What is the average value of f on [0,2] for f(x) = x^2?

What is the average value of f on [0,2] for f(x) = x^2?

• A. 4/3
• B. 2
• C. 1
• D. 8/3

Answer: (A)

The average value of a function on an interval is the height that would give the same total when spread evenly over that interval. Mathematically, on [a,b] it’s (1/(b−a)) ∫_a^b f(x) dx. For f(x) = x^2 on [0,2], compute the integral: ∫_0^2 x^2 dx = x^3/3 from 0 to 2 = 8/3. The interval length is 2, so the average value is (1/2) × (8/3) = 4/3. So the average value is 4/3. Intuitively, the area under the curve from 0 to 2 is 8/3, and spreading that area over a width of 2 gives a constant height of 4/3, which also lies between f(0)=0 and f(2)=4 as expected.

The average value of a function on an interval is the height that would give the same total when spread evenly over that interval. Mathematically, on [a,b] it’s (1/(b−a)) ∫_a^b f(x) dx.

For f(x) = x^2 on [0,2], compute the integral: ∫_0^2 x^2 dx = x^3/3 from 0 to 2 = 8/3. The interval length is 2, so the average value is (1/2) × (8/3) = 4/3.

So the average value is 4/3. Intuitively, the area under the curve from 0 to 2 is 8/3, and spreading that area over a width of 2 gives a constant height of 4/3, which also lies between f(0)=0 and f(2)=4 as expected.