Definite integral as the limit of a sum - Practice Questions & MCQ

Let f(x) be a continuous real-valued function defined on the closed interval [a, b] which is divided into n parts as shown in the figure.

Each subinterval denoted as $\left[x_0, x_1\right],\left[x_1, x_2\right], \ldots\left[x_{i-1}, x_i\right], \ldots,\left[x_{n-1}, x_n\right]$ having equal width $\frac{b-a}{n}$ where, $x_0=a$ and $x_n=b$.
$
\Rightarrow \quad x_i-x_{i-1}=\frac{b-a}{n} \quad \text { for } i=1,2,3, \ldots, n
$

We denote the width of each subinterval with the notation $\Delta x$,
so $\quad \Delta x=\frac{b-a}{n}$ and $x_i=x_0+i \Delta x$
On each subinterval, $\left[x_{i-1}, x_i\right]$ (for $\left.i=1,2,3, \ldots, n\right)$ a rectangle is constructed with width $\Delta \mathrm{x}$ and height equal to $f\left(x_{i-1}\right.$ ), which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is $f\left(x_{i-1}\right) \Delta x$. Adding the areas of all these rectangles, we get an approximate value for A
$
\begin{aligned}
A \approx L_n & =f\left(x_0\right) \Delta x+f\left(x_1\right) \Delta x+\cdots+f\left(x_{n-1}\right) \Delta x \\
& =\sum_{i=1}^n f\left(x_{i-1}\right) \Delta x
\end{aligned}
$

L_n to denote that this is a left-endpoint approximation of A using n subintervals.

The second method for approximating the area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

This time the height of the rectangle is determined by the function value f(x_i) at the right endpoint of the subinterval. Then, the area of each rectangle is f(x_i)Δx, and the approximation for A is given by

$\begin{aligned} A \approx R_n & =f\left(x_1\right) \Delta x+f\left(x_2\right) \Delta x+\cdots+f\left(x_n\right) \Delta x \\ & =\sum_{i=1}^n f\left(x_i\right) \Delta x\end{aligned}$

As n → ∞ strips become narrower and narrower, it is assumed that the limiting values of L_n and R_n are the same in both cases and the common limiting value is the required area under the curve.

Symbolically, we write

$
\begin{aligned}
& \lim _{n \rightarrow \infty} \mathrm{~L}_n=\lim _{n \rightarrow \infty} \mathrm{R}_{\mathrm{n}}=\text { area of the region }=\int_a^b f(x) d x \\
& \int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{i=1}^n f\left(x_i\right) \Delta x=\lim _{n \rightarrow \infty} \sum_{i=1}^n\left(\frac{b-a}{n}\right) f\left(a+\left(\frac{b-a}{n}\right) i\right)
\end{aligned}
$
where, $\quad \Delta x=\frac{b-a}{n}$ and $x_i=x_0+\Delta x . i$

NOTE:

If a = 0, b = 1, then 

$
\int_0^1 f(x) d x=\lim _{n \rightarrow \infty} \sum_{i=0}^{n-1} \frac{1}{n} f\left(\frac{i}{n}\right)
$
2. From the definition of definite integral, we have $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=\varphi(n)}^{\psi(n)} f\left(\frac{i}{n}\right)=\int_a^b f(x) d x$, where
(i) $\quad \Sigma$ is replaced by $\int \operatorname{sign}$
(ii) $\frac{i}{n}$ is replaced by $x$
(iii) $\frac{1}{n}$ is replaced by $d x$
(iv) To obtain the limits of integration, we use $\mathrm{a}=\lim _{\mathrm{n} \rightarrow \infty} \frac{\phi(\mathrm{n})}{\mathrm{n}}$ and $\mathrm{b}=\lim _{\mathrm{n} \rightarrow \infty} \frac{\psi(\mathrm{n})}{\mathrm{n}}$       

 For example:

 $
\lim _{n \rightarrow \infty} \sum_{r=1}^{p . n} \frac{1}{n} f\left(\frac{r}{n}\right)=\int_\alpha^\beta f(x) d x
$
where, $\alpha=\lim _{n \rightarrow \infty} \frac{r}{n}=0$ ( as $r=1$ ) and $\quad \beta=\lim _{n \rightarrow \infty} \frac{r}{n}=p($ as $r=p n)$