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irina1246 [14]
3 years ago
7

The average value of the function F (T) =(t-4)^2 on [0,9] is

Mathematics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

7

Step-by-step explanation:

Average value of a function F(t) on interval [a,b]

is \frac{1}{b-a} \int_a^b F(x)dx

So let's plug it in!

a=0

b=9

F(x)=(x-4)^2

To integrate (x-4)^2 just use power rule for integration.

So this what we get

\frac{1}{9-0} \frac{(x-4)^3}{3}|_0^9

\frac{1}{9-0} [\frac{(9-4)^3}{3}-\frac{(0-4)^3}{3}]

\frac{1}{9} [\frac{5^3}{3}-\frac{(-4)^3}{3}]

\frac{1}{9} \cdot \frac{125+64}{3}

\frac{1}{9} \cdot \frac{189}{3}

7

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