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

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  • Last Updated : 20 May, 2021
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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) = ex.

=> 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) = ex.

=> 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)

= ea (eb-a −1)

= eb − ea

Therefore, the value of\int_{a}^{b}e^xdx  as limit of sum is eb − ea.

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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