{"id":626500,"date":"2023-09-12T04:10:44","date_gmt":"2023-09-12T04:10:44","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626500"},"modified":"2023-09-12T04:11:31","modified_gmt":"2023-09-12T04:11:31","slug":"kc-sinha-exercise-21-2-mathematics-solution-class-12-chapter-21-nishchit-samakalano-ke-gundharm","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-2-mathematics-solution-class-12-chapter-21-nishchit-samakalano-ke-gundharm\/","title":{"rendered":"KC Sinha: Exercise 21.2- Mathematics Solution Class 12 Chapter 21 \u0928\u093f\u0936\u094d\u091a\u093f\u0924 \u0938\u092e\u0915\u0932\u0928\u094b\u0902 \u0915\u0947 \u0917\u0941\u0923\u0927\u0930\u094d\u092e"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

(i)<\/strong> $\\int_{0}^{2} x d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=0, b=2<\/p>\n\n\n\n

nh=2-0<\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=x<\/p>\n\n\n\n

$\\int_{0}^{2} x d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{x=1}^{n} f(0+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} rh$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h^2 \\sum_{r=1}^{n} r$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h^{2} \\frac{n(n+1)}{2}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\times h \\frac{n(n+1)}{2}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{nh(nh+h)}{2}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{2(2+h)}{2}$<\/p>\n\n\n\n

$=\\frac{2(2+0)}{2}$<\/p>\n\n\n\n

=2<\/p>\n\n\n\n

(ii)<\/strong> $\\int_{0}^{5}(x+1) d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=0, nh=5-0=5<\/p>\n\n\n\n

f(x)=x+1<\/p>\n\n\n\n

$\\int_{0}^{5}(x+1) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(0+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(r h+1)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\sum_{r=1}^{n} r h+\\sum_{r=1}^{n} 1\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[h\\sum_{r=1}^{n} r+n\\right]$<\/p>\n\n\n\n

$=\\lim_{h\\rightarrow 0} h\\left[h \\frac{n(n+1)}{2}+n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left(h^{2} \\frac{n(n+1)}{2}+n h\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left(\\frac{nh(nh+h)}{2}+n h\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left(\\frac{5(5+h)}{2}+5\\right)$<\/p>\n\n\n\n

$=\\frac{5(5+0)}{2}+5=\\frac{25}{2}+5=\\frac{35}{2}$<\/p>\n\n\n\n

(iii)<\/strong> $\\int_{0}^{2}(x+4) d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=0, b=2, nh=2<\/p>\n\n\n\n

f(x)=x+4<\/p>\n\n\n\n

$\\int_{0}^{2}(x+4) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(0+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\sum_{r=1}^{n} f(rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(rh+4)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\sum_{r=1}^{n} rh+\\sum_{r=1}^{n} 4\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[h\\sum_{r=1}^{n} r+4n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left(\\frac{hn(n+1)}{2}+4 n\\right]$<\/p>\n\n\n\n

$=\\lim_{h\\rightarrow 0}\\left[h^{2} \\frac{n(n+1)}{2}+4 nh\\right]$<\/p>\n\n\n\n

$=\\lim_{h\\rightarrow 0} \\left[ \\frac{nh(nh+h)}{2}\\right]+4nh$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{2(2+h)}{2}+4 \\times 2\\right]$<\/p>\n\n\n\n

=2+0+8<\/p>\n\n\n\n

=10<\/p>\n\n\n\n

(v)<\/strong> $\\int_{-1}^{1}(x+3) dx$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=-1, b=1<\/p>\n\n\n\n

nh=1-(-1)<\/p>\n\n\n\n

f(x)=x+3<\/p>\n\n\n\n

$\\int_{-1}^{1}(x+3) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(-1+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(-1+rh+3)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(rh+2)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\sum_{r=1}^{n} rh+\\sum_{r=1}^{n} 2\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[h \\frac{n(n+1)}{2}+2 n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{nh(nh+h)}{2}+2 nh\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{2(2+h)}{2}+2 \\times 2\\right]$<\/p>\n\n\n\n

