Area under curve by integration method
Brief note on definite integrals
Definite integrals are used to find the area under the given limits ‘a’ and ‘b’ for a differential derivative function. Though many rules are there to determine definite integrals for every specific function, basic algebraic function is given.
For example, if ‘y’ as a function of ‘x’ or f(x) is given as,
where ‘a’ is lower limit, ‘b’ is upper limit and n ≠ –1.
Find the definite integral between the limits 1 to 4 for the function:
y = 2x2 + 3x.
On solving this, the value is 64.5.
Now applying this concept to find the area under a curve.
The given data (x and y) and the corresponding curve is given below:
x y
0.0 0.000
0.1 0.146
0.2 0.284
0.3 0.414
0.4 0.536
0.5 0.650
0.6 0.756
0.7 0.854
0.8 0.944
0.9 1.026
1.0 1.100
1.1 1.166
1.2 1.224
1.3 1.274
1.4 1.316
1.5 1.350
1.6 1.376
1.7 1.394
1.8 1.404
1.9 1.406
2.0 1.400
2.1 1.386
2.2 1.364
2.3 1.334
2.4 1.296
2.5 1.250
2.6 1.196
2.7 1.134
2.8 1.064
2.9 0.986
3.0 0.900
3.1 0.806
3.2 0.704
3.3 0.594
3.4 0.476
3.5 0.350
3.6 0.216
3.7 0.074
3.8 -0.076
The data fits in the following equation:
y = -0.4x2 + 1.5x.
Integrate from x = 0 to x = 3.8,
= 3.51373
So, value of area under the curve is 3.51373.