Definite Integral - Calculus

Definite integral as the limit of a sum

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b] and F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then

$\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

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 Δx 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 s $f\left(x_{i-1}\right) \Delta x$. Adding the areas of all these rectangles, we get an approximate value for A

$\begin{aligned} A \approx L_n & =f\left(x_0\right) \Delta x+f\left(x_1\right) \Delta x+\cdots+f\left(x_{n-1}\right) \Delta x \\ & =\sum_{i=1}^n f\left(x_{i-1}\right) \Delta x\end{aligned}$

$\mathrm{L}_n$ 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 $\mathrm{n} \rightarrow \infty$ 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

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

NOTE:

1. If $\mathrm{a}=0, \mathrm{~b}=1$, 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 $\int$ 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:

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

Recommended Video Based on Definite Integral as the Limit of a Sum

Solved Example Based on Definite integral as the limit of a sum:

Example 1: $\int_0^3\{x\} d x$

1) 2

2) 3/2

3) 4

4) 5

Solution

As we learned

Definite Integrals as the limit of a sum -

$\int_0^l f(x) d x=\lim _{x \rightarrow \infty} \sum \frac{1}{x} f\left(\frac{r}{x}\right)$

Or

$\int_a^b f(x) d x=\lim _{x \rightarrow \infty} h \sum_{r=0}^x f(a+r h)$

- wherein

Where $f(x)$ is a continuous function in $[0, l]$

Where $h=\frac{b-a}{x}$ And $f(x)$ is continuous in $[a, b]$

$\int_0^3\{x\}=\int_0^1\{x\} d x+\int_1^2\{x\} d x+\int_2^3\{x\} d x$

$=3 / 2$

Example 2: $\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\ldots \ldots \ldots \ldots \ldots+\frac{n}{(2 n-1)^2}\right]$ is equal to:

1) $\frac{1}{2}$
2) 1
3) $\frac{1}{3}$
4) $\frac{1}{4}$

Solution

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\ldots+\frac{n}{(2 n-1)^2}\right]$

$\begin{aligned} & =\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^2}=\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{n^2+2 n r+r^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \frac{1}{(r / n)^2+2(r / n)+1}\end{aligned}$

$=\int_0^1 \frac{\mathrm{dx}}{(\mathrm{x}+1)^2}=\left[\frac{-1}{(\mathrm{x}+1)}\right]_0^1=\frac{1}{2}$

Hence, the answer is the option (1).

Example 3: $\left.\lim _{n \rightarrow \infty}\left(\frac{(n+1)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}\right)+\frac{(n+2)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+\ldots \ldots \ldots+\frac{(2 n)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}\right)$ is equal to:

1) $\frac{3}{4}(2)^{\frac{4}{3}}-\frac{3}{4}$
2) $\frac{4}{3}(2)^{\frac{4}{3}}$
3) $\frac{3}{4}(2)^{\frac{4}{3}}-\frac{4}{3}$
4) $\frac{4}{3}(2)^{\frac{3}{4}}$

Solution

$\lim _{n \rightarrow \infty}\left(\frac{(n+1)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+\frac{(n+2)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+\ldots \ldots \ldots \ldots+\frac{(2 n)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}\right)$

$=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(1+\frac{1}{n}\right)^{\frac{1}{3}}+\left(1+\frac{2}{n}\right)^{\frac{1}{3}}+\ldots \ldots \ldots+\left(1+\frac{n}{n}\right)^{\frac{1}{3}}\right]$

$\begin{aligned} & =\frac{1}{n} \sum_{r=1}^n\left(1+\frac{r}{n}\right)^{\frac{1}{3}} \\ & =\int_0^1(1+x)^{\frac{1}{3}} d x=\frac{3}{4}\left(2^{\frac{4}{3}}-1\right)\end{aligned}$

Hence, the answer is option (1).

Example 4: $\lim _{n \rightarrow \infty}\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\frac{n^2}{\left(n^2+9\right)(n+3)}+\ldots+\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)$

1) $\frac{\pi}{8}+\frac{1}{4} \log _e 2$
2) $\frac{\pi}{4}+\frac{1}{8} \log _e 2$
3) $\frac{\pi}{4}-\frac{1}{8} \log _{\mathrm{e}} 2$
4) $\frac{\pi}{8}+\log _c \sqrt{2}$

Solution:

$\lim _{n \rightarrow \infty}\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\cdots+\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)$

$=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n^2}{\left(n^2+r^2\right)(n+r)}$

$=\int_0^1 \frac{1}{\left(1+x^2\right)(1+x)} d x$

$\begin{aligned} & =\frac{1}{2} \int_0^1 \frac{1}{1+x} d x+\frac{1}{2} \int_b^1 \frac{d x}{1+x^2}-\frac{1}{4} \int_0^1 \frac{2 x}{\left(1+x^2\right)} d x \\ & =\frac{1}{2}[\ln (1+x)]_0^1+\frac{1}{2}\left[\tan ^{-1} x\right]_0^1-\frac{1}{4}\left[\ln \left(1+x^2\right)\right]_0^1 \\ & =\frac{1}{2} \ln 2+\frac{\pi}{8}-\frac{1}{4} \ln 2 \\ & =\frac{\pi}{8}+\frac{1}{4} \log e^2\end{aligned}$

Hence, the answer is the option (1).

Example 5: If $\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)$$=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^k+2^k+3^k+\ldots+n^k\right]$,

then the integral value of $k$ is equal to _______.

1) 5

2) 9

3) 8

4) 6

Solution

$\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n k+1}[n k \cdot n+1+2+\cdots+n]$

$=\lim _{\mathrm{n} \rightarrow \infty} \frac{(\mathrm{n}+1)^{\mathrm{k}-1}}{\mathrm{n}^{\mathrm{k}+1}} \cdot\left[\mathrm{n}^2 \mathrm{k}+\frac{(\mathrm{n}(\mathrm{n}+1)}{2}\right]$

$\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1} \cdot n^2\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}{n k+1}$

$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)$

$\Rightarrow\left(k+\frac{1}{2}\right)$

RHS

$\Rightarrow \lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{nk}+1}\left(1^{\mathrm{k}}+2^{\mathrm{k}}+\cdots+\mathrm{h}^{\mathrm{k}}\right)=\frac{1}{\mathrm{k}+1}$

LHS=RHS

$\Rightarrow \mathrm{k}+\frac{1}{2}=33 \cdot \frac{1}{\mathrm{k}+1}$

$\begin{aligned} & \Rightarrow(2 k+1)(k+1)=66 \\ & \Rightarrow(k-5)(2 k+13)=0 \\ & \Rightarrow k=5 \text { or } \frac{13}{9}\end{aligned}$

Hence, the answer is the (5).