1. **State the problem:** We want to evaluate the definite integral $$\int_0^4 x^3(4x - x^2)^{1/2} \, dx.$$\n\n2. **Simplify the integrand:** Notice that $$4x - x^2 = x(4 - x),$$ so:\n$$x^3 (4x - x^2)^{1/2} = x^3 (x(4 - x))^{1/2} = x^3 \cdot x^{1/2} (4 - x)^{1/2} = x^{3 + \frac{1}{2}} (4 - x)^{1/2} = x^{\frac{7}{2}} (4 - x)^{1/2}.$$\n\n3. **Use substitution:** Let $$t = 4 - x.$$ Then $$dt = -dx$$ and the limits change:\nWhen $$x = 0\Rightarrow t = 4,$$\nWhen $$x = 4\Rightarrow t = 0.$$\nAlso, $$x = 4 - t.$$\n\nRewriting the integral in terms of $$t$$:\n$$\int_0^4 x^{\frac{7}{2}} (4 - x)^{\frac{1}{2}} dx = \int_{t=4}^{0} (4 - t)^{\frac{7}{2}} t^{\frac{1}{2}} (-dt) = \int_0^4 (4 - t)^{\frac{7}{2}} t^{\frac{1}{2}} dt.$$\n\n4. **Evaluate the integral:** To make it easier, perform the substitution $$t = 4u$$ with $$u \in [0,1].$$ Then $$dt = 4 du,$$ and:\n$$\int_0^4 (4 - t)^{\frac{7}{2}} t^{\frac{1}{2}} dt = \int_0^1 (4-4u)^{\frac{7}{2}} (4u)^{\frac{1}{2}} 4 du = \int_0^1 (4(1-u))^{\frac{7}{2}} (4u)^{\frac{1}{2}} 4 du.$$\nSimplify powers of 4:\n$$= \int_0^1 4^{\frac{7}{2}} (1-u)^{\frac{7}{2}} \cdot 4^{\frac{1}{2}} u^{\frac{1}{2}} \cdot 4 du = 4^{\frac{7}{2} + \frac{1}{2} + 1} \int_0^1 (1-u)^{\frac{7}{2}} u^{\frac{1}{2}} du.$$\nCalculate $$4^{\frac{7}{2} + \frac{1}{2} + 1} = 4^{4 + 1} = 4^6 = 4096.$$\nSo the integral is:\n$$4096 \int_0^1 u^{\frac{1}{2}} (1-u)^{\frac{7}{2}} du.$$\n\n5. **Use Beta function:** Recall the Beta function $$B(p,q) = \int_0^1 t^{p-1} (1-t)^{q-1} dt = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}.$$\nHere we have $$p-1 = \frac{1}{2} \Rightarrow p = \frac{3}{2},$$ and $$q-1 = \frac{7}{2} \Rightarrow q = \frac{9}{2}.$$\nThus:\n$$\int_0^1 u^{\frac{1}{2}} (1-u)^{\frac{7}{2}} du = B\left(\frac{3}{2}, \frac{9}{2}\right) = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{9}{2}\right)}{\Gamma\left(\frac{3}{2} + \frac{9}{2}\right)} = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{9}{2}\right)}{\Gamma(6)}.$$\n\n6. **Evaluate Gamma functions:**\n- $$\Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \sqrt{\pi}.$$\n- $$\Gamma\left(\frac{9}{2}\right) = \Gamma\left(4 + \frac{1}{2}\right) = \frac{7 \cdot 5 \cdot 3 \cdot 1 \sqrt{\pi}}{2^4} = \frac{105 \sqrt{\pi}}{16}.$$\n- $$\Gamma(6) = 5! = 120.$$\n\nSubstitute these back:\n$$B\left(\frac{3}{2}, \frac{9}{2}\right) = \frac{\frac{1}{2} \sqrt{\pi} \cdot \frac{105 \sqrt{\pi}}{16}}{120} = \frac{\frac{105}{32} \pi}{120} = \frac{105 \pi}{32 \cdot 120} = \frac{105 \pi}{3840} = \frac{7 \pi}{256}.$$\n\n7. **Final answer:** The original integral is:\n$$4096 \cdot \frac{7 \pi}{256} = \frac{4096}{256} \cdot 7 \pi = 16 \cdot 7 \pi = 112 \pi.$$\n\n**Therefore:** $$\boxed{\int_0^4 x^3 (4x - x^2)^{1/2} dx = 112 \pi}.$$