Applications of Integration

Average Value of a Function

Find the average value of f(x) = x² on [0, 3]: f_avg = (1/3) ∫_0^3 x² dx = 9/3 = 3.

Example

Divide the definite integral by the interval length to get the average value.

highlighted = computed this step

Step 1 — Set up

Set up 1 over 3 times the integral from 0 to 3.

favg=13∫03x2 dxf_{\text{avg}}=\frac{ 1 }{ 3 }\int_{ 0 }^{ 3 } x^{2} \,dx

Step 2 — Antiderivative

Find an antiderivative.

F(x)=x33F(x)= \hlmath{\frac{x^{3}}{3}}

Step 3 — Integral value

Use FTC on 0 to 3 to get integral value 9.

∫03x2 dx=9\int_{ 0 }^{ 3 } x^{2} \,dx= \hl{9}

Step 4 — Interval length

Compute the interval length 3 minus 0 equals 3.

3−0=33 - 0 = \hl{3}

Step 5 — Average value

Divide 9 by 3 to get average value 3.

9÷3=39 \div 3 = \hl{3}
average-value The average value of f on [a, b] is f_avg = (1/(b-a)) ∫_a^b f(x) dx = (F(b) - F(a)) / (b - a).