Evaluate the following definite integrals as limits of sums:
Question 23. \int_{0}^{4}(x+e^{2x})dx
Solution:
We have,
I =
\int_{0}^{4}(x+e^{2x})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 = 4 and f(x) = x + e2x.
=> h = 4/n
=> nh = 4
So, we get,
I =
\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)] =
\lim_{h\to0}h[1+[h+e^{2h}]+[2h+e^{4h}]+...+[(n-1)h+e^{2(n-1)h}]] =
\lim_{h\to0}h[h(1+2+3+...+(n-1))+(1+e^{2h}+e^{4h}+...e^{2(n-1)h})] =
\lim_{h\to0}h[h(\frac{n(n-1)}{2})+(\frac{(e^{2h})^2-1}{e^{2h}-1})] =
\lim_{h\to0}h[h(\frac{n(n-1)}{2})+(\frac{e^{2nh}-1}{e^{2h}-1})] =
\lim_{h\to0}h^2\left[(\frac{n(n-1)}{2})+\left(\frac{e^{8}-1}{2(\frac{e^{2h}-1}{h})}\right)\right] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}[\frac{16}{n^2}(\frac{n(n-1)}{2})+(\frac{e^{8}-1}{2})] =
\lim_{n\to\infty}[\frac{8n^2}{n^2}(1-\frac{1}{n})+(\frac{e^{8}-1}{2})] =
8+(\frac{e^{8}-1}{2}) =
\frac{15+e^{8}}{2} Therefore, the value of
\int_{0}^{4}(x+e^{2x})dx as limit of sum is\frac{15+e^{8}}{2} .
Question 24. \int_{0}^{2}(x^2+x)dx
Solution:
We have,
I =
\int_{0}^{2}(x^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 = 0, b = 2 and f(x) = x2 + x.
=> 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[0+(h^2+h)+[(2h)^2+2h]+...+[(n-1)h)^2+(n-1)h]] =
\lim_{h\to0}h[h^2(1^2+2^2+3^2+...(n-1)^2)+h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[h^2(\frac{n(n-1)(2n-1)}{6})+h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{2}{n}[\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{2}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{2n^2}{n^2}(1-\frac{1}{n})] =
\frac{8}{3}+2 =
\frac{14}{3} Therefore, the value of
\int_{0}^{2}(x^2+x)dx as limit of sum is\frac{14}{3} .
Question 25. \int_{0}^{2}(x^2+2x+1)dx
Solution:
We have,
I =
\int_{0}^{2}(x^2+2x+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 + 2x + 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+2h+1)+[(2h)^2+2(2h)+1]+...[(n-1)^2+2(n-1)+1] =
\lim_{h\to0}h[n+h^2(1^2+2^2+3^2+...(n-1)^2)+2h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[n+h^2(\frac{n(n-1)(2n-1)}{6})+2h(\frac{n(n-1)}{2})] 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})+\frac{4}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[2+\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{4n^2}{n^2}(1-\frac{1}{n})] =
2+\frac{8}{3}+4 =
\frac{26}{3} Therefore, the value of
\int_{0}^{2}(x^2+2x+1)dx as limit of sum is\frac{26}{3} .
Question 26. \int_{0}^{3}(2x^2+3x+5)dx
Solution:
We have,
I =
\int_{0}^{3}(2x^2+3x+5)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) = 2x2 + 3x + 5.
=> 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[5+(2h^2+3h+5)+[2(2h)^2+3(2h)+5]+...+[2(n-1)^2h^2+3((n-1)h)+5]] =
\lim_{h\to0}h[5n+2h^2(1^2+2^2+3^2+...+(n-1)^2)+3h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[5n+2h^2(\frac{n(n-1)(2n-1)}{6})+3h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{3}{n}[5n+\frac{18}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{9}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[15+\frac{9n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{27n^2}{2n^2}(1-\frac{1}{n})] = 15 + 18 +
\frac{27}{2} =
\frac{93}{2} Therefore, the value of
\int_{0}^{3}(2x^2+3x+5)dx as limit of sum is\frac{93}{2} .
Question 27. \int_{a}^{b}xdx
Solution:
We have,
I =
\int_{a}^{b}xdx 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 = a, b = b and f(x) = x.
=> h =
\frac{b-a}{n} => nh = b − a
So, we get,
I =
\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] =
\lim_{h\to0}h[a+(a+h)+(a+2h)+...+(a+(n-1)h)] =
\lim_{h\to0}h[na+h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[na+h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{b-a}{n}[na+\frac{b-a}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}(b-a)[a+\frac{b-a}{n}(\frac{n-1}{2})] =
\lim_{n\to\infty}(b-a)[a+(b-a)(\frac{1-\frac{1}{n}}{2})] =
(b-a)[a+\frac{b-a}{2}] =
\frac{(b-a)(b+a)}{2} =
\frac{b^2-a^2}{2} Therefore, the value of
\int_{a}^{b}xdx as limit of sum is\frac{b^2-a^2}{2} .
Question 28. \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{25n^2}{2n^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 29. \int_{2}^{3}x^2dx
Solution:
We have,
I =
\int_{2}^{3}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 = 2, b = 3 and f(x) = x2.
=> 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[4+(2+h)^2+(2+2h)^2+...+(2+(n-1)h)^2] =
\lim_{h\to0}h[4+[2^2+2.h+h^2]+[2^2+2.2h+(2h)^2]+...] =
\lim_{h\to0}h[4n+h^2(1^2+2^2+...+(n-1)^2)+4h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[4n+h^2(\frac{n(n-1)(2n-1)}{6})+4h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{1}{n}[4n+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{4}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[4+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{2n^2}{n^2}(1-\frac{1}{n})] =
4+\frac{2}{6}+2 =
4+\frac{1}{3}+2 =
\frac{19}{3} Therefore, the value of
\int_{2}^{3}x^2dx as limit of sum is\frac{19}{3} .
