In the previous section, we saw the definite integral as limit of sum. We saw a solved example also. In this section, we will see a few more solved examples.
Solved Example 23.70
Evaluate $\small{\int_0^5{\left[x+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^5{\left[x+1 \right]dx}}$ is the area bounded by the four items:
♦ The line y = f(x) = x+1.
♦ The vertical line x = 0
♦ The vertical line x = 5
♦ The horizontal line y = 0 (x-axis)
•
A rough sketch is shown in the fig.23.5 below. Our task is to find the area of the violet shaded area.
 |
| Fig.23.5 |
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
•
In our present case, a = 0 and b = 5
So $\small{h~=~\frac{b-a}{n}~=~\frac{5-0}{n}~=~\frac{5}{n}}$
3. Now the formula becomes:
$\small{\int_0^5{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{5-0}{n} \left[f(a+\frac{5}{n})~+~f(a+\frac{10}{n})~+~~.~.~.~+f(a+\frac{5n}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{5}{n} \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$
4. Let us determine the quantity inside the square brackets:
$\small{ \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$
5. So we have to do three summations:
(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}~=~na~=~n(0)~=~0}$
(ii) $\small{\frac{5}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$
$\small{~~~~~=~\frac{5}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{5(n+1)}{2}~=~\frac{5n}{2}~+~\frac{5}{2}}$
(Here we use the technique of arithmetic progression)
(iii) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n(1)~=~n}$
6. So the total of three summations is:
$\small{0~+~\frac{5n}{2}~+~\frac{5}{2}~+~n}$
7. Now the limit in step (3) can be calculated:
$\small{\lim_{n\rightarrow \infty} \frac{5}{n} \left[\frac{5n}{2}~+~\frac{5}{2}~+~n \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \left[\frac{25}{2}~+~\frac{25}{2n}~+~5 \right]~=~\frac{35}{2}}$
Solved Example 23.71
Evaluate the definite integral $\small{\int_0^2{\left[x^2+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[x^2+1 \right]dx}}$ is the area bounded by the four items:
♦ The line y = f(x) = x2+1.
♦ The vertical line x = 0
♦ The vertical line x = 2
♦ The horizontal line y = 0 (x-axis)
•
A rough sketch is shown in the fig.23.6 below. Our task is to find the area of the violet shaded area.
 |
| Fig.23.6 |
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$
3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \bigg[\big[(a+\frac{2}{n})^2~+~1\big]~+~\big[(a+\frac{4}{n})^2~+~1 \big]~+~\big[(a+\frac{6}{n})^2~+~1 \big]~+~~.~.~.}$
$\small{~~~~~~~~~~.~.~.~+~\big[(a+\frac{2n}{n})^2~+~1 \big] \bigg]}$
4. Let us determine the quantity inside the large square brackets:
$\small{~=~ \bigg[\big[a^2 + \frac{4a}{n}+\frac{2^2}{n^2}~+~1\big]~+~\big[a^2 + \frac{8a}{n}+\frac{4^2}{n^2}~+~1 \big]~+~\big[a^2 + \frac{12a}{n}+\frac{6^2}{n^2}~+~1 \big]~+~ }$
$\small{~~~~~~~~.~.~.~+~\big[a^2 + \frac{2(2an)}{n}+\frac{2^2\,n^2}{n^2}~+~1 \big] \bigg]}$
5. So we have to do four summations:
(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(0^2)~=~0}}$
(ii) $\small{\frac{4a}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)~~~~~=~0, ~~~\because a~=~0}$
(iii) $\small{\frac{2^2}{n^2}\left(1^2~+~2^2~+~3^2~+~.~.~.~ \text{n terms} \right)}$
$\small{~~~~~=~\frac{2^2}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{4n}{3}~+~2~+~\frac{2}{3n}}$
(Here we use the technique that we saw in section 9.5)
(iv) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n (1)~=~n}$
6. So the total of 4 summations is:
$\small{0+0+\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n}$
8. Now the limit in step (3) can be calculated:
$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \left[\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \left[\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}+2 \right]~=~\frac{8}{3}+2~=~\frac{14}{3}}$
Solved Example 23.72
Evaluate the definite integral $\small{\int_0^2{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[e^x \right]dx}}$ is the area bounded by the four items:
♦ The curve y = f(x) = ex.
♦ The vertical line x = 0 (y-axis)
♦ The vertical line x = 2
♦ The horizontal line y = 0 (x-axis)
•
A rough sketch is shown in the fig.23.7 below. Our task is to find the area of the violet shaded area.
 |
| Fig.23.7 |
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$
3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$
4. Let us determine the quantity inside the large square brackets:
$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$
$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$
Here $\small{e^a~=~1~~\because a = 0}$
• First term = $\small{e^{\frac{2}{n}}}$
• Common ratio $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$
6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$
• So we get:
Sum of all terms inside the brackets
$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$
7. So the limit in (3) becomes:
$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$
$\small{~=~\lim_{n\rightarrow \infty} \Bigg[\frac{e^{\frac{2}{n}} \left({e^{2}}~-~1 \right)}{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$
$\small{~=~\frac{e^{0} \left({e^{2}}~-~1 \right)}{1} ~=~e^2 ~-~1}$
• Here we use two facts:
(i) $\small{\lim_{n\rightarrow \infty} \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$
(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$
So $\small{\lim_{n\rightarrow \infty} \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty} \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$
Solved Example 23.73
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ is the area bounded by the four items:
♦ The curve y = f(x) = ex.
