Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 2

Evaluate the following definite integrals as limits of sums:

Question 12. $\int_{0}^{2}(x^2+4)dx$

Solution:

We have,
I = $\int_{0}^{2}(x^2+4)dx$
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 = 2 and f(x) = x² + 4.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[4+(h^2+4)+((2h)^2+4)+...+(((n-1)h)^2+4)]$
= $\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[4n+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[4n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[8+\frac{4n^3}{3n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= 8 + $\frac{4(2)}{3}$
= 8 + $\frac{8}{3}$
= $\frac{32}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+4)dx$ as limit of sum is $\frac{32}{3}$ .

Question 13. $\int_{1}^{4}(x^2-x)dx$

Solution:

We have,
I = $\int_{1}^{4}(x^2-x)dx$
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 = 4 and f(x) = x²− x.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[(1^2-1)+[(1+h)^2-(1+h)]+[(1+2h)^2-(1+2h)]+...+[(1+(n-1)h)^2-(1+(n-1)h)]]$
= $\lim_{h\to0}h[0+[h+h^2]+[2h+(2h)^2]+...+[(n-1)h+((n-1)h)^2]]$
= $\lim_{h\to0}h[h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[\frac{3}{n}(\frac{n(n-1)}{2})+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[\frac{9n^2}{2n^2}(1-\frac{1}{n})+\frac{3n^3}{2n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= $\frac{9}{2}+3$
= $\frac{15}{2}$
Therefore, the value of $\int_{1}^{4}(x^2-x)dx$ as limit of sum is $\frac{15}{2}$ .

Question 14. $\int_{0}^{1}(3x^2+5x)dx$

Solution:

We have,
I = $\int_{0}^{1}(3x^2+5x)dx$
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 = 0, b = 1 and f(x) = 3x² + 5x.
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[0+[3h^2+5h]+[3(2h)^2+5(2h)]+...+[3((n-1)h)^2+5((n-1)h)]$
= $\lim_{h\to0}h[3h^2(1^2+2^2+3^2+...+(n-1)^2)+5h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[3h^2(\frac{n(n-1)(2n-1)}{6})+5h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[\frac{3}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{5}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[\frac{n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{5n^2}{2n^2}(1-\frac{1}{n})]$
= 1 + $\frac{5}{2}$
= $\frac{7}{2}$
Therefore, the value of $\int_{0}^{1}(3x^2+5x)dx$ as limit of sum is $\frac{7}{2}$ .

Question 15. $\int_{0}^{2}e^xdx$

Solution:

We have,
I = $\int_{0}^{2}e^xdx$
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 = 0, b = 2 and f(x) = e^x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+e^h+e^{2h}+...+e^{(n-1)h}]$
= $\lim_{h\to0}h[\frac{(e^h)^n-1}{e^h-1}]$
= $\lim_{h\to0}h[\frac{e^{nh}-1}{e^h-1}]$
= $\lim_{h\to0}h[\frac{e^{2}-1}{e^h-1}]$
= $\lim_{h\to0}\left(\frac{e^{2}-1}{\frac{e^h-1}{h}}\right)$
= e² − 1
Therefore, the value of $\int_{0}^{2}e^xdx$ as limit of sum is e² − 1.

Question 16. $\int_{a}^{b}e^xdx$

Solution:

We have,
I = $\int_{a}^{b}e^xdx$
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) = e^x.
=> h = $\frac{b-a}{n}$
=> nh = b − a
So, we get,
I = $\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$
= $\lim_{h\to0}h[e^a+e^{a+h}+e^{a+2h}+...+e^{a+(n-1)h}]$
= $\lim_{h\to0}he^a[1+e^{h}+e^{2h}+...+e^{(n-1)h}]$
= $\lim_{h\to0}he^a[\frac{(e^h)^n-1}{e^h-1}]$
= $\lim_{h\to0}he^a[\frac{e^{nh}-1}{e^h-1}]$
= $\lim_{h\to0}he^a[\frac{e^{b-a}-1}{e^h-1}]$
= $\lim_{h\to0}e^a\left(\frac{e^{b-a}-1}{\frac{e^h-1}{h}}\right)$
= e^a (e^b-a−1)
= e^b − e^a
Therefore, the value of $\int_{a}^{b}e^xdx$ as limit of sum is e^b − e^a.

Question 17. $\int_{a}^{b}cosxdx$

Solution:

We have,
I = $\int_{a}^{b}cosxdx$
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) = cos x.
=> h = $\frac{b-a}{n}$
=> nh = b − a
So, we get,
I = $\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$
= $\lim_{h\to0}h[cosa+cos(a+h)+cos(a+2h)+...+cos(a+(n-1)h)]$
= $\lim_{h\to0}h\left[\frac{cos(a+(n-1)\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}h\left[\frac{cos(a+\frac{nh}{2}-\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}h\left[\frac{cos(a+\frac{b-a}{2}-\frac{h}{2})sin\frac{b-a}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}\left[\frac{\frac{h}{2}}{sin\frac{h}{2}}×2cos(a+\frac{b-a}{2}-\frac{h}{2})(sin\frac{b-a}{2})\right]$
= $\lim_{h\to0}\left[2cos(a+\frac{b-a}{2})sin(\frac{b-a}{2})\right]$
= $2cos(\frac{a+b}{2})sin(\frac{b-a}{2})$
= sin b − sin a
Therefore, the value of $\int_{a}^{b}cosxdx$ as limit of sum is sin b − sin a.

