Question

Question 11 or 12 whichever u can answer see photo

Answer 1

$-\frac{216}{5115}$

Explanation

(11)

$f(t)=\frac{t^2}{t^3+4}$

the average value of a function is given by:

$average=\frac{F(b)-f(a)}{b-a}$

Step 1

evaluate the function in the given limits

a)t=3

$\begin{gathered} f(t)=\frac{t^2}{t^3+4} \\ f(3)=\frac{3^2}{3^3+4} \\ f(3)=\frac{9}{27+4}=\frac{9}{31} \\ f(3)=\frac{9}{31} \\ so \\ a=3,\text{ f\lparen a\rparen=}\frac{9}{31} \end{gathered}$

b) t= 6

$\begin{gathered} f(t)=\frac{t^2}{t^3+4} \\ f(6)=\frac{6^2}{6^3+4} \\ f(6)=\frac{36}{216+4}=\frac{36}{220}=\frac{18}{110}=\frac{9}{55} \\ f(6)=\frac{9}{55} \\ b=6\text{ and F\lparen b\rparen=}\frac{9}{55} \end{gathered}$

Step 2

now, replace in the formula

$\begin{gathered} average=\frac{F(b)-f(a)}{b-a} \\ average=\frac{\frac{9}{55}-\frac{9}{31}}{6-3} \\ average=\frac{\frac{9}{55}-\frac{9}{31}}{3}=\frac{-\frac{216}{1705}}{\frac{3}{1}}=-\frac{216}{5115} \end{gathered}$

.so the answer is

$-\frac{216}{5115}$

I hope this helps you