Accumulation and the fundamental theorem

What this concept is

A definite integral is the limit of sums of value times interval width. It represents signed accumulation. If f is continuous, A(x)=∫ from a to x of f(t)dt satisfies A′(x)=f(x). If F′=f on the interval, ∫ from a to b of f equals F(b)−F(a).

Why it matters here

This connects measured rates to totals and later connects density to probability. A rate in bytes/second integrated over seconds gives bytes.

Concrete example

For f(t)=2t on [0,3], an antiderivative is t² and the integral is 9. With n right rectangles, Δt=3/n and the sum is 18/n²·n(n+1)/2=9(1+1/n), tending to 9. The finite rectangle sum overestimates this increasing function. For f(t)=t on [−1,1], signed area is zero although total geometric area is 1.

Common confusion / misconception

An indefinite integral is a family F+C; a definite integral is a number with units. Area cancellation is legitimate for net flow but not for total distance.

How to use it

Draw interval and sign. Check continuity before the elementary theorem, obtain an antiderivative, then subtract endpoint values in order. Reverse limits to negate the result. A variable upper bound g(x) adds a chain factor: d/dx ∫ from 0 to x² of exp(t)dt = exp(x²)·2x.

Check yourself

If inflow minus outflow is −2 litres/minute for 3 minutes, what is net change? Can the remaining volume be inferred without initial volume?

Mini drill or application

Integrate 3t²−2 on [0,2] and verify by differentiating your antiderivative. Answer: 4. State what extra operation would give total absolute accumulation instead.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.