Question 18. $\int_{0}^{\frac{\pi}{2}}sinxdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}sinxdx$
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 = $\frac{\pi}{2}$ and f(x) = sin x.
=> h = $\frac{\pi}{2n}$
=> nh = $\frac{2}{\pi}$
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[sin0+sinh+sin2h+...+sin(n-1)h]$
= $\lim_{h\to0}h\frac{sin(\frac{(n-1)h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$
= $\lim_{h\to0}h\frac{sin(\frac{nh}{2}-\frac{h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$
= $\lim_{h\to0}h\frac{sin(\frac{\pi}{4}-\frac{h}{2})sin(\frac{\pi}{4})}{sin\frac{h}{2}}$
= $2(\frac{1}{2})$
= 1
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}sinxdx$ as limit of sum is 1.

Question 19. $\int_{0}^{\frac{\pi}{2}}cosxdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}cosxdx$
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 = $\frac{\pi}{2}$ and f(x) = cos x.
=> h = $\frac{\pi}{2n}$
=> nh = $\frac{2}{\pi}$
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[cos0+cosh+cos2h+...+cos(n-1)h]$
= $\lim_{h\to0}h[\frac{cos(\frac{nh}{2}-\frac{h}{2})cos\frac{nh}{2}}{cos\frac{h}{2}}]$
= $\lim_{h\to0}h[\frac{cos(\frac{\pi}{4}-\frac{h}{2})cos\frac{\pi}{4}}{cos\frac{h}{2}}]$
= $2(\frac{1}{2})$
= 1
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}cosxdx$ as limit of sum is 1.

Question 20. $\int_{1}^{4}(3x^2+2x)dx$

Solution:

We have,
I = $\int_{1}^{4}(3x^2+2x)dx$
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 =1, b = 4 and f(x) = 3x² + 2x.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(0)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[(3+2)+[3(1+h)^2+2(1+h)]+[3(1+2h)^2+2(1+2h)]+...+[3(1+(n-1)h)^2+2(1+(n-1)h)]]$
= $\lim_{h\to0}h[5n+8h(1+2+3...+(n-1))+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[5n+8h(\frac{n(n-1)}{2})+3h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[5n+\frac{24}{n}(\frac{n(n-1)}{2})+\frac{27}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[15+\frac{36n^2}{n^2}(1-\frac{1}{n})+\frac{27n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 15 + 36 + 27
= 78
Therefore, the value of $\int_{1}^{4}(3x^2+2x)dx$ as limit of sum is 78.

Question 21. $\int_{0}^{2}(3x^2-2)dx$

Solution:

We have,
I = $\int_{0}^{2}(3x^2-2)dx$
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 =0, b = 2 and f(x) = 3x² − 2.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[-2+[3h^2-2]+[3(2h)^2-2]+...+[3((n-1)h)^2-2]]$
= $\lim_{h\to0}h[-2n+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[-2n+3h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[-2n+\frac{12}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[-4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= −4 + 8
= 4
Therefore, the value of $\int_{0}^{2}(3x^2-2)dx$ as limit of sum is 4.

Question 22. $\int_{0}^{2}(x^2+2)dx$

Solution:

We have,
I = $\int_{0}^{2}(x^2+2)dx$
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 =0, b = 2 and f(x) = x² + 2.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[2+(h^2+2)+[((2h)^2+2)]+...+[((n-1)h)^2+2]]$
= $\lim_{h\to0}h[2n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[2n+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[2+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= 4 + $\frac{4(2)}{3}$
= 4 + $\frac{8}{3}$
= $\frac{20}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+2)dx$ as limit of sum is $\frac{20}{3}$ .

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Last Updated : 20 May, 2021

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Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 2

Evaluate the following definite integrals as limits of sums:

Question 12. $\int_{0}^{2}(x^2+4)dx$

Question 13. $\int_{1}^{4}(x^2-x)dx$

Question 14. $\int_{0}^{1}(3x^2+5x)dx$

Question 15. $\int_{0}^{2}e^xdx$

Question 16. $\int_{a}^{b}e^xdx$

Question 17. $\int_{a}^{b}cosxdx$

Question 18. $\int_{0}^{\frac{\pi}{2}}sinxdx$

Question 19. $\int_{0}^{\frac{\pi}{2}}cosxdx$

Question 20. $\int_{1}^{4}(3x^2+2x)dx$

Question 21. $\int_{0}^{2}(3x^2-2)dx$

Question 22. $\int_{0}^{2}(x^2+2)dx$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 2

Evaluate the following definite integrals as limits of sums:

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 12. $\int_{0}^{2}(x^2+4)dx$

Question 13. $\int_{1}^{4}(x^2-x)dx$

Question 14. $\int_{0}^{1}(3x^2+5x)dx$

Question 15. $\int_{0}^{2}e^xdx$

Question 16. $\int_{a}^{b}e^xdx$

Question 17. $\int_{a}^{b}cosxdx$

Question 18. $\int_{0}^{\frac{\pi}{2}}sinxdx$

Question 19. $\int_{0}^{\frac{\pi}{2}}cosxdx$

Question 20. $\int_{1}^{4}(3x^2+2x)dx$

Question 21. $\int_{0}^{2}(3x^2-2)dx$

Question 22. $\int_{0}^{2}(x^2+2)dx$