If you have a triangular park and want to plant grass in that. If cost of grass is $10 per sq meter, you need to estimate the cost for your park, which is let us pretend a right angle triangle, with sides as 4meters, and hypotenuse as 4*sqrt(2) = 5.656meters.
If you know the area you can estimate the cost. You are told that area is (1/2) of 4*4 = 8 sq meter and cost is $80.. If you believe what you are told, you will go ahead and find a suitable contractor. If you do not believe the statement, that area is 8 sq meter, you could plan to roughly verify that. You can subdivide the park into many "segments", and find the area of each. If the segments are 100, then each segment will be 4cm wide, and will be having a length that increases from almost zero on one side to full 4 meters on other end. The fifty-th segment will have a length of 2 meters. So the calculation will go as follows. The first will have a length of about 4cms, second about 8 cms, third 12cms, 25th will have a length of 25*4 cms = 1 meter etc. Each of this has an area of 4cms*length. The first will have 16sqcm, second will be 32 sq cm, 3rd will be 12*4 = 48 sq cm and so on. So you sum up all these!! The last will be 400*4 = 1600 sq cm. The total area is in sq cm, 16+32+48+64+....+1584+1600 = 16*(1+2+3+...+100) = 16*101*(100/2) = 80800 sq cm or 8.08 sq cm.
If you forget about the 8 sqcm difference, you now have confirmation that the area you were "told" is very nearly true.
What you have done is integral of 0 to 100, and got it as 100^2/2 = 5000 and then multipled by 16 to get 80000sq cm = 8 sq meter. Integration is all about summation. You need to place in the integral an appropriate equation that represents the function you have. In the case of a triangle of the type seen, it is length which increases linearly from 0 to 4 meters. It will be integral of t.dt. This integral evalues to t^2/2. The limits are 0 and 100. So intergal evaluated from 0 to 100 results in (100^2)/2
You can repeat with segments each of them 2 cm wide. You will have 200 segments, and area will be (2^2)*200^2/2 = 80000. Intergal evaluated from 0 to 200 = (200^2)/2
You can now see what happens if say the hypotenuse which is a straight line is replaced by an arc of a circle.(area of a segment of a circle, perhaps one quadrant.). For one quadrant, you should find that area calculated thus is (pi.4^2)/4 , one fourth of area of circle, which is pi.r^2.