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

Question

Emily Ammerman · Accepted Answer

#55.107# square units ( 3 .d.p.)

Since we need the area between the x axis and the curve, in the interval #[2,4]#, our integral will be:

#int_(2)^(4)(sqrt(x+3)-x^3) dx=2/3(x+3)^(3/2)-1/4x^4#

#Area=[2/3(x+3)^(3/2)-1/4x^4]^(4)-[2/3(x+3)^(3/2)-1/4x^4]_(2)#

Plugging in upper and lowers bounds:

#Area=[2/3(4+3)^(3/2)-1/4(4)^4]^(4)-[2/3(2+3)^(3/2)-1/4(2)^4]_(2)#

#Area=[14/3sqrt(7)-64]-[10/3sqrt(5)-4]~~-55.107#

Area #= 55.107# square units.

GRAPH:

Samantha Adkison · Answer

undefined 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 $[2, 4]$:

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

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

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

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

$$
\approx (-49.02) - 0.97
$$

$$
\approx -50
$$

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 $-50$.