Question

Definite integral as the limit of sum expressed as

• A. $\underset{a}{\overset{b}{\int }}f(x)dx=(b-a)\underset{h\to 0}{\lim }\frac{1}{n}[f(a)+f(a+h)+\dots .+f$
• B. $\underset{a}{\overset{b}{\int }}f(x)dx=\frac{(b-a)}{n}\underset{h\to \infty }{\lim }\frac{1}{n}[f(a)\times f(a+h)\dots ..\times f$
• C. $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{h\to 0}{\lim }h[f(a)+f(a+h)+\dots +f(a+(n-1)h]$
• D. None of the above

Answer 1

Answer: (C)

Let $f$ be a continuous function on close interval $[a,b]$. Assume that all the values taken by the function are non-negative, so the graph of the function is a curve above the $x$-axis.
The definite integral $\underset{a}{\overset{b}{\int }}f(x)dx$ is the area bounded by the curve $y=f(x)$, the ordinates $x=a,x=b$ and $x$-axis. To evaluate this area, consider the region PRSQP between this curve, $x$-axis and the ordinates $x=a$ and $x=b$

Divide the interval $[a,b]$ into $n$ equal subintervals denoted by $\left[x_0,x_1\right],\left[x_1,x_2\right]\dots \left[x_{r-1},x_r\right],\dots ,\left[x_{n-1},x_n\right]$, Where $x_0=a_1$ $x_1=a,+h,x_2=a+2h,\dots$
$x_r=a+rh$ and $x_n=b=a+nh$ or $n=\frac{b-a}{h}$. We note that as $n\to \infty ,h\to 0$
The region $PRSQP$ under consideration is the sum of $n$ subregions, where each subregion is defined on subinetervals $\left[x_{r-1},x_r\right],r=12,3,\dots ,n$. From fig. we have area of the rectangle $(ABLC)<$ area of the region $(ABDCA)<$ area of the rectangle $(ABDM)....$(i)
Evidently as $x_r-x_{r-1}\to 0$, i.e., $h\to 0$ all the three areas shown in (1) become nearly equal to each other, Now we form the following sums.
$S_n=h\left[f\left(x_0\right)+\dots +f\left(x_{n-1}\right)\right]=h\sum _{r=1}^{n-1}f\left(x_r\right)......$(ii)
and $S_n=h\left[f\left(x_1\right)+f\left(x_2\right)+\dots +f\left(x_n\right)\right]=h\sum _{r=1}^{n}f\left(x_r\right)......$(iii)
Here, $S_n$ and $S_n$ denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals $\left[x_{r-1},x_r\right]$ for $r=1,2,3,\dots n$ respectively.
In view of the inequality (i) for an arbitrary subinterval $\left[x_{r-1},x_r\right]$, we have
$S_n<$ area of the region $PRSQP<S_n$...(iv)
As $n\to \infty$ strip become narrower and narrower, it is assumed that the limiting values of (ii) and (iii) are the same in both cases and the common limiting value is the required area under the curve. Symbolically, we write
$\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }{S}_n=$ area of the region $PRSQP$
$=\underset{a}{\overset{b}{\int }}f(x)dx.....$(v)
It follows that this ares is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangle above the curve. For the sake of convenience, we shall take rectangles with height equal to that of the curve at the left hand edge of subinterval., Thus, we rewrite (v) as
$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{h\to \infty }{\lim }h[f(a)+f(a+h)+\dots +f\{a+(n-1)h\}]$
or $\underset{a}{\overset{b}{\int }}f(x)dx=(b-a)\underset{n\to \infty }{\lim }\frac{1}{h}[f(a)+f(a+h)$
$+\dots +f(a+(n-1)h]......$(vi)
where, $h=\frac{b-a}{n}\to 0$ as $n\to \infty$
The above expression (vi) is known as the definition of definite integral as the limit of sum.