=2+0+4<\/p>\n\n\n\n

=6<\/p>\n\n\n\n

(ix)<\/strong> $\\int_{a}^{b} x dx$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

f(x)=x ; nh=b-a<\/p>\n\n\n\n

$\\int_{a}^{b} f(x) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(a+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\sum_{r=1}^{n} a+\\sum_{r=1}^{n} rh\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\operatorname{an}+h \\frac{n(n-1)}{2}\\right]$<\/p>\n\n\n\n

$=\\lim_{h\\rightarrow 0}\\left[a nh+\\frac{ nh(nh+h)}{2}\\right)$<\/p>\n\n\n\n

$=\\lim_{h \\rightarrow 0}\\left[a(b-a)+\\frac{(b-a)(b-a+h)}{2}\\right]$<\/p>\n\n\n\n

$=\\left[a(b-a)+\\frac{b-a)(b-a)}{2}\\right]$<\/p>\n\n\n\n

$=(b-a)\\left[a+\\frac{b-a}{2}\\right]$<\/p>\n\n\n\n

$=(b-a)\\left[\\frac{2a+b-a}{2}\\right]$<\/p>\n\n\n\n

$=\\frac{(b-a)(b+a)}{2}$<\/p>\n\n\n\n

$=\\frac{b^{2}-a^{2}}{2}$<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

$\\int_{1}^{2} x^{2} d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=1 ,b=2<\/p>\n\n\n\n

nh=1, <\/p>\n\n\n\n

f(x)=x2<\/sup><\/p>\n\n\n\n

$\\int_{1}^{2} f(x) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h\\rightarrow 0} h \\sum_{r=1}^{n}f(1+x h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(1+r h)^{2}$<\/p>\n\n\n\n

$=\\lim_{h\\rightarrow 0} h \\sum_{r=1}^{n}\\left(1+r^{2}h^2+2 rh\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\sum_{r=1}^{n}1+h\\sum_{r=1}^{n} r^{2}+2 h \\sum_{r=1}^{n} r$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[n+h^{2} \\frac{n(n+1)(2 n+1)}{6}+2h\\frac{n(n+1)}{2}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[n h+\\frac{nh(nh+h)(2nh+h)}{6}+n h(nh+h)\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[1+\\frac{1(1+h)(2\\times 1+h)}{6}+1(1+b)\\right]$<\/p>\n\n\n\n

$=1+\\frac{2}{6}+1$<\/p>\n\n\n\n

$=2+\\frac{1}{3}=\\frac{7}{3}$<\/p>\n\n\n\n

(v)<\/strong> $\\int_{1}^{3}\\left(2 x^{2}+5\\right) d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=1, b=3, <\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=2x2<\/sup>+5<\/p>\n\n\n\n

$\\int_{1}^{3}\\left(2 x^{2}+5\\right) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}f(1+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2(1+rh)^{2}+5\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2\\left(1+r^{2} h^{2}+2 r h\\right)+5\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2+2 r^2 h^{2}+4rh+5\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2 r^{2} h^{2}+4rh+7\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[2 h\\sum_{r=1}^{n} r^{2}+4h\\sum_{r=1}^{n} r+\\sum_{r=1}^{n} 7\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[2h^{2} \\frac{n(n+1)(2 n+1)}{6}+4 h \\frac{n(n+1)}{2}+7 n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{n h(n h+h)(2nh+h)}{3}+2 n h(n h+h)+7 n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{2(2+h)(2\\times 2+4)}{3}+2 \\times 2(2+h)+7 \\times 2\\right]$<\/p>\n\n\n\n

$=\\frac{2(2+0)(4+0)}{3}+4(2+0)+14$<\/p>\n\n\n\n

$=\\frac{16}{3}+8+14$<\/p>\n\n\n\n

$=\\frac{16}{3}+22$<\/p>\n\n\n\n

$=\\frac{16+66}{3}=\\frac{82}{3}$<\/p>\n\n\n\n

(vi)<\/strong> $\\int_{1}^{3}\\left(x^{2}+x\\right) d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=1, b=3,<\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=x2<\/sup>+x<\/p>\n\n\n\n

$\\int_{1}^{3} \\left(x^{2}+x\\right) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(1+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[(1+r h)^{2}+1+r h\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[1+r^2h^{2}+2rh+1+x h\\right]$<\/p>\n\n\n\n

$=\\lim _{x \\rightarrow 0} h \\sum_{r=1}^{n}\\left[r^{2} h^{2}+3rh+1\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[h^{2} \\sum_{r=1}^{n} r^{2}+3 h \\sum_{r=1}^{n} r+\\sum_{r=1}^{n} 1\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[h^{2} \\frac{n(n+1)(2n+1)}{6}+3 h \\frac{n(n+1)+n}{2}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{n h\\left(nh+h\\right)(2nh+h)}{6}+3 n \\frac{h(nh+h)}{2}+nh\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{2(2+h)(2 \\cdot 2+h)}{6}+\\frac{3 \\cdot 2(2+h)}{2}+2\\right]$<\/p>\n\n\n\n

$=\\frac{(2+0)(4+0)}{3}+3(2+0)+2$<\/p>\n\n\n\n

$=\\frac{8}{3}+6+2$<\/p>\n\n\n\n

$=\\frac{8}{3}+8$<\/p>\n\n\n\n

$=\\frac{8+24}{3}=\\frac{32}{3}$<\/p>\n\n\n\n

(xiii)<\/strong> $\\int_{3}^{1}\\left(2 x^{2}+5 x\\right) d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=3, b=1,<\/p>\n\n\n\n

nh=-2<\/p>\n\n\n\n

f(x)=2x2<\/sup>+5x<\/p>\n\n\n\n

$\\int_{3}^{1}\\left(2 x^{2}+5 x\\right) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(3+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\sum_{r=1}^{n}\\left[2(3+r h)^{2}+5(3+r h)\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2\\left(9+r^{2}h^2+6rh)+15+5 rh\\right]\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[18+2 r^{2}h^2+12rh+15+5rh\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left[2 \\pi^{2} h^{2}+17 r h+33\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[2 h^{2} \\sum_{r=1}^{n} r^{2}+\\operatorname{17h} \\sum_{r=1}^{n} r+\\sum_{r=1}^{n} 33\\right]$<\/p>\n\n\n\n

$=\\lim_{h \\rightarrow 0} h\\left[2 h^{2} \\frac{n(n+1)(2 n+1)}{6}+17 h \\frac{n(n+1)}{2}+33 n\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{nh(nh+h)(2 n h+h)}{3}+\\frac{17}{2} n h(nh+h)+33 n h\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0}\\left[\\frac{-2(-2+h)(-4+h)}{3}+\\frac{17}{2}(-2)(-2+0)+33(-2)\\right]$<\/p>\n\n\n\n

$=\\frac{-2(-2+0)(-4+0)}{3}+\\frac{17}{2} \\times 4-66)$<\/p>\n\n\n\n

$=\\frac{-16}{3}+34-66$<\/p>\n\n\n\n

$=-\\frac{16}{3}-32$<\/p>\n\n\n\n

$=-\\frac{16}{3}-32$<\/p>\n\n\n\n

$=\\frac{-16-96}{3}=\\frac{-112}{3}$<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

$\\int_{0}^{2} e^{x} d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=0, b=2<\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=ex<\/sup><\/p>\n\n\n\n

