Application of Antiderivatives

# Area as a Definite Integral

Before Leibnitz and Newton, the computation of the area under the curve of a function was really a challenge for mathematicians. The great Greek mathematician Archimedes solved the problem for some special curves by using some special methods. So, the areas under the graphs of some simple curves such as linear, quadratic, cubic and of higher powers were only known. [Polynomial Equation]

Lastly, Leibnitz and other mathematicians brought the problem into a right track by considering the area under the curve as an infinite sum of infinitely thin rectangles. Leibnitz introduced the integral sign in 1676 by elongating S which stands for ‘sum’. The subject concerning this theory was decided by the agreement between Bernoulli and Leibnitz to be called ‘Integral Calculus‘ in 1696, while the term ‘Integral’ was itself introduced by Bernoulli in 1690.

## Determination of Area as an Infinite Sum of Infinitely Thin Rectangles

Let us consider an example. We shall find the area included between the curve $y=x^2$, the x-axis and the ordinates $x=0$ and $x=a$. Draw the parabola $y=x^2$ and divide the the interval $(0,a)$ into $n$ subintervals of equal length $h=\frac{a}{n}$. Draw $n$ rectangles as shown in the figure.

All the shaded rectangles such as $ABCD$ lie under the curve and constitute a set of rectangles inscribed in the parabolic region whose area we have to find out. Let $s_n$ be the sum of the areas of these inscribed rectangles. When $x=h$, $y=x^2$ gives $y=h^2$, therefore $s_n=0+h(h)^2+h(2h)^2+…+h[(n-1)h]^2$ $=h^3(1^2+2^2+…+(n-1)^2)$ $=\frac{a^3}{n^3}\left(\frac{1}{6}(n-1)n(2n-1)\right)$ $=\frac{1}{6}a^3\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)$ $\therefore \lim_{n\to\infty}s_n=\frac{1}{3}a^3$ [Parabola]

Now, consider the rectangles of the type $ABC’D’$. They constitute a set of rectangles circumscribing the parabolic region. Let $S_n$ be the sum of the areas of these circumscribing rectangles. $\therefore S_n=h(h)^2+h(2h)^2+h(3h)^2+…+h(nh)^2$ $=h^3(1^2+2^2+…+n^2)$ $=\frac{a^3}{n^3}\frac{1}{6}n(n+1)(2n+1)$ $=\frac{1}{6}a^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)$ $\therefore S_n=\frac{1}{3}a^3$

If $A$ be the area of the parabolic region, then, we have, $\lim_{n\to\infty}s_n≤A≤\lim_{n\to\infty}S_n$ $\frac{1}{3}a^3≤A≤\frac{1}{3}a^3$ $\therefore A=\frac{1}{3}a^3$ Hence, using this method, we can easily find the area of a plane region.

## Definite Integral as the Limit of a Sum

If $ƒ(x)$ be a function continuous in an interval $[a,b]$ be divided into $n$ equal parts with each of length $h$, so that $nh=b-a$, then $\lim_{h\to 0}h[ƒ(a)+ƒ(a+h)+ƒ(a+2h)+…$$+ƒ(a+\overline{n-1}h)]$ is known as the definite integral of $ƒ(x)$ with respect to $x$, the range is from $a$ to $b$ and is written as $\int_a^b ƒ(x)dx$. $\therefore \int_a^b ƒ(x)dx=\lim_{h\to 0}h[ƒ(a)+ƒ(a+h)$$+ƒ(a+2h)+…+ƒ(a+\overline{n-1}h)]$ This relation can also be written as $\int_a^b ƒ(x)dx=\lim_{h\to 0}h[ƒ(a+h)+ƒ(a+2h)$$+ƒ(a+3h)+…+ƒ(a+nh)]$

This relation gives the definition of an integral as the limit of a sum. It shows that the integral gives the area of the plane region under given conditions. Here, $a$ is the lower limit and $b$ is the upper limit. If $a=0$, then, $\int_0^bƒ(x)dx=\lim_{h\to 0}h[ƒ(h)+ƒ(2h)+ƒ(3h)$$+…+ƒ(nh)]$ where $nh=b-0=b$.

