Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App
geeksforgeeks
Browser
Continue

Related Articles

Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 1

Improve Article
Save Article
Like Article
author
gurjotloveparmar
proficient
156 published articles
Improve Article
Save Article
Like Article

Evaluate the following definite integrals as limits of sums:

Question 1. \int_{0}^{3}(x+4)dx

Solution:

We have,

I =\int_{0}^{3}(x+4)dx

We know,\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 0, b = 3 and f(x) = x + 4.

=> h = 3/n

=> nh = 3

So, we get,

I =\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]

=\lim_{h\to0}h[4+(h+4)+(2h+4)+...+(n-1)h+4]

=\lim_{h\to0}h[4n+h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[4n+h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{3}{n}[4n+\frac{3}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}\left[12+\frac{9}{2}(1-\frac{1}{n})\right]

= 12 + \frac{9}{2}

=\frac{33}{2}

Therefore, the value of\int_{0}^{3}(x+4)dx as limit of sum is\frac{33}{2} .

Question 2. \int_{0}^{2}(x+3)dx

Solution:

We have,

I =\int_{0}^{2}(x+3)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 0, b = 2 and f(x) = x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]

=\lim_{h\to0}h[3+(h+3)+(2h+3)+...+(n-1)h+3]

=\lim_{h\to0}h[3n+h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[3n+h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[3n+\frac{2}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[6+2(1-\frac{1}{n})]

= 6 + 2

= 8

Therefore, the value of\int_{0}^{2}(x+3)dx as limit of sum is 8.

Question 3. \int_{1}^{3}(3x-2)dx

Solution:

We have,

I =\int_{1}^{3}(3x-2)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 1, b = 3 and f(x) = 3x − 2.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]

=\lim_{h\to0}h[1+[3(1+h)-2]+[3(1+2h)-2]+...+[3(1+(n-1)h)-2]]

=\lim_{h\to0}h[n+3h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[n+3h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[n+\frac{6}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[2+6(1-\frac{1}{n})]

= 2 + 6

= 8

Therefore, the value of\int_{1}^{3}(3x-2)dx as limit of sum is 8.

Question 4. \int_{-1}^{1}(x+3)dx

Solution:

We have,

I =\int_{-1}^{1}(x+3)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = −1, b = 1 and f(x) = x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(-1)+f(-1+h)+f(-1+2h)+...+f(-1+(n-1)h)]

=\lim_{h\to0}h[2+(2+h)+(2+2h)+...+((n-1)h+2)]

=\lim_{h\to0}h[2n+h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[2n+h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[2n+\frac{2}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[4+2(1-\frac{1}{n})]

= 4 + 2

= 6

Therefore, the value of\int_{-1}^{1}(x+3)dx as limit of sum is 6.

Question 5. \int_{0}^{5}(x+1)dx

Solution:

We have,

I =\int_{0}^{5}(x+1)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 0, b = 5 and f(x) = x + 1.

=> h = 5/n

=> nh = 5

So, we get,

I =\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]

=\lim_{h\to0}h[1+(h+1)+(2h+1)+...+((n-1)h+1)]

=\lim_{h\to0}h[n+h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[n+h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{5}{n}[n+\frac{5}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[5+\frac{25}{2}(1-\frac{1}{n})]

= 5 +\frac{25}{2}

=\frac{35}{2}

Therefore, the value of\int_{0}^{5}(x+1)dx as limit of sum is\frac{35}{2} .

Question 6. \int_{1}^{3}(2x+3)dx

Solution:

We have,

I =\int_{1}^{3}(2x+3)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , whereh = \frac{b-a}{n}

Here a = 1, b = 3 and f(x) = 2x + 3.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]

=\lim_{h\to0}h[2+3+[2(1+h)+3]+[2(1+2h)+3]+...+[2(1+(n-1)h+3)]]

=\lim_{h\to0}h[5+(5+2h)+(5+4h)+...+(5+2(n-1)h)]

=\lim_{h\to0}h[5n+2h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[5n+2h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[5n+\frac{4}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[10+4(1-\frac{1}{n})]

= 10 + 4

= 14

Therefore, the value of\int_{1}^{3}(2x+3)dx as limit of sum is 14.

Question 7. \int_{3}^{5}(2-x)dx

Solution:

We have,

I =\int_{3}^{5}(2-x)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 3, b = 5 and f(x) = 2 − x.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(3)+f(3+h)+f(3+2h)+...+f(3+(n-1)h)]

=\lim_{h\to0}h[(2-3)+(2-(3+h))+f(2-(3+2h))+...+f(2-(3+(n-1)h))]

=\lim_{h\to0}h[-1+(-1-h)+(-1-2h)+...+(-1-(n-1)h)]

=\lim_{h\to0}h[-n-h(1+2+3+...+(n-1))]

=\lim_{h\to0}h[-n-h(\frac{n(n-1)}{2})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[-n-\frac{2}{n}(\frac{n(n-1)}{2})]

=\lim_{n\to\infty}[-2-2(1-\frac{1}{n})]

= –2 – 2

= –4

Therefore, the value of\int_{3}^{5}(2-x)dx as limit of sum is –4.

