c1 - average value of a function: problem 3\n(1 point)\na car drives down a road in such a way that its…

c1 - average value of a function: problem 3\n(1 point)\na car drives down a road in such a way that its velocity (in m/s) at time ( t ) (seconds) is\n v(t)=2 t^{1 / 2}+5 \nfind the cars average velocity (in m/s) between ( t = 4 ) and ( t = 8 ).\nanswer =\nnote: you are in the reduced scoring period. all work counts for 50% of the original.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

c1 - average value of a function: problem 3\n(1 point)\na car drives down a road in such a way that its velocity (in m/s) at time ( t ) (seconds) is\n v(t)=2 t^{1 / 2}+5 \nfind the cars average velocity (in m/s) between ( t = 4 ) and ( t = 8 ).\nanswer =\nnote: you are in the reduced scoring period. all work counts for 50% of the original.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

Answer

Answer:

$\frac{1}{8 - 4}\int_{4}^{8}(2t^{\frac{1}{2}} + 5)dt=\frac{1}{4}\left(\frac{4}{3}t^{\frac{3}{2}}+5t\right)\big|_{4}^{8}=\frac{1}{4}\left[\left(\frac{4}{3}(8)^{\frac{3}{2}}+5\times8\right)-\left(\frac{4}{3}(4)^{\frac{3}{2}}+5\times4\right)\right]=\frac{1}{4}\left[\left(\frac{4}{3}\times16\sqrt{2}+40\right)-\left(\frac{4}{3}\times8 + 20\right)\right]=\frac{1}{4}\left(\frac{64\sqrt{2}}{3}+40-\frac{32}{3}-20\right)=\frac{1}{4}\left(\frac{64\sqrt{2}- 32}{3}+20\right)=\frac{16\sqrt{2}-8}{3}+5=\frac{16\sqrt{2}-8 + 15}{3}=\frac{16\sqrt{2}+7}{3}\approx\frac{16\times1.414+7}{3}=\frac{22.624 + 7}{3}=\frac{29.624}{3}\approx9.87$

Explanation:

Step1: Recall the formula for the average value of a function

The formula for the average value of a function (y = v(t)) over the interval ([a,b]) is (v_{avg}=\frac{1}{b - a}\int_{a}^{b}v(t)dt). Here, (a = 4), (b = 8) and (v(t)=2t^{\frac{1}{2}}+5).

Step2: Integrate the function

Use the power - rule for integration (\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C(n\neq - 1)). (\int(2t^{\frac{1}{2}}+5)dt=2\int t^{\frac{1}{2}}dt+5\int dt). (2\times\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+5t+C=\frac{4}{3}t^{\frac{3}{2}}+5t+C).

Step3: Evaluate the definite integral

(\int_{4}^{8}(2t^{\frac{1}{2}}+5)dt=\left[\frac{4}{3}t^{\frac{3}{2}}+5t\right]_{4}^{8}). (=\left(\frac{4}{3}(8)^{\frac{3}{2}}+5\times8\right)-\left(\frac{4}{3}(4)^{\frac{3}{2}}+5\times4\right)). Since (8^{\frac{3}{2}}=(2^{3})^{\frac{3}{2}}=2^{4}\sqrt{2}=16\sqrt{2}) and (4^{\frac{3}{2}}=(2^{2})^{\frac{3}{2}}=8). (=\left(\frac{64\sqrt{2}}{3}+40\right)-\left(\frac{32}{3}+20\right)=\frac{64\sqrt{2}-32}{3}+20).

Step4: Calculate the average value

(v_{avg}=\frac{1}{8 - 4}\int_{4}^{8}(2t^{\frac{1}{2}}+5)dt=\frac{1}{4}\left(\frac{64\sqrt{2}-32}{3}+20\right)). (=\frac{16\sqrt{2}-8}{3}+5=\frac{16\sqrt{2}+7}{3}\approx9.87).