question let the region r be the area enclosed by the function f(x)=3x² and g(x)=4x. if the region r is the…

Explanation:

Step1: Find intersection points

Set $3x^{2}=4x$, then $3x^{2}-4x = 0$, $x(3x - 4)=0$. So $x = 0$ and $x=\frac{4}{3}$.

Step2: Determine the length of the leg

The length of the leg of the isosceles - right triangle cross - section, $l$, is $l = 4x-3x^{2}$ (since $g(x)\geq f(x)$ on the interval $[0,\frac{4}{3}]$).

Step3: Find the area of the cross - section

The area of an isosceles right triangle with leg length $l$ is $A(x)=\frac{1}{2}l^{2}$. Substituting $l = 4x - 3x^{2}$, we get $A(x)=\frac{1}{2}(4x - 3x^{2})^{2}=\frac{1}{2}(16x^{2}-24x^{3}+9x^{4})$.

Step4: Calculate the volume using integration

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b=\frac{4}{3}$, so $V=\int_{0}^{\frac{4}{3}}\frac{1}{2}(16x^{2}-24x^{3}+9x^{4})dx$. [ \begin{align*} V&=\frac{1}{2}\left[\frac{16}{3}x^{3}-6x^{4}+\frac{9}{5}x^{5}\right]_{0}^{\frac{4}{3}}\ &=\frac{1}{2}\left(\frac{16}{3}\times\left(\frac{4}{3}\right)^{3}-6\times\left(\frac{4}{3}\right)^{4}+\frac{9}{5}\times\left(\frac{4}{3}\right)^{5}\right)\ &=\frac{1}{2}\left(\frac{16}{3}\times\frac{64}{27}-6\times\frac{256}{81}+\frac{9}{5}\times\frac{1024}{243}\right)\ &=\frac{1}{2}\left(\frac{1024}{81}-\frac{1536}{81}+\frac{9216}{1215}\right)\ &=\frac{1}{2}\left(\frac{1024\times15 - 1536\times15+9216}{1215}\right)\ &=\frac{1}{2}\left(\frac{15360-23040 + 9216}{1215}\right)\ &=\frac{1}{2}\times\frac{1536}{1215}\ &=\frac{512}{810}\approx0.632 \end{align*} ]

Answer:

$0.632$