Free HSC maths sample lesson

Picture the region under a curve sliced into many thin vertical strips, each one almost a rectangle. A strip sitting at position x has height f(x) (the value of the curve there) and a tiny width, so its area is roughly height times width. The area under the whole curve is the accumulated total of all these strips. Hold onto that picture: area under a curve is a sum of thin rectangular strips, and even if you forget every formula you can rebuild the idea from it.

Try it with numbers before any integral. To estimate the area under y=x^2+1 from x=0 to x=2, cover the region with two rectangles of width 1. Using the left edge of each strip the heights are f(0)=1 and f(1)=2, giving an estimate of 1+2=3 square units; using the right edge the heights are f(1)=2 and f(2)=5, giving 2+5=7. The true area lies between these, and the two estimates disagree only because the rectangles are too wide.

Now make the strips thinner. Each strip has area f(x)\,dx, where dx is its width, and the total area is the sum of all these f(x)\,dx pieces across the interval. As the strips get thinner the staircase of rectangles hugs the curve more closely, and in the limit the sum becomes exact. That limiting sum is precisely what the definite integral \int_a^b f(x)\,dx means: the accumulated area of infinitely many infinitely thin strips, and it uses the same height-times-width idea as the area of a single rectangle.

We do not add the strips one at a time. Because f(x) is the rate at which area builds up as x increases, an antiderivative F (a function whose rate is f) tracks the running total of area. Evaluating F(b)-F(a) then gives the area accumulated from x=a to x=b: the running total at the end minus the running total at the start. This is why finding an antiderivative and substituting the bounds computes the area, and it is differentiation run in reverse.

One warning about sign. Where the curve dips below the x-axis the strips there have negative height f(x), so their f(x)\,dx contributions are negative and that part of the integral subtracts instead of adds. The raw integral therefore reports signed area, which can be smaller than the true geometric area or even negative. Geometric area is never negative, so for a below-axis region take the absolute value, and if the curve crosses the axis inside the interval, split at the crossing as in the signed area lesson before adding the pieces.