval #[0, 2sqrt3]#?

Make a trig substitution to simplify the denominator, then evaluate as normal

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Here,

So,

NOTE :

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To find the definite integral of ( \frac{x^3}{\sqrt{16 - x^2}} ) in the interval ( [0, 2\sqrt{3}] ), you can use the substitution method.

Let ( u = 16 - x^2 ), then ( du = -2x , dx ).

This implies ( dx = -\frac{du}{2x} ).

Substitute ( u ) and ( dx ) into the integral:

[ \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16 - x^2}} , dx = -\frac{1}{2} \int_{16}^{4} \frac{1}{\sqrt{u}} , du ]

Now, integrate with respect to ( u ):

[ -\frac{1}{2} \int_{16}^{4} \frac{1}{\sqrt{u}} , du = -\frac{1}{2} \left[ 2\sqrt{u} \right]_{16}^{4} ]

[ = -\left[ \sqrt{4} - \sqrt{16} \right] ]

[ = -(2 - 4) ]

[ = 2 ]

So, the definite integral of ( \frac{x^3}{\sqrt{16 - x^2}} ) in the interval ( [0, 2\sqrt{3}] ) is ( 2 ).

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