Exercise 21.2
Question 1
(i) $\int_{0}^{2} x d x$
Sol :
a=0, b=2
nh=2-0
nh=2
f(x)=x
$\int_{0}^{2} x d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+rh)$
$=\lim _{h \rightarrow 0} h \sum_{x=1}^{n} f(0+rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} rh$
$=\lim _{h \rightarrow 0} h^2 \sum_{r=1}^{n} r$
$=\lim _{h \rightarrow 0} h^{2} \frac{n(n+1)}{2}$
$=\lim _{h \rightarrow 0} h \times h \frac{n(n+1)}{2}$
$=\lim _{h \rightarrow 0} \frac{nh(nh+h)}{2}$
$=\lim _{h \rightarrow 0} \frac{2(2+h)}{2}$
$=\frac{2(2+0)}{2}$
=2
(ii) $\int_{0}^{5}(x+1) d x$
Sol :
a=0, nh=5-0=5
f(x)=x+1
$\int_{0}^{5}(x+1) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(0+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(r h+1)$
$=\lim _{h \rightarrow 0} h\left[\sum_{r=1}^{n} r h+\sum_{r=1}^{n} 1\right]$
$=\lim _{h \rightarrow 0}\left[h\sum_{r=1}^{n} r+n\right]$
$=\lim_{h\rightarrow 0} h\left[h \frac{n(n+1)}{2}+n\right]$
$=\lim _{h \rightarrow 0}\left(h^{2} \frac{n(n+1)}{2}+n h\right)$
$=\lim _{h \rightarrow 0}\left(\frac{nh(nh+h)}{2}+n h\right)$
$=\lim _{h \rightarrow 0}\left(\