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What is the average value of 3x^(2)+4x on the interval 2 <= x <= 6 ?

What is the average value of 3x2+4x 3 x^{2}+4 x on the interval 2≤x≤6 2 \leq x \leq 6 ?

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Full solution

Q. What is the average value of 3x2+4x 3 x^{2}+4 x on the interval 2≤x≤6 2 \leq x \leq 6 ?
  1. Average Value Formula: To find the average value of a function f(x)f(x) over the interval [a,b][a, b], we use the formula:\newlineAverage value = 1(b−a)∫abf(x) dx\frac{1}{(b-a)} \int_{a}^{b} f(x) \, dx\newlineHere, f(x)=3x2+4xf(x) = 3x^2 + 4x, a=2a = 2, and b=6b = 6.
  2. Definite Integral Calculation: First, we need to find the definite integral of f(x)f(x) from 22 to 66.∫26(3x2+4x) dx\int_{2}^{6} (3x^2 + 4x) \, dxThis requires us to integrate the function term by term.
  3. Term by Term Integration: The integral of 3x23x^2 with respect to xx is x3x^3, and the integral of 4x4x with respect to xx is 2x22x^2. So, ∫\int of (3x2+4x) dx=x3+2x2(3x^2 + 4x) \, dx = x^3 + 2x^2.
  4. Evaluation at Limits: Now we evaluate the antiderivative at the upper and lower limits of the interval and subtract: 63+2⋅626^3 + 2\cdot 6^2 - 23+2⋅222^3 + 2\cdot 2^2
  5. Upper Limit Calculation: Calculating the values:\newline(63+2×62)=(216+2×36)=(216+72)=288(6^3 + 2\times6^2) = (216 + 2\times36) = (216 + 72) = 288\newline(23+2×22)=(8+2×4)=(8+8)=16(2^3 + 2\times2^2) = (8 + 2\times4) = (8 + 8) = 16
  6. Subtract Lower Limit: Subtract the lower limit evaluation from the upper limit evaluation: 288−16=272288 - 16 = 272
  7. Average Value Calculation: Now, we divide this result by (b−a)(b - a) to find the average value:\newlineAverage value = 272(6−2)\frac{272}{(6 - 2)}
  8. Denominator Calculation: Calculating the denominator: 6−2=46 - 2 = 4
  9. Division Result: Dividing the result by the interval length:\newlineAverage value = 2724\frac{272}{4}
  10. Final Average Value: Simplifying the fraction:\newlineAverage value = 6868

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