Question

homework5: problem 33
(1 point) find the average value of ( f(x) = x^2 ) on the interval 1, 6.
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Explanation:

Step1: Recall the average value formula

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by \( \frac{1}{b - a} \int_{a}^{b} f(x) dx \). Here, \( a = 1 \), \( b = 6 \), and \( f(x)=x^{2} \).

Step2: Compute the integral of \( f(x) \)

First, find the antiderivative of \( x^{2} \). The antiderivative of \( x^{n} \) is \( \frac{x^{n + 1}}{n+1} \), so for \( n = 2 \), the antiderivative \( F(x)=\frac{x^{3}}{3} \).
Then, evaluate the definite integral \( \int_{1}^{6} x^{2} dx=F(6)-F(1)=\frac{6^{3}}{3}-\frac{1^{3}}{3}=\frac{216}{3}-\frac{1}{3}=\frac{215}{3} \).

Step3: Compute the average value

Using the average value formula, \( \frac{1}{6 - 1}\times\frac{215}{3}=\frac{1}{5}\times\frac{215}{3}=\frac{43}{3}\approx14.333\cdots \)

Answer:

\(\frac{43}{3}\) (or approximately \(14.33\) if a decimal approximation is preferred)