$\\int_{0}^{2} e^{x} d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(x h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\operatorname{h} \\sum_{r=1}^{n} e^{r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\left(e^{h}\\right)+e^{2 h}+e^{3 h}+e^{4 h}+\\ldots+e^{nh}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{h}\\left(\\frac{\\left(e^{h}\\right)^{n}-1}{e^{h}-1}\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{h}\\left(\\frac{e^{n h}-1}{e^{h}-1}\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} {h}e^h \\frac{\\left(e^{2}-1\\right)}{\\left(\\frac{e^{h}-1}{h}\\right)\\times h}$<\/p>\n\n\n\n

=e0<\/sup>(e2<\/sup>-1)=e2<\/sup>-1<\/p>\n\n\n\n

(ii)<\/strong> $\\int_{1}^{3} e^{-x} d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=1, b=3<\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=e-x<\/sup><\/p>\n\n\n\n

$\\int_{1}^{3} e^{-\\pi} d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(1+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{-(1+r h)}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{-1-r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n}\\left({e}^{-1}\\right) e^{-rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{h}{e} \\sum_{r=1}^{n} e^{-rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{h}{e}\\left[e^{-h}+e^{-2 h}+e^{-3 h}+\\cdots+e^{-n h}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{h}{e} e^{-h \\left(\\frac{(e^{-h})^{n}-1}{e^{-h}-1}\\right)}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{h}{e} e^{-h}\\left(\\frac{e^{-n h}-1}{e^{-h}-1}\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} \\frac{h}{e} e^{-h}\\left(\\frac{e^{-2}-1}{\\left(\\frac{e^{-1}-1}{-h}\\right)(-h)}\\right)$<\/p>\n\n\n\n

$=-\\lim _{h \\rightarrow 0} \\frac{1}{e} e^{-h}\\left(e^{-2}-1\\right)$<\/p>\n\n\n\n

$=-\\left(e^{-3}-e^{-1}\\right)$<\/p>\n\n\n\n

=e-1<\/sup>-e-2<\/sup><\/p>\n\n\n\n

$=\\frac{1}{e}-\\frac{1}{e^{3}}$<\/p>\n\n\n\n

$=\\frac{e^{2}-1}{e^{3}}$<\/p>\n\n\n\n

(iii)<\/strong> $\\int_{a}^{b} e^{x} dx$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

nh=b-a<\/p>\n\n\n\n

f(x)=ex<\/sup><\/p>\n\n\n\n

$\\int_{a}^{b} f(x) d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{a+r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{a} \\cdot e^{rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{a} \\sum_{r=1}^{n} e^{r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{a}\\left[e^{h}+e^{2 h}+e^{3 h}+\\cdots+e^{n h}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} he^{a}e^h{\\left(\\frac{e^{\\operatorname{nh}}-1}{\\frac{e^h-1}{h}\\times h}\\right)}$<\/p>\n\n\n\n

=ea<\/sup>(=eb-a<\/sup>-1)<\/p>\n\n\n\n

=ea+b-a<\/sup>-ea<\/sup><\/p>\n\n\n\n

=eb<\/sup>-ea<\/sup><\/p>\n\n\n\n

(iv)<\/strong> $\\int_{0}^{1} e^{2-3 x} d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=0, b=1<\/p>\n\n\n\n

nh=1<\/p>\n\n\n\n

f(x)=e2-3x<\/sup><\/p>\n\n\n\n

$\\int_{0}^{1} e^{2-3x} d x=\\lim_{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{2-3r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} e^{2} \\cdot e^{-3r h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{2} \\sum_{r=1}^{n} e^{-3rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} he^{2}\\left[e^{-3(h)}+e^{-3(2 h)}+e^{-3(3h)}+\\cdots \\cdot e^{-3nh}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h e^{x} e^{-3h}\\left(\\frac{e^{-3 n h}-1}{\\frac{e^{-3h}-1}{-3 h}\\times (-3 h)} \\right)$<\/p>\n\n\n\n

$=-\\frac{1}{3} \\lim _{h \\rightarrow 0} e^{x} e^{-3 h}\\left(e^{-3}-1\\right)$<\/p>\n\n\n\n

