Analyzing the Setup
The Greatest Integer Function, denoted as [x], behaves like a staircase. It remains constant at an integer k for all x in the interval [k,k+1).
When evaluating the integral ∫1a[x]f′(x)dx, the discontinuity of [x] prevents direct integration. We must respect its piecewise nature to solve the problem effectively.
Breaking the Barrier
Let n=[a] be the largest integer less than or equal to a. We split the integral from 1 to a into discrete, manageable intervals: [1,2),[2,3),…,[n,a].
This transforms the integral into a sum of smaller, friendly components:
∫1a[x]f′(x)dx=∫121⋅f′(x)dx+∫232⋅f′(x)dx+⋯+∫nan⋅f′(x)dx
The Mechanics of Integration
In each sub-interval, [x] is a constant. We can pull these constants out of the integral signs. Applying the Fundamental Theorem of Calculus, where the integral of f′(x) is f(x), we obtain:
1[f(2)−f(1)]+2[f(3)−f(2)]+3[f(4)−f(3)]+⋯+n[f(a)−f(n)]
This step successfully converts a calculus problem into an algebraic one.
The Telescoping Magic
Expanding the brackets, we get:
f(2)−f(1)+2f(3)−2f(2)+3f(4)−3f(3)+⋯+nf(a)−nf(n)
By grouping terms with the same function value f(k), we observe a pattern. For any integer k where 1<k≤n, the term f(k) appears with a coefficient of (k−1)−k=−1.
The expression simplifies significantly as the intermediate terms cancel out:
nf(a)−f(1)−f(2)−f(3)−⋯−f(n)
Final Result
Factoring out the negative sign, we arrive at the final expression. Substituting n=[a] back into the equation, we obtain the elegant result:
By breaking the problem down, we have uncovered the underlying structure. Whenever you encounter a step function, remember to partition the domain, integrate the pieces, and allow the telescoping sum to reveal the solution.