Average Value Calculator
Free average value calculator for finding the mean value of functions over intervals using definite integrals. Get step-by-step solutions with integration techniques and geometric interpretations. Perfect for calculus students learning integral applications.
Last updated: December 15, 2024
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Use x² for squared, x³ for cubed, etc.
Average Value Result
Average Value:
3.000000
Definite Integral:
9.000000
Interval Length:
3.000000
Sample Points:
(0, 0)
(0.43, 0.18)
(0.86, 0.73)
(1.29, 1.65)
(1.71, 2.94)
(2.14, 4.59)
(2.57, 6.61)
(3, 9)
Step-by-Step Solution:
Average Value Tips:
- • Formula: Average = (1/(b-a)) × ∫ₐᵇ f(x) dx
- • Represents the "height" of a rectangle with same area as the integral
- • Can be visualized as a horizontal line cutting the area in half
- • Useful for finding mean values in physics and statistics
- • Always positive for positive functions on positive intervals
Average Value Formula & Methods
Formula
(1/(b-a)) × ∫ₐᵇ f(x) dx
Definite integral divided by interval length
Interpretation
Rectangle height = area ÷ width
Height of rectangle with same area as curve
Theorem
f(c) = average value
Function attains average value at some point c
Example
f(x) = x² → ∫x²dx = x³/3
Use power rule for integration
Example
f(x) = sin(x) → ∫sin(x)dx = -cos(x)
Use trigonometric integral formulas
Example
f(x) = eˣ → ∫eˣdx = eˣ
Exponential functions integrate to themselves
Quick Example Result
Average value of f(x) = x² on [0, 3]:
Average Value
3
Definite Integral
9
Interval Length
3
How to Find Average Value
The average value of a function represents the mean height of the function over a specified interval. This concept is fundamental in calculus applications and connects to the Mean Value Theorem for Integrals, providing insights into function behavior and area calculations.
The Average Value Process
This systematic approach ensures accurate calculation of mean values for any integrable function.
Geometric Interpretation
Geometrically, the average value represents the height of a rectangle with width (b-a) that has the same area as the area under the curve y = f(x) from x = a to x = b. This rectangle has area equal to the definite integral, and its height is the average value. It's like finding the "average height" of the function.
- Area under curve = ∫ₐᵇ f(x) dx
- Rectangle area = height × width = average value × (b-a)
- Setting them equal: average value × (b-a) = ∫ₐᵇ f(x) dx
- Therefore: average value = (1/(b-a)) × ∫ₐᵇ f(x) dx
Sources & References
- Calculus: Early Transcendentals - James Stewart (9th Edition)Comprehensive coverage of average value and integral applications
- Thomas' Calculus - Weir, Hass, Giordano (14th Edition)Detailed explanations of Mean Value Theorem and applications
- Khan Academy - Average Value of FunctionsVideo tutorials and practice problems on average value calculations
Need help with other calculus topics? Check out our derivative calculator and area between curves calculator.
Get Custom Calculator for Your PlatformAverage Value Example
Given Function:
Solution Steps:
- Step 1: Identify the function and interval
- f(x) = x², [0, 3]
- Step 2: Calculate the definite integral
- ∫ₐᵇ x² dx = [x³/3]ₐᵇ
- = (3³/3) - (0³/3)
- = (27/3) - (0/3)
- = 9 - 0
- = 9
- Step 3: Find the interval length
- b - a = 3 - 0 = 3
- Step 4: Apply the average value formula
- Average Value = (1/(b-a)) × ∫ₐᵇ f(x) dx
- = (1/3) × 9
- = 3
Final Answer:
Average Value
3
Definite Integral
9
Interval Length
3
Linear Function
f(x) = x on [1, 4]
Average = 2.5
Cubic Function
f(x) = x³ on [0, 2]
Average = 2
Frequently Asked Questions
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