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

  • Last Updated : 20 May, 2021

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}.

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