determine whether the integral is convergent or divergent. if it is convergent, evaluate it. (if the…

determine whether the integral is convergent or divergent. if it is convergent, evaluate it. (if the quantity diverges, enter diverges.)\n$$\\int_{-2}^{14} \\frac{8}{\\sqrt4{x + 2}} dx$$

determine whether the integral is convergent or divergent. if it is convergent, evaluate it. (if the quantity diverges, enter diverges.)\n$$\\int_{-2}^{14} \\frac{8}{\\sqrt4{x + 2}} dx$$

Answer

Explanation:

Step1: Substitute (t = x + 2)

When (x=-2), (t = 0); when (x = 14), (t=16). And (dt=dx). The integral becomes (\int_{0}^{16}\frac{8}{t^{\frac{1}{4}}}dt).

Step2: Use the integral formula (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1))

We know that (\int\frac{8}{t^{\frac{1}{4}}}dt=8\int t^{-\frac{1}{4}}dt). By the formula (\int t^{n}dt=\frac{t^{n + 1}}{n + 1}+C) ((n=-\frac{1}{4})), we have (8\times\frac{t^{-\frac{1}{4}+1}}{-\frac{1}{4}+1}+C=8\times\frac{t^{\frac{3}{4}}}{\frac{3}{4}}+C=\frac{32}{3}t^{\frac{3}{4}}+C).

Step3: Evaluate the definite - integral

(\left[\frac{32}{3}t^{\frac{3}{4}}\right]_{0}^{16}=\frac{32}{3}\times16^{\frac{3}{4}}-\frac{32}{3}\times0^{\frac{3}{4}}). Since (16^{\frac{3}{4}}=(2^{4})^{\frac{3}{4}}=2^{3} = 8), then (\frac{32}{3}\times8-0=\frac{256}{3}).

Answer:

(\frac{256}{3})