### $y=x; x=0, x=a$

Here, $ƒ(x)=x$ and $nh=a-0=a$. Using, $\int_0^a ƒ(x)dx=\lim_{h\to 0}h[ƒ(h)+ƒ(2h)+ƒ(3h)$$+…+ƒ(nh)]$ We have, $\int_0^a xdx=\lim_{h\to 0} h[h+2h+3h+…+nh]$ $=\lim_{h\to 0}h^2[1+2+3+…+n]$ $=\lim_{h\to 0}h^2.n\left(\frac{n+1}{2}\right)$ $=\lim_{h\to 0}nh\left(\frac{nh+h}{2}\right)$ $=\lim_{h\to 0}a\left(\frac{a+h}{2}\right)=\frac{a^2}{2}$

### $y=2x^2-3; x=0, x=a$

Here, $ƒ(x)=2x^2-3$ and $nh=a-0=a$. Using, $\int_0^a ƒ(x)dx=\lim_{h\to 0}h[ƒ(h)+ƒ(2h)+ƒ(3h)$$+…+ƒ(nh)]$ We have, $\int_0^a(2x^2-3)dx=\lim_{h\to 0}[(2h^2-3)+(2.2^2h^2-3)$ $+(2.3^2h^2-3)+…+(2.n^2h^2-3)]$ $=\lim_{h\to 0}h[2h^2(1^2+2^2+3^2+…+n^2)-3n]$ $=\lim_{h\to 0}\left[2h^3.\frac{n(n+1)(2n+1)}{6}-3nh\right]$ $=\lim_{h\to 0}\left(\frac{1}{3}.nh(nh+h)(2nh+h)-3nh\right]$ $=\frac{1}{3}a(a+0)(2a+0)-3a$ $=\frac{2}{3}a^3-3a$

### $y=4ax^2; x=0, x=a$

Here, $ƒ(x)=4ax^2$ and $nh=a-0=a$. Using, $\int_0^a ƒ(x)dx=\lim_{h\to 0}h[ƒ(h)+ƒ(2h)+ƒ(3h)$$+…+ƒ(nh)]$ We have, $\lim_{h\to 0}h[4ah^2+4a(2h)^2+4a(3h)^2+…+4a(nh)^2]$ $=\lim_{h\to 0}h.4ah^2[1^2+2^2+3^2+…+n^2]$ $=\lim_{h\to 0}h.4ah^2×\frac{n(n+1)(2n+1)}{6}$ $=\lim_{h\to 0}\frac{2a}{3}nh(nh+h)(2nh+h)$ $=\lim_{h\to 0}\frac{2a}{3}a(a+h)(2a+h)$ $=\frac{2a}{3}a(a+0)(2a+0)=\frac{4a^4}{3}$

### $y=e^{3x}; x=0, x=b$

Here, $ƒ(x)=e^{3x}$ and $nh=b-0=b$. Using, [\int_0^b ƒ(x)dx=\lim_{h\to 0}h[ƒ(h)+ƒ(2h)+ƒ(3h)][+…+ƒ(nh)]] We have, $\int_0^be^{3x}dx=\lim_{h\to 0}h[e^{3h}+e^{3.2h}+e^{3.3h}+…+e^{3.nh}$ $=\lim_{h\to 0}h\frac{e^{3h}(e^{3nh}-1)}{(e^{3h}-1)}$ $=\lim_{h\to 0}\frac{he^{3h}(e^{3b}-1)}{(e^{3h}-1)}\text{ }\left(\frac{0}{0}\text{ form}\right)$ Hence, applying L Hospital’s Rule, we get $\lim_{h\to 0}\frac{(e^{3h}+3e^{3h}h)(e^{3b}-1)}{3e^{3h}}$ $=\frac{(e^0+0)(e^{3b}-1)}{3e^0}=\frac{e^{3b}-1}{3}$

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