# What is the net area between #f(x) = xsqrt(x^2-1) + 6# and the x-axis over #x in [2, 3 ]#?

The are can be calculated solving the definite integral:

Calculate the two integrals separately:

So:

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To find the net area between ( f(x) = x\sqrt{x^2-1} + 6 ) and the x-axis over ( x ) in ( [2, 3] ), we first need to integrate the function within that interval, and then take the absolute value of the result.

The integral of ( f(x) ) over the interval ( [2, 3] ) is:

[ \int_{2}^{3} (x\sqrt{x^2-1} + 6) , dx ]

To integrate this, you can use u-substitution. Let ( u = x^2 - 1 ), then ( du = 2x , dx ).

Substituting ( u = x^2 - 1 ) and ( du = 2x , dx ) into the integral, we get:

[ \frac{1}{2} \int_{u(2)}^{u(3)} \sqrt{u} , du ]

[ = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_{3}^{8} ]

[ = \frac{1}{3}(8\sqrt{8} - 3\sqrt{3}) ]

[ = \frac{8\sqrt{8} - 3\sqrt{3}}{3} ]

[ \approx 15.242 ]

Since we're interested in the net area, we take the absolute value of this result.

So, the net area between ( f(x) = x\sqrt{x^2-1} + 6 ) and the x-axis over ( x ) in ( [2, 3] ) is approximately ( 15.242 ) square units.

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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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