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Find the average value of f(x) = x^2 on [0, 3]

Worked through step by step. More calculus.

Answer
3\boxed{3}

By the Mean Value Theorem for Integrals, the function f(x)=x2f(x) = x^2 must attain this average value of 33 on the interval [0,3][0, 3], which happens when x=3≈1.732x = \sqrt{3} \approx 1.732.

Problem: Find the average value of f(x)=x2f(x) = x^2 on the interval [0,3][0, 3].

Steps

  1. Average value formula
    favg=13−0∫03x2 dxf_{\text{avg}} = \frac{1}{3 - 0} \int_{0}^{3} x^2 \, dx
  2. Simplify the fraction
    favg=13∫03x2 dxf_{\text{avg}} = \frac{1}{3} \int_{0}^{3} x^2 \, dx
  3. Integrate using the power rule
    favg=13[x33]03f_{\text{avg}} = \frac{1}{3} \left[ \frac{x^3}{3} \right]_{0}^{3}
  4. Evaluate at the upper and lower limits
    favg=13(333−033)f_{\text{avg}} = \frac{1}{3} \left( \frac{3^3}{3} - \frac{0^3}{3} \right)
  5. Simplify the expression
    favg=13(273)=13(9)=3f_{\text{avg}} = \frac{1}{3} \left( \frac{27}{3} \right) = \frac{1}{3} (9) = 3

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