Integrals Beyond Area
The same slicing logic measures physical work, typical values, and the length of a curve.
Work Done
Work is force times distance, but when force varies with position you slice the path instead. Each small stretch contributes force times dx, and the integral sums them. Pumping water out of a tank uses the same idea, with force replaced by the weight of each thin layer of water.
W = \int_a^b F(x)\,dxAverage Value
The average of a function over an interval is its integral divided by the interval's width. It is the constant height with the same total accumulation, so it answers what the function is typically doing. Stretching a spring from 0 to a meters by Hooke's law averages the force at half its final value, which is why the work is k a squared over 2.
f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dxArc Length
Approximate a curve by tiny straight segments, each with length the square root of one plus the slope squared times dx. The integral of that expression is the exact arc length. For y = x to the two thirds from 0 to 1, the antiderivative works out to 27 over 40 plus the square root of 9 over 4, a finite and tidy length.
L = \int_a^b \sqrt{1 + \bigl(f'(x)\bigr)^2}\,dxFlashcard
Work under a variable force
Check your understanding
A spring follows Hooke's law with force kx. Why does stretching it a meters require k a squared over 2 of work rather than k a squared?