$=-\\frac{1}{3} e^{2\\left(e^{-3}-1\\right)}$<\/p>\n\n\n\n

$=-\\frac{1}{3}\\left(e^{-1}-e^{2}\\right)$<\/p>\n\n\n\n

$=\\frac{e^{2}-e^{-1}}{3}$<\/p>\n\n\n\n

(v)<\/strong> $\\int_{2}^{4} 2^{x} d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

a=2,b=4<\/p>\n\n\n\n

nh=2<\/p>\n\n\n\n

f(x)=2x<\/sup><\/p>\n\n\n\n

$\\int_{2}^{4} 2^{x} d n=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(2+rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} 2^{2+rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} 2^{2} \\cdot 2^{2 h}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} 4 h \\sum_{r=1}^{n} 2^{rh}$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} 4 h\\left[2^{h}+2^{2 h}+2^{2 h}+\\cdots-+2^{nh}\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} 4h2^{h}\\left(\\frac{2^{nh}-1}{\\frac{2^{nh}-1}{h}\\times h}\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} 4 \\cdot 2^h{\\left(\\frac{2^{2}-1}{\\log 2}\\right)}$<\/p>\n\n\n\n

$=\\frac{4 \\times 3}{\\log 2}$<\/p>\n\n\n\n

$=\\frac{12}{\\log 2}$<\/p>\n\n\n\n

(vii)<\/strong> $\\int_{0}^{\\pi \/2} \\cos x d x$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\sin (a+h)+\\sin (a+2 h)+\\sin (a+3 h)+\\cdots+\\sin \\left(a+n^{h}\\right)=\\frac{\\sin \\left(\\frac{n h}{2}\\right)}{\\sin \\left(\\frac{h}{2}\\right)} \\sin \\left(\\frac{(a+h)+(a+n h)}{2}\\right)$<\/p>\n\n\n\n

$\\cos (a+h)+\\cos (a+2 h)+\\cos (a+3h)+-\\cdots+\\cos \\left(a+n^{h}\\right)=\\frac{\\sin \\left(\\frac{n h}{2}\\right)}{\\sin \\left(\\frac{h}{2}\\right)} \\cos \\left(\\frac{(a+h)+\\left(a+n^{h}\\right)}{2}\\right)$<\/p>\n\n\n\n

a=0, $b=\\frac{\\pi}{2},n h=\\frac{\\pi}{2}$<\/p>\n\n\n\n

f(x)=cosx <\/p>\n\n\n\n

$\\int_{0}^{\\pi\/2} \\cos d n=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+r h)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(rh)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} \\cos rh$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\cos h+\\cos 2h+\\cos 3h+\\cos 4h+\\dots+\\cos rh\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h \\frac{\\sin \\left(\\frac{n h}{2}\\right)}{\\sin \\left(\\frac{h}{2}\\right)} \\cos \\left(\\frac{h+nh}{2}\\right)$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} h\\left[\\frac{\\sin \\pi \/4}{\\frac{\\sin \\frac{h}{2}}{\\frac{h}{2}}\\times \\frac{h}{2}} \\sin \\left(\\frac{h+\\pi\/2}{2}\\right)\\right]$<\/p>\n\n\n\n

$=\\lim _{h \\rightarrow 0} 2\\left[\\frac{1}{\\sqrt2} \\sin \\left(\\frac{h+\\pi \/2}{2}\\right)\\right]$<\/p>\n\n\n\n

$=\\quad 2\\left[\\frac{1}{\\sqrt{2}} \\sin \\frac{\\pi}{4}\\right]$<\/p>\n\n\n\n

$=2 \\frac{1}{\\sqrt2} \\cdot \\frac{1}{\\sqrt2}$<\/p>\n\n\n\n

$=2 \\times \\frac{1}{2}=1$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

$\\lim _{n \\rightarrow \\infty}\\left(\\frac{1}{n^{2}}+\\frac{2}{n^{2}}+\\frac{3}{n^{2}}+\\cdots+\\frac{n-1}{n^{2}}\\right)$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{n-1} \\frac{r}{n^{2}}$<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{n-1}\\left(\\frac{r}{n}\\right) \\cdot \\frac{1}{n}$<\/p>\n\n\n\n