♦ The vertical line x = −1
♦ The vertical line x = 1
♦ The horizontal line y = 0 (x-axis)
• A rough sketch is shown in the fig.23.8 below. Our task is to find the area of the violet shaded area.
 |
| Fig.23.8 |
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = -1 and b = -1
So $\small{h~=~\frac{b-a}{n}~=~\frac{1-(-1)}{n}~=~\frac{2}{n}}$
3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$
4. Let us determine the quantity inside the large square brackets:
$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$
$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$
Here $\small{e^a~=~\frac{1}{e}~~\because a = -1}$
• First term = $\small{e^{\frac{2}{n}}}$
• Common ratio $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$
6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$
• So we get:
Sum of all terms inside the brackets
$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$
7. So the limit in (3) becomes:
$\small{\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{1}{e} \right) \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{{e^{2}}~-~1}{e} \right) \big[\frac{e^{\frac{2}{n}}}{e^{\frac{2}{n}}~-~1} \big]}$
$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty} \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$
$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty} \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$
$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right) \Big[\frac{1}{1} \Big]}$
$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right) ~=~e~-~\frac{1}{e}}$
• Here we use two facts:
(i) $\small{\lim_{n\rightarrow \infty} \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$
(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$
So $\small{\lim_{n\rightarrow \infty} \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty} \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$
Solved Example 23.74
Evaluate the definite integral $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ is the area bounded by the four items:
♦ The curve y = f(x) = x+e2x.
♦ The vertical line x = 0 (y-axis)
♦ The vertical line x = 4
♦ The horizontal line y = 0 (x-axis)
• A rough sketch is shown in the fig.23.9 below. Our task is to find the area of the violet shaded area.
 |
| Fig.23.9 |
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 4
So $\small{h~=~\frac{b-a}{n}~=~\frac{4-0}{n}~=~\frac{4}{n}}$
3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \left[f(a+\frac{4}{n})~+~f(a+\frac{8}{n})~+~f(a+\frac{12}{n})~+~~.~.~.~+f(a+(n)\frac{4}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \bigg[\big[a+\frac{4}{n}~+~e^{2(a+\frac{4}{n})}\big]~+~\big[a+\frac{8}{n}~+~e^{2(a+\frac{8}{n})}\big]~+~\big[a+\frac{12}{n}~+~e^{2(a+\frac{12}{n})}\big]~+~~.~.~.~}$
$\small{~~~~~~~~~.~.~.~+~\big[a+(n)\frac{4}{n}~+~e^{2(a+(n)\frac{4}{n})}\big]\bigg]}$
4. Let us determine the quantity inside the large square brackets. We have to do three summations:
(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}}$
$~=~\small{n a~=~n(0)~=~0}$
(ii) $\small{\frac{4}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$
$\small{~~~~~=~\frac{4}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{4(n+1)}{2}~=~2n~+~2}$
(Here we use the technique of arithmetic progression)
(iii) $\small{e^{2(a+\frac{4}{n})}~+~e^{2(a+\frac{8}{n})}~+~e^{2(a+\frac{12}{n})}~+~.~.~.~ \text{n terms}}$
$\small{~=~e^{2a}~\times~e^{\frac{8}{n}}~+~e^{2a}~\times~e^{\frac{16}{n}}~+~e^{2a}~\times~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms}}$
$\small{~=~e^{2a}\left(e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} \right)}$
$\small{~=~e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} }$
$\small{~~~~~~\because e^{2a}~=~e^{2(0)}~=~e^0~=~1}$
• So we have a geometric progression.
• First term = $\small{e^{\frac{8}{n}}}$
• Common ratio $\small{~r~=~\frac{e^{\frac{16}{n}}}{e^{\frac{8}{n}}}~=~\frac{e^{\frac{24}{n}}}{e^{\frac{16}{n}}}~=~e^{\frac{8}{n}}}$
• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$
$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$
• So we get:
Sum of all terms of the G.P
$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$
$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$
• So we get:
Sum of all terms inside the brackets
$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$
5. So the limit in (3) becomes:
$\small{\lim_{n\rightarrow \infty} \frac{4}{n} \big[0~+~2n\,+\,2~+~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \big[2n\,+\,2 \big]~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~8~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$
6. In the above expression, we need to evaluate the limit for the second term. It can be done as follows:
$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}}~\times~\left(\frac{e^8}{e^\frac{8}{n}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8~-~e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{1 }{e^{\frac{8}{n}}~-~1} \big]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$
$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty} \bigg[\frac{1 }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \bigg]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty} \big[\frac{e^{\frac{8}{n}} }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \big]}$
$\small{~=~\frac{e^8}{2}\, \bigg[\frac{1}{1} \bigg]~-~\frac{1}{2}\, \big[\frac{1}{1} \big]~=~\frac{e^8~-~1}{2}}$
7. So the limit in (3) becomes:
$\small{8~+~\frac{e^8~-~1}{2}~=~\frac{16~+~e^8~-~1}{2}~=~\frac{15~+~e^8}{2}}$
• Here we use two facts:
(i) $\small{\lim_{n\rightarrow \infty} \left[e^{\frac{8}{n}} \right]~=~e^{\frac{8}{\infty}}~=~e^0~=~1}$
(ii) Let $\small{\frac{8}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$
So $\small{\lim_{n\rightarrow \infty} \Big[\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}\Big]~=~\lim_{n\rightarrow \infty} \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$
The link below gives a few more solved examples:
Exercise 23.8
We
have completed a discussion on the definite integral as limit of sum.
In the next section, we will see fundamental theorem of calculus.
Previous
Contents
Copyright©2025 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment