Find the number(s) a such that the average value of the function $f(x)=50-28 x+3 x^{2}$ on the interval $[0, a]$ is equal to 10.
A) $a=10$
B) $a=-4$
C) $a=-10$
D) $a=4$
E) $a=-10$.

Question

Find the number(s) a such that the average value of the function $f(x)=50-28 x+3 x^{2}$ on the interval $[0, a]$ is equal to 10. 
A) $a=10$ 
B) $a=-4$ 
C) $a=-10$ 
D) $a=4$ 
E) $a=-10$.

Quizplus · Accepted Answer

The average value of a function $f(x)$ on the interval $[a, b]$ is given by $$\frac{1}{b-a}\int_a^b f(x) \ dx$$So, the average value of $f(x)=50-28 x+3 x^{2}$ on the interval $[0, a]$ is$$\frac{1}{a-0}\int_0^a (50-28 x+3 x^{2}) \ dx = \frac{1}{a}\left[50x-14 x^2+x^3\right]_0^a = \frac{50a-14 a^2+a^3}{a} = a^2-14 a+50$$Setting this equal to 10, we get$$a^2-14 a+50 = 10$$$$a^2-14 a+40 = 0$$$$(a-10)(a-4) = 0$$So, $a=10$ or $a=4$.

Find the Number(s) a Such That the Average Value of the Function