find the total area bounded by the x - axis and the curve y = f(x) on the indicated interval. enter your…

Explanation:

Step1: Recall area - under - curve formula

The area (A) between the (x) - axis and the curve (y = f(x)) on the interval ([a,b]) is given by (A=\int_{a}^{b}|f(x)|dx). Since (f(x)=\frac{5}{\sqrt{8x + 8}}>0) for (x\in[4,7]), then (A=\int_{4}^{7}\frac{5}{\sqrt{8x + 8}}dx).

Step2: Use substitution

Let (u = 8x+8), then (du=8dx) and (dx=\frac{1}{8}du). When (x = 4), (u=8\times4 + 8=40); when (x = 7), (u=8\times7+8 = 64). The integral becomes (\int_{4}^{7}\frac{5}{\sqrt{8x + 8}}dx=\frac{5}{8}\int_{40}^{64}u^{-\frac{1}{2}}du).

Step3: Integrate using power rule

The power - rule for integration is (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). For (n=-\frac{1}{2}), (\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2u^{\frac{1}{2}}+C). So (\frac{5}{8}\int_{40}^{64}u^{-\frac{1}{2}}du=\frac{5}{8}\times2\left[u^{\frac{1}{2}}\right]_{40}^{64}).

Step4: Evaluate the definite integral

(\frac{5}{4}\left[u^{\frac{1}{2}}\right]_{40}^{64}=\frac{5}{4}(\sqrt{64}-\sqrt{40})=\frac{5}{4}(8 - 2\sqrt{10})=\frac{5(4-\sqrt{10})}{2}\approx\frac{5(4 - 3.162)}{2}=\frac{5\times0.838}{2}=2.095).

Answer:

(\frac{5(4-\sqrt{10})}{2}\approx2.095)