# What is the net area between #f(x) = sqrt(x+3)-x^3 # and the x-axis over #x in [2, 4 ]#?

Since we need the area between the x axis and the curve, in the interval

Plugging in upper and lowers bounds:

Area

GRAPH:

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To find the net area between the curve (f(x) = \sqrt{x+3} - x^3) and the x-axis over the interval ([2, 4]), integrate the absolute value of (f(x)) over that interval and subtract the area under the x-axis:

[ \text{Net Area} = \int_{2}^{4} |\sqrt{x+3} - x^3| , dx - \int_{2}^{4} |\text{x-axis}| , dx ]

[ \text{Net Area} = \int_{2}^{4} (\sqrt{x+3} - x^3) , dx - 0 ]

Now, integrate the function (f(x)) over the interval ([2, 4]):

[ \int_{2}^{4} (\sqrt{x+3} - x^3) , dx = \left[\frac{2}{3}(x+3)^{\frac{3}{2}} - \frac{1}{4}x^4\right]_{2}^{4} ]

[ = \left[\frac{2}{3}(4+3)^{\frac{3}{2}} - \frac{1}{4}(4)^4\right] - \left[\frac{2}{3}(2+3)^{\frac{3}{2}} - \frac{1}{4}(2)^4\right] ]

[ = \left[\frac{2}{3}(7)^{\frac{3}{2}} - 64\right] - \left[\frac{2}{3}(5)^{\frac{3}{2}} - 8\right] ]

[ = \left[\frac{2}{3}(7)^{\frac{3}{2}} - 64\right] - \left[\frac{2}{3}(5)^{\frac{3}{2}} - 8\right] ]

[ \approx \left[14.98 - 64\right] - \left[8.97 - 8\right] ]

[ \approx (-49.02) - 0.97 ]

[ \approx -50 ]

So, the net area between the curve (f(x) = \sqrt{x+3} - x^3) and the x-axis over the interval ([2, 4]) is approximately (-50).

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