1. **State the problem:** We have the curve $$y=\frac{32}{x^2}+3x-8$$ for $$x>0$$ and point $$P(4,6)$$ lies on this curve. The line $$l$$ is the normal to the curve at $$P$$. The region $$R$$ is bounded by the curve $$C$$, the normal line $$l$$, the vertical line $$x=2$$, and the x-axis. We need to find the area of region $$R$$. 2. **Verify point P on curve:** Substitute $$x=4$$ into $$y=\frac{32}{x^2}+3x-8$$: $$y=\frac{32}{4^2}+3(4)-8=\frac{32}{16}+12-8=2+12-8=6$$ So, point $$P(4,6)$$ lies on the curve. 3. **Find the slope of the tangent at P:** Differentiate $$y$$ with respect to $$x$$: $$y=32x^{-2}+3x-8$$ $$\frac{dy}{dx}=-64x^{-3}+3=-\frac{64}{x^3}+3$$ At $$x=4$$: $$\frac{dy}{dx}=-\frac{64}{4^3}+3=-\frac{64}{64}+3=-1+3=2$$ So, slope of tangent at $$P$$ is $$2$$. 4. **Find slope of normal line $$l$$:** Normal slope is negative reciprocal of tangent slope: $$m_{normal}=-\frac{1}{2}$$ 5. **Equation of normal line $$l$$ at $$P(4,6)$$:** Using point-slope form: $$y-6=-\frac{1}{2}(x-4)$$ Simplify: $$y-6=-\frac{1}{2}x+2$$ $$y=-\frac{1}{2}x+8$$ 6. **Find intersection of normal line $$l$$ with x-axis:** Set $$y=0$$: $$0=-\frac{1}{2}x+8$$ $$\frac{1}{2}x=8$$ $$x=16$$ So, normal line intersects x-axis at $$x=16$$. 7. **Region $$R$$ boundaries:** - Curve $$C$$ from $$x=2$$ to $$x=4$$ - Vertical line $$x=2$$ - Normal line $$l$$ from $$x=4$$ to $$x=16$$ - x-axis from $$x=2$$ to $$x=16$$ 8. **Find y-values at $$x=2$$ on curve:** $$y=\frac{32}{2^2}+3(2)-8=\frac{32}{4}+6-8=8+6-8=6$$ 9. **Area of region $$R$$ is sum of two parts:** - Area under curve $$C$$ from $$x=2$$ to $$x=4$$ above x-axis - Area under normal line $$l$$ from $$x=4$$ to $$x=16$$ above x-axis 10. **Calculate area under curve $$C$$ from $$2$$ to $$4$$:** $$A_1=\int_2^4 \left(\frac{32}{x^2}+3x-8\right) dx$$ Calculate integral: $$\int \frac{32}{x^2} dx=32\int x^{-2} dx=32(-x^{-1})=-\frac{32}{x}$$ $$\int 3x dx=\frac{3x^2}{2}$$ $$\int -8 dx=-8x$$ So, $$A_1=\left[-\frac{32}{x}+\frac{3x^2}{2}-8x\right]_2^4$$ Evaluate at $$4$$: $$-\frac{32}{4}+\frac{3(4)^2}{2}-8(4)=-8+\frac{48}{2}-32=-8+24-32=-16$$ Evaluate at $$2$$: $$-\frac{32}{2}+\frac{3(2)^2}{2}-8(2)=-16+\frac{12}{2}-16=-16+6-16=-26$$ So, $$A_1=(-16)-(-26)=10$$ 11. **Calculate area under normal line $$l$$ from $$4$$ to $$16$$:** $$A_2=\int_4^{16} \left(-\frac{1}{2}x+8\right) dx$$ Calculate integral: $$\int -\frac{1}{2}x dx=-\frac{1}{4}x^2$$ $$\int 8 dx=8x$$ So, $$A_2=\left[-\frac{1}{4}x^2+8x\right]_4^{16}$$ Evaluate at $$16$$: $$-\frac{1}{4}(16)^2+8(16)=-\frac{1}{4}(256)+128=-64+128=64$$ Evaluate at $$4$$: $$-\frac{1}{4}(4)^2+8(4)=-\frac{1}{4}(16)+32=-4+32=28$$ So, $$A_2=64-28=36$$ 12. **Total area of region $$R$$:** $$A=A_1+A_2=10+36=46$$ **Final answer:** $$\boxed{46}$$ unit$^2$