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Average Value of a Function on an Interval

AP Calculus AB· difficulty 2/5

Average value of f(x)=x2f(x) = x^2 on [0,3][0, 3] is

  • A

    33

    check_circle
  • B

    92\tfrac{9}{2}

  • C

    11

  • D

    33

Explanation

13∫03x2 dx=13⋅9=3\dfrac{1}{3}\int_0^3 x^2\,dx = \dfrac{1}{3} \cdot 9 = 3.

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