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find the average of y=x2y=x^2 on [0;6][0;6]

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Q. find the average of y=x2y=x^2 on [0;6][0;6]
  1. Average Value Formula: To find the average value of a function on an interval [a,b][a, b], we use the formula: Average=1(ba)×abf(x)dx\text{Average} = \frac{1}{(b-a)} \times \int_{a}^{b} f(x) \, dx. Here, f(x)=x2f(x) = x^2, a=0a = 0, and b=6b = 6.
  2. Calculate Definite Integral: First, calculate the definite integral of f(x)=x2f(x) = x^2 from 00 to 66.06x2dx=[x33]06\int_{0}^{6} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{6}.
  3. Subtract Limits: Plug in the upper limit of the integral and then the lower limit, and subtract the two.\newline[63/3][03/3][6^3/3] - [0^3/3] = (216/3)(0/3)(216/3) - (0/3) = 72072 - 0 = 7272.
  4. Divide by Interval Length: Now, divide the result of the integral by the length of the interval, which is 60=66 - 0 = 6.\newlineAverage = (1/6)72=12(1/6) * 72 = 12.

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