Question 30. \int_{1}^{3}(x^2+x)dx
Solution:
We have,
I =
\int_{1}^{3}(x^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 = 1, b = 3 and f(x) = x2 + x.
=> 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+[(1+h)^2+(1+h)]+[(1+2h)^2+(1+2h)]+...] =
\lim_{h\to0}h[2n+h^2(1^2+2^2+3^2+...+(n-1)^2)+3h(1+2+3+...+(n-1))] =
\lim_{h\to0}h[2n+h^2(\frac{n(n-1)(2n-1)}{6})+3h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{2}{n}[2n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[4+\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{6n^2}{n^2}(1-\frac{1}{n})] =
4+\frac{8}{3}+6 =
\frac{38}{3} Therefore, the value of
\int_{1}^{3}(x^2+x)dx as limit of sum is\frac{38}{3} .
Question 31. \int_{0}^{2}(x^2-x)dx
Solution:
We have,
I =
\int_{0}^{2}(x^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 = 0, b = 2 and f(x) = x2 − x.
=> 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[0+[h^2-h]+[(2h)^2-2h]+...] =
\lim_{h\to0}h[h^2(1^2+2^2+...+(n-1)^2)-h(1+2+...+(n-1))] =
\lim_{h\to0}h[h^2(\frac{n(n-1)(2n-1)}{6})-h(\frac{n(n-1)}{2})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{2}{n}[\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})-\frac{2}{n}(\frac{n(n-1)}{2})] =
\lim_{n\to\infty}[\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})-\frac{2n^2}{n^2}(1-\frac{1}{n})] =
\frac{8}{3}-2 =
\frac{2}{3} Therefore, the value of
\int_{0}^{2}(x^2-x)dx as limit of sum is\frac{2}{3} .
Question 32. \int_{1}^{3}(2x^2+5x)dx
Solution:
We have,
I =
\int_{1}^{3}(2x^2+5x)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) = 2x2 + 5x.
=> 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[7+[2(1+h)^2+5(1+h)]+[2(1+2h)^2+5(1+2h)]+...] =
\lim_{h\to0}h[7n+9h(1+2+3+...+(n-1))+2h^2(1^2+2^2+3^2+...(n-1)^2)] =
\lim_{h\to0}h[7n+9h(\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{2}{n}[7n+\frac{18}{n}(\frac{n(n-1)}{2})+\frac{8}{n^2}(\frac{n(n-1)(2n-1)}{6})] =
\lim_{n\to\infty}[14+\frac{18n^2}{n^2}(1-\frac{1}{n})+\frac{8n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})] = 14 + 18 +
\frac{16}{3} =
\frac{112}{3} Therefore, the value of
\int_{1}^{3}(2x^2+5x)dx as limit of sum is\frac{112}{3} .
Question 33. \int_{1}^{3}(3x^2+1)dx
Solution:
We have,
I =
\int_{1}^{3}(3x^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 = 3 and f(x) = 3x2 + 1.
=> 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[4+[3(1+h)^2+1]+[3(1+2h)^2+1]+...] =
\lim_{h\to0}h[4n+6h(1+2+3+...(n-1))+3h^2(1^2+2^2+3^2+...(n-1)^2)] =
\lim_{h\to0}h[4n+6h(\frac{n(n-1)}{2})+3h^2(\frac{n(n-1)(2n-1)}{6})] Now if h −> 0, then n −> ∞. So, we have,
=
\lim_{n\to\infty}\frac{2}{n}[4n+\frac{12}{n}(\frac{n(n-1)}{2})+\frac{12}{n^2}(\frac{n(n-1)(2n-1)}{6})] =
\lim_{n\to\infty}[8+\frac{12n^2}{n^2}(1-\frac{1}{n})+\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})] = 8 + 12 + 8
= 28
Therefore, the value of
\int_{1}^{3}(3x^2+1)dx as limit of sum is 28.
Practice Questions on Definite Integrals
1. Evaluate: ∫[0 to π/2] (sin x)/(1 + cos x) dx
2. Calculate: ∫[0 to 1] x^2 / (1 + x^3) dx
3. Find the value of: ∫[0 to π/4] tan x dx
4. Evaluate: ∫[0 to 1] dx / (1 + x^2)^2
5. Compute: ∫[0 to π/2] cos^3 x dx
6. Calculate: ∫[0 to 1] x ln(1 + x) dx
7. Evaluate: ∫[0 to π/2] sin^2 x cos^2 x dx
8. Find the value of: ∫[1 to e] (ln x)^2 dx
9. Compute: ∫[0 to 1] x^2 e^x dx
10. Evaluate: ∫[0 to π/4] sec^3 x dx
Summary
Chapter 20 of RD Sharma's Class 12 mathematics textbook focuses on Definite Integrals. Exercise 20.5 | Set 3 deals with more advanced techniques for evaluating definite integrals. Key points covered in this exercise set include:
- Integration of trigonometric functions and their combinations
- Definite integrals involving algebraic and transcendental functions
- Application of substitution method in definite integrals
- Integration by parts for definite integrals
- Handling integrals with logarithmic and exponential functions
- Techniques for evaluating integrals with rational functions
- Use of trigonometric identities to simplify integrands
- Recognizing and applying standard integral formulas