Compute the average value of any function over an interval using definite integration. Adjustable precision, symbolic results + numeric approximation with interactive graph.
The average value of a function \(f(x)\) over the interval \([a,b]\) is defined by the definite integral. It generalizes the concept of an average of a finite set of numbers to a continuum.
Mean Value Theorem for Integrals:
If f is continuous on [a,b], there exists at least one number c in [a,b] such that:
\( f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \)
i.e., the function attains its average value at some point in the interval.
| Function f(x) | Indefinite Integral ∫ f(x) dx | Example |
|---|---|---|
| \(x^n\) (n ≠ -1) | \(\frac{x^{n+1}}{n+1} + C\) | ∫ x² dx = x³/3 |
| \(1/x\) | \(\ln|x| + C\) | ∫ 1/x dx = ln|x| |
| \(e^x\) | \(e^x + C\) | ∫ eˣ dx = eˣ |
| \(\sin x\) | \(-\cos x + C\) | ∫ sin x dx = -cos x |
| \(\cos x\) | \(\sin x + C\) | ∫ cos x dx = sin x |
| \(\sec^2 x\) | \(\tan x + C\) | ∫ sec² dx = tan x |
| \(\frac{1}{\sqrt{1-x^2}}\) | \(\arcsin x + C\) | |
| \(\frac{1}{1+x^2}\) | \(\arctan x + C\) |
Area equivalence: The area under the curve over [a,b] equals the area of a rectangle of width (b-a) and height equal to the average value.
Mean Value Theorem: For continuous functions, the horizontal line at the average value will always intersect the curve at least once.
Connection to averages: If you sample the function at infinitely many points, the arithmetic mean converges to the average value.
Calculator Features:
x^2
avg on [0,2] = 4/3
sin(x)
avg on [0,π] = 2/π ≈ 0.6366
exp(x)
avg on [0,1] = e-1 ≈ 1.718
1/x
avg on [1,e] = 1/(e-1)
sqrt(x)
avg on [1,4] = 14/9 ≈ 1.5556