I don't know that the equation that I am going to explain below is correct or not, and this is why I am asking this question.
So, I have found out that area under the curve could be found out by using this equation: $$ \displaystyle\text{Area = }\sum_{a=0}^{n} \frac{a^3-2a^2}{n} $$
where $n$ is the number of smallest possible rectangular strips after the division of the area under the curve.
Here's I got that formula:
Suppose a graph for $y=x^2$ .
We know that the over-estimate area of the strip is
$$ \displaystyle\text{Over-Est. Area = }\frac{a}{n}\cdot\Delta a^2\text{=}\frac{a\cdot\Delta a^2}{n} $$
And under-estimate area is
$$ \displaystyle\text{Under-Est. Area = }\frac{a}{n}\cdot a^2\text{=}\frac{a^3}{n} $$
So we can say that: $$ \displaystyle\sum_{a=0}^{n}\frac{a\cdot\Delta a^2}{n} < \text{Area of Curve} < \sum_{a=0}^{n}\frac{a^3}{n} $$
And so: $$ \displaystyle\text{Area = }\sum_{a=0}^{n}\frac{a^3}{n} - \sum_{a=0}^{n}\frac{a\cdot\Delta a^2}{n} $$ $$ = \sum_{a=0}^{n} (\frac{a^3}{n} - \frac{a\cdot\Delta a^2}{n}) = \sum_{a=0}^{n} (\frac{a^3}{n} - \frac{a\cdot 2a}{n}) $$ $$ = \sum_{a=0}^{n} (\frac{a^3}{n} - \frac{2a^2}{n}) = \sum_{a=0}^{n} \frac{a^3-2a^2}{n} $$
So is it the right thing that I have derived? If not, then please point out where I have done wrong. Thanks. :)
This is not correct. First of all, subtracting the underestimate from the overestimate does not produce the area under the curve. It instead produces an error estimate for the two estimates of the area. Second, even if you take the arithmetic mean of the two, like I am guessing you intended, it only produces yet another estimate called the trapezoidal rule. However, if you take the limit of any of these approximations as $n$ goes to $\infty$, it produces a riemann sum, which is, in fact, the exact area under the curve. (Incidentally, riemann sums actually are used to define definite integrals.)