Let f(x)=2x, then, if we take the integral for 0≤x≤10, the area under the curve is in fact the area of the right triangle with a base length of 10 units and a height of 20, making its area 100 square units.
If we approximate the area under f(x) by slicing it into thin rectangles of width h units, then each rectangle will have an area hf(x). ∑hf(x) will approximate to the actual area 100. If that approximation is not to exceed 5% of 100, then 95≤∑hf(x)≤105.
The left rectangular areas will be hf(h)+hf(2h)+hf(3h)+hf(4h)+... The sum will be greater than the actual area. If we'd taken the right rectangles, the sum would be less than the actual area. f(x) is "trapped" by the rectangular shapes.
When f(x)=2x, this sum becomes h(2h)+h(4h)+h(6h)+h(8h)+..., that is:
2h2(1+2+3+4+...).
But what is the last term n in the summation? n=10/h, so the sum is:
2h2(n(n+1)/2)=nh2(n+1).
So 95≤nh2(n+1)≤105, 95≤10h(10/h+1)≤105, 95≤100+10h≤105,
-5≤10h≤5, -0.5≤h≤0.5. Since h>0 (the rectangle width), 0<h≤0.5.
Therefore, to be 95% accurate in the estimation of the area h cannot exceed 0.5 units in the specific case where f(x)=2x..