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Average value of a function on an interval.

Created by @nathanedwards
 at November 1st 2023, 2:40:53 am.
AP_Calculus_AB-Questions
#average-value
#definite-integral
#ap-calculus-ab

AP Calculus AB Exam Question

Consider the function f(x) defined by:

f(x)=2x3−5x2+3x+4 f(x) = 2x^3 - 5x^2 + 3x + 4

over the interval [0,2].

(a) Find the average value of f(x) over the interval [0,2].
(b) Find the x-coordinate(s) at which the average value of f(x) occurs over the interval [0,2].

Answer

(a) The average value of a function f(x) over an interval [a,b] is given by the formula:

Average value of f(x)=1b−a∫abf(x) dx \text{{Average value of }} f(x) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx

For our function f(x) and the interval [0,2], we have:

Average value of f(x)=12−0∫02(2x3−5x2+3x+4) dx \text{{Average value of }} f(x) = \frac{1}{2-0} \int_{0}^{2} (2x^3 - 5x^2 + 3x + 4) \, dx

Integrating term by term, we get:

12[24x4−53x3+32x2+4x]02 \frac{1}{2} \left[ \frac{2}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 4x \right]_0^2

Simplifying and substituting the upper and lower limits, we have:

12[24(24)−53(23)+32(22)+4(2)]−12[24(04)−53(03)+32(02)+4(0)] \frac{1}{2} \left[ \frac{2}{4}(2^4) - \frac{5}{3}(2^3) + \frac{3}{2}(2^2) + 4(2) \right] - \frac{1}{2} \left[ \frac{2}{4}(0^4) - \frac{5}{3}(0^3) + \frac{3}{2}(0^2) + 4(0) \right]
12[24(16)−53(8)+32(4)+8]−12[24(0)−53(0)+32(0)+0] \frac{1}{2} \left[ \frac{2}{4}(16) - \frac{5}{3}(8) + \frac{3}{2}(4) + 8 \right] - \frac{1}{2} \left[ \frac{2}{4}(0) - \frac{5}{3}(0) + \frac{3}{2}(0) + 0 \right]
12[162−403+6+8]−12[0−0+0+0] \frac{1}{2} \left[ \frac{16}{2} - \frac{40}{3} + 6 + 8 \right] - \frac{1}{2} \left[ 0 - 0 + 0 + 0 \right]
12[162−403+6+8] \frac{1}{2} \left[ \frac{16}{2} - \frac{40}{3} + 6 + 8 \right]
12[8−403+6+8] \frac{1}{2} \left[ 8 - \frac{40}{3} + 6 + 8 \right]
12[243−403+18+24] \frac{1}{2} \left[ \frac{24}{3} - \frac{40}{3} + 18 + 24 \right]
12[24−40+54+723] \frac{1}{2} \left[ \frac{24 - 40 + 54 + 72}{3} \right]
12[1103] \frac{1}{2} \left[ \frac{110}{3} \right]
1106 \frac{110}{6}
Average value of f(x)=553 \text{{Average value of }} f(x) = \frac{55}{3}

Hence, the average value of f(x) over the interval [0,2] is 553\frac{55}{3}.

(b) To find the x-coordinate(s) at which the average value of f(x) occurs over the interval [0,2], we set up the equation:

12−0∫02(2x3−5x2+3x+4) dx=553 \frac{1}{2-0} \int_{0}^{2} (2x^3 - 5x^2 + 3x + 4) \, dx = \frac{55}{3}
12[24x4−53x3+32x2+4x]02=553 \frac{1}{2} \left[ \frac{2}{4}x^4 - \frac{5}{3}x^3 + \frac{3}{2}x^2 + 4x \right]_0^2 = \frac{55}{3}

Applying the Fundamental Theorem of Calculus, we have:

12[24(24)−53(23)+32(22)+4(2)]−12[24(04)−53(03)+32(02)+4(0)]=553 \frac{1}{2} \left[ \frac{2}{4}(2^4) - \frac{5}{3}(2^3) + \frac{3}{2}(2^2) + 4(2) \right] - \frac{1}{2} \left[ \frac{2}{4}(0^4) - \frac{5}{3}(0^3) + \frac{3}{2}(0^2) + 4(0) \right] = \frac{55}{3}

Simplifying and evaluating the integrals, we have:

12[24(16)−53(8)+32(4)+8]−12[24(0)−53(0)+32(0)+0]=553 \frac{1}{2} \left[ \frac{2}{4}(16) - \frac{5}{3}(8) + \frac{3}{2}(4) + 8 \right] - \frac{1}{2} \left[ \frac{2}{4}(0) - \frac{5}{3}(0) + \frac{3}{2}(0) + 0 \right] = \frac{55}{3}
12[162−403+6+8]−12[0−0+0+0]=553 \frac{1}{2} \left[ \frac{16}{2} - \frac{40}{3} + 6 + 8 \right] - \frac{1}{2} \left[ 0 - 0 + 0 + 0 \right] = \frac{55}{3}
12[243−403+18+24]=553 \frac{1}{2} \left[ \frac{24}{3} - \frac{40}{3} + 18 + 24 \right] = \frac{55}{3}
12[24−40+54+723]=553 \frac{1}{2} \left[ \frac{24 - 40 + 54 + 72}{3} \right] = \frac{55}{3}
12[1103]=553 \frac{1}{2} \left[ \frac{110}{3} \right] = \frac{55}{3}
1106=553 \frac{110}{6} = \frac{55}{3}

Therefore, the x-coordinate(s) at which the average value of f(x) occurs over the interval [0,2] is 2.

In summary, the average value of the function f(x) over the interval [0,2] is 553\frac{55}{3}, and it occurs at the x-coordinate 2.

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