Question 8. \int_{0}^{2}(x^2+1)dx

Solution:

We have,

I =\int_{0}^{2}(x^2+1)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 0, b = 2 and f(x) = x2 + 1.

=> h = 2/n

=> nh = 2

So, we get,

I =\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]

=\lim_{h\to0}h[1+(h^2+1)+((2h)^2+1)+...+(((n-1)h)^2+1)]

=\lim_{h\to0}h[n+h^2(1^2+2^2+3^2+...+(n-1)^2)]

=\lim_{h\to0}h[n+h^2\frac{n(n-1)(2n-1)}{6}]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{2}{n}[n+\frac{4}{n^2}\frac{n(n-1)(2n-1)}{6}]

=\lim_{n\to\infty}[2+\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})]

=2+\frac{4×2}{3}

=2+\frac{8}{3}

=\frac{14}{3}

Therefore, the value of\int_{0}^{2}(x^2+1)dx as limit of sum is\frac{14}{3} .

Question 9. \int_{1}^{2}x^2dx

Solution:

We have,

I =\int_{1}^{2}x^2dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 1, b = 2 and f(x) = x2.

=> h = 1/n

=> nh = 1

So, we get,

I =\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]

=\lim_{h\to0}h[1+(1+h)^2+(1+2h)^2+...+(1+(n-1)h)^2]

=\lim_{h\to0}h[1+(1+2h+h^2)+(1+4h+4h^2)+...+(1+2(n-1)h+(n-1)^2h^2)]

=\lim_{h\to0}h[n+2h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]

=\lim_{h\to0}h[n+2h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{1}{n}[n+\frac{2}{n}(\frac{n(n-1)}{2})+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})]

=\lim_{n\to\infty}[1+(1-\frac{1}{n})+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})]

= 1 + 1 +\frac{2}{6}

= 1 + 1 +\frac{1}{3}

=\frac{7}{3}

Therefore, the value of\int_{1}^{2}x^2dx as limit of sum is\frac{7}{3} .

Question 10. \int_{2}^{3}(2x^2+1)dx

Solution:

We have,

I =\int_{2}^{3}(2x^2+1)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 2, b = 3 and f(x) = 2x2 + 1.

=> h = 1/n

=> nh = 1

So, we get,

I =\lim_{h\to0}h[f(2)+f(2+h)+f(2+2h)+...+f(2+(n-1)h)]

=\lim_{h\to0}h[9+[2(2+h)^2+1]+[2(2+2h)^2+1]+...+[2(2+(n-1)h)^2+1]]

=\lim_{h\to0}h[9n+8h(1+2+3+...+(n-1))+2h^2(1^2+2^2+3^2+...(n-1)^2)]

=\lim_{h\to0}h[9n+8h(\frac{n(n-1)}{2})+2h^2(\frac{n(n-1)(2n-1)}{6})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{1}{n}[9n+\frac{8}{n}(\frac{n(n-1)}{2})+\frac{2}{n^2}(\frac{n(n-1)(2n-1)}{6})]

=\lim_{n\to\infty}[9+\frac{4n^2}{n^2}(1-\frac{1}{n})+\frac{n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})]

= 9 + 4 +\frac{2}{3}

=\frac{41}{3}

Therefore, the value of\int_{2}^{3}(2x^2+1)dx as limit of sum is\frac{41}{3} .

Question 11. \int_{1}^{2}(x^2-1)dx

Solution:

We have,

I =\int_{1}^{2}(x^2-1)dx

We know,

\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] , where h =\frac{b-a}{n}

Here a = 1, b = 2 and f(x) = x2 − 1.

=> h = 1/n

=> nh = 1

So, we get,

I =\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]

=\lim_{h\to0}h[(1^2-1)+[(1+h)^2-1]+[(1+2h)^2-1]+...+[(1+(n-1)h)^2-1)]

=\lim_{h\to0}h[0+2h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]

=\lim_{h\to0}h[2h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]

Now if h −> 0, then n −> ∞. So, we have,

=\lim_{n\to\infty}\frac{1}{n}[\frac{2}{n}(\frac{n(n-1)}{2})+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})]

=\lim_{n\to\infty}[\frac{n^2}{n^2}(1-\frac{1}{n})+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})]

= 1 +\frac{2}{6}

= 1 +\frac{1}{3}

=\frac{4}{3}

Therefore, the value of \int_{1}^{2}(x^2-1)dx as limit of sum is \frac{4}{3}.


My Personal Notes arrow_drop_up
Last Updated : 20 May, 2021
Like Article
Save Article
Similar Reads
1. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 3
2. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 2
3. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 1
4. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 2
5. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 3
6. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 1
7. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 2
8. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 3
9. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 1
10. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 2
Related Tutorials
1. Linear Algebra
2. Maths
3. Physics
4. Chemistry
5. CBSE Class 12 Syllabus 2023-2024
Previous
Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part B
Next
Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 2
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
gurjotloveparmar
gurjotloveparmar
Vote for difficulty
Article Tags :
Report Issue
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox