Click on the graph to increase/decrease the intervels. Press on the mouse button near the origin in the graph area and the tangents will be extended.
Applet shows the relation between integration and differentiation for a definite integral.
Imagine the graph represents the position of a body moving along the x- axis with varying velocity. If you wanted the displacement between 1 s and 3 s, you could note the values of x coordinate (represented on the y axis) at these times and the difference would be the sought displacement. But let us get it by a different method.
Draw a tangent at t=1 to the curve. The slope of this would be the velocity at the moment. Applet shows this tangent when the number of intervals is set to 1. The displacement of the body treating the velocity as constant at this value for the entire duration of 2 s is the length of the blue line. The red line indicates the error in the evaluation. That is the difference between the values x(3) - x(1) is the sum of the lengths of the red and blue segments. The displacement so evaluated can be represented as
= v(1) * 2;
where v(1) is the velocity at t=1 s, got by drawing a tangent at x=1 to the curve, the the 2 s duration is represented by the length of the green line and is the length of the blue line. The ratio of the blue line's length to the green line's length is the slope of the black tangent line and is equal to v(1).
If the number of intervals is increased to 2, you would draw tangents at t=1 and t=2 and treat the velocities for the intervals as constant and equal to the value at the begining of the interval for the one second duration, you would get the displacement as sum of two terms. The two blue segments represent these terms. It can be seen that the sum of the two red segments, which is represents the error has become smaller. The displacement so evaluated is
= v(1) * 1+v(2)*1, where v(1) and v(2) are the velocities at t=1 and t=2 seconds.
Increasing the number of intervals would decrease the error, and for a sufficiently large number of intervals the error would become inconsequential and can be taken as zero.
= v(1) * 0.5+v(1.5)*0.5 + v(2)*0.5+v(2.5)*0.5; for four intevals
= v(1) * 0.25+v(1.25)*0.25 + ......+v(2.75)*0.25; for eight intevals
=v(t)*, where can be so chosen to obtain to the desired degree of accuracy and v(t) is the velocity at the begining of the different intervals of time, got by finding slopes of x-t graph. Thus v(t) is nothing but the derivative of x(t).
If the intervals are made very small this evaluation is and this is equal to x(3)-x(1);
What this would mean is if you wanted to evaluate you would find a function x(t) whoose derivative is v(t) and its value at the upper limit minus its value at the lower limit would give us the value of the integral. Generalising this