Integral Calculus, Adding and Scaling Integrals

Adding and Scaling Integrals

If f() is integrable over [a,b] and [b,c], then f is integrable over [a,c], and the integral is the sum of the two smaller integrals. Let q bound the function f over [a,c]. Given ε, find δ such that the Riemann sums over [a,b] and [b,c] are always within ε/3 of their respective limits. Also, restrict δ so that 2qδ < ε/3. Given any net over [a,c] of granularity δ, add the point b to the net. This changes the Riemann sum by less than ε/3. Now the Riemann sums over [a,b] and [b,c] are also within ε/3 of their limits. The entire Riemann sum is within ε of the sum of the two integrals, and that completes the proof.

This theorem is true even if a b and c are not in order. If c is less than b, then adding the integral over [b,c] subtracts the integral over [c,b], neatly cutting off the right end of the interval [a,b]. Verify this for all configurations of a b and c.

Note that this holds for indefinite integrals, e.g. [a,b] and [b,c). In fact c might be +∞. The theorem is true for the intervals [a,b] + [b,x] for all x, so it holds for the indefinite integral as x approaches c.

We can compute the integral of f+g as the integral of f plus the integral of g. This is a straightforward exercise in limits, and it can be extended to indefinite integrals. Similarly, the integral of a constant c times a function f is c times the integral of f. Combining these results, the integral of a linear combination of functions is the linear combination of the individual integrals.

Verify this theorem when the functions and the scalars are complex numbers, or vectors.