$=\\int_{0}^{1} xdx$<\/p>\n\n\n\n

$=\\frac{\\left[x^{2}\\right]_{0}^{1}}{2}$<\/p>\n\n\n\n

$=\\frac{1-0}{2}=\\frac{1}{2}$<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

$\\lim _{n \\rightarrow \\infty}\\left[\\frac{1}{n+0}+\\frac{1}{n+1}+\\cdots+\\frac{1}{n+5 n}\\right]$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{5n} \\frac{1}{n+r}$<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{n=1}^{5 n} \\frac{1}{n+\\left(\\frac{r}{h}\\right)}$<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{5n}\\left(\\frac{1}{1+\\frac{r}{n}}\\right)\\frac{1}{n}$<\/p>\n\n\n\n

$=\\int_{0}^{5} \\frac{1}{1+x} d x$<\/p>\n\n\n\n

$=[\\log |1+x|]_{0}^{5}$<\/p>\n\n\n\n

=log6-log1<\/p>\n\n\n\n

=log6<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

$\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{n-1} \\frac{1}{\\sqrt{4 n^{2}-r^{2}}}$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{n-1} \\frac{1}{n \\sqrt{4-(\\frac{r}{n})^{2}}}$<\/p>\n\n\n\n

$=\\lim _{n \\rightarrow \\infty} \\sum_{r=1}^{n-1} \\frac{1}{n} \\frac{1}{\\sqrt{4-(\\frac{r}{n}})^{2}}$<\/p>\n\n\n\n

$=\\int_{0}^{1} \\frac{1}{\\sqrt{4-x^{2}}} d x$<\/p>\n\n\n\n

$=\\left[\\sin \\frac{x}{2}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\sin^{-1} \\frac{1}{2}-\\sin ^{-1} \\frac{0 }{2}$<\/p>\n\n\n\n

$=\\frac{\\pi}{6}-0$<\/p>\n\n\n\n

$=\\frac{\\pi}{6}$<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 12 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 (i) $\\int_{0}^{2} x d x$ Sol : a=0, b=2 nh=2-0 nh=2 f(x)=x $\\int_{0}^{2} x d x=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(a+rh)$ $=\\lim _{h \\rightarrow 0} h \\sum_{x=1}^{n} f(0+rh)$ $=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} f(rh)$ $=\\lim _{h \\rightarrow 0} h \\sum_{r=1}^{n} rh$ $=\\lim _{h \\rightarrow 0} h^2 \\sum_{r=1}^{n} r$ $=\\lim _{h \\rightarrow 0} […]<\/p>\n","protected":false},"author":302,"featured_media":626507,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[25],"tags":[],"boards":[],"class_list":["post-626500","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 21.2- Mathematics Solution Class 12 Chapter 21 \u0928\u093f\u0936\u094d\u091a\u093f\u0924 \u0938\u092e\u0915\u0932\u0928\u094b\u0902 \u0915\u0947 \u0917\u0941\u0923\u0927\u0930\u094d\u092e - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 (i) $int_{0}^{2} x d x$ Sol : a=0, b=2 nh=2-0 nh=2 f(x)=x $int_{0}^{2} x d x=lim _{h rightarrow 0} h sum_{r=1}^{n} f(a+rh)$ $=lim\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-2-mathematics-solution-class-12-chapter-21-nishchit-samakalano-ke-gundharm\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 21.2- Mathematics Solution Class 12 Chapter 21 \u0928\u093f\u0936\u094d\u091a\u093f\u0924 \u0938\u092e\u0915\u0932\u0928\u094b\u0902 \u0915\u0947 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content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" 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