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Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.5 | Set 3

  • Last Updated : 20 May, 2021

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.




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