Question:
Let the function f(x) be defined as[ f(x) = \int_0^x (t^2 + 3t) dt.]
(a) Find the average value of f(x) over the interval [2, 5].
(b) Use the Mean Value Theorem for Integrals to find a value c in the interval [2, 5] such that f'(c) = average value of f(x) over [2, 5].
(c) Verify your answer to part (b) by calculating f'(x) and the average value of f(x) over [2, 5].
Answer:
(a) To find the average value of f(x) over the interval [2, 5], we first need to evaluate the definite integral and then divide the result by the length of the interval.
The integral of f(x) is given by:
f ( x ) = ∫ 0 x ( t 2 + 3 t ) d t = [ t 3 3 + 3 t 2 2 ] 0 x = x 3 3 + 3 x 2 2 − 0 = x 3 3 + 3 x 2 2 \begin{align*}
f(x) &= \int_0^x (t^2 + 3t) dt \\
& = \left[\frac{t^3}{3} + \frac{3t^2}{2}\right]_0^x \\
& = \frac{x^3}{3} + \frac{3x^2}{2} - 0 \\
& = \frac{x^3}{3} + \frac{3x^2}{2}
\end{align*} f ( x ) = ∫ 0 x ( t 2 + 3 t ) d t = [ 3 t 3 + 2 3 t 2 ] 0 x = 3 x 3 + 2 3 x 2 − 0 = 3 x 3 + 2 3 x 2 Now, we can calculate the average value by dividing the integral of f(x) over the interval [2, 5] by the length of the interval:
Average value of f ( x ) = 1 5 − 2 ∫ 2 5 ( t 2 + 3 t ) d t = 1 3 ∫ 2 5 ( t 2 + 3 t ) d t = 1 3 [ t 3 3 + 3 t 2 2 ] 2 5 = 1 3 ( 125 3 + 75 2 − ( 8 3 + 6 ) ) = 1 3 ( 221 6 ) = 221 18 \begin{align*}
\text{Average value of } f(x) &= \frac{1}{5 - 2} \int_2^5 (t^2 + 3t) dt \\
& = \frac{1}{3} \int_2^5 (t^2 + 3t) dt \\
& = \frac{1}{3} \left[\frac{t^3}{3} + \frac{3t^2}{2}\right]_2^5 \\
& = \frac{1}{3} \left(\frac{125}{3} + \frac{75}{2} - \left(\frac{8}{3} + 6\right)\right) \\
& = \frac{1}{3} \left(\frac{221}{6}\right) \\
& = \frac{221}{18}
\end{align*} Average value of f ( x ) = 5 − 2 1 ∫ 2 5 ( t 2 + 3 t ) d t = 3 1 ∫ 2 5 ( t 2 + 3 t ) d t = 3 1 [ 3 t 3 + 2 3 t 2 ] 2 5 = 3 1 ( 3 125 + 2 75 − ( 3 8 + 6 ) ) = 3 1 ( 6 221 ) = 18 221 Therefore, the average value of f(x) over the interval [2, 5] is 221 18 \frac{221}{18} 18 221
(b) According to the Mean Value Theorem for Integrals, there exists a value c in the interval [2, 5] such that:
f ′ ( c ) = average value of f ( x ) over [2, 5] . f'(c) = \text{average value of } f(x) \text{ over [2, 5]}. f ′ ( c ) = average value of f ( x ) over [2, 5] . To find c, we need to find f'(x) and set it equal to the average value obtained in part (a):
f ′ ( x ) = 221 18 f'(x) = \frac{221}{18} f ′ ( x ) = 18 221 Taking the derivative of f(x) with respect to x:
f ′ ( x ) = d d x ( x 3 3 + 3 x 2 2 ) f'(x) = \frac{d}{dx} \left(\frac{x^3}{3} + \frac{3x^2}{2}\right) f ′ ( x ) = d x d ( 3 x 3 + 2 3 x 2 ) f ′ ( x ) = x 2 + 3 x f'(x) = x^2 + 3x f ′ ( x ) = x 2 + 3 x Setting this equal to the average value:
x 2 + 3 x = 221 18 x^2 + 3x = \frac{221}{18} x 2 + 3 x = 18 221 We can solve this quadratic equation to find the value of x:
18 x 2 + 54 x − 221 = 0 18x^2 + 54x - 221 = 0 18 x 2 + 54 x − 221 = 0 We can either factor this quadratic or use the quadratic formula. Solving using the quadratic formula:
x = − 54 ± 5 4 2 − 4 ( 18 ) ( − 221 ) 2 ( 18 ) x = \frac{-54 \pm \sqrt{54^2 - 4(18)(-221)}}{2(18)} x = 2 ( 18 ) − 54 ± 5 4 2 − 4 ( 18 ) ( − 221 ) x = − 54 ± 174 36 x = \frac{-54 \pm 174}{36} x = 36 − 54 ± 174 We'll take the positive value for x:
x = − 54 + 174 36 x = \frac{-54 + 174}{36} x = 36 − 54 + 174 x = 120 36 x = \frac{120}{36} x = 36 120 x = 10 3 x = \frac{10}{3} x = 3 10 Therefore, according to the Mean Value Theorem for Integrals, there exists a value c = 10 3 c = \frac{10}{3} c = 3 10 f ′ ( c ) = 221 18 f'(c) = \frac{221}{18} f ′ ( c ) = 18 221
(c) We can verify our answer to part (b) by calculating f'(x) and the average value of f(x) over [2, 5].
Taking the derivative of f(x):
f ′ ( x ) = x 2 + 3 x f'(x) = x^2 + 3x f ′ ( x ) = x 2 + 3 x Now, let's calculate f'(c) at c = 10 3 c = \frac{10}{3} c = 3 10
f ′ ( 10 3 ) = ( 10 3 ) 2 + 3 ( 10 3 ) = 100 9 + 90 9 = 190 9 f'\left(\frac{10}{3}\right) = \left(\frac{10}{3}\right)^2 + 3\left(\frac{10}{3}\right) = \frac{100}{9} + \frac{90}{9} = \frac{190}{9} f ′ ( 3 10 ) = ( 3 10 ) 2 + 3 ( 3 10 ) = 9 100 + 9 90 = 9 190 Comparing this value with the average value obtained in part (a), f ′ ( c ) = 221 18 f'(c) = \frac{221}{18} f ′ ( c ) = 18 221
Therefore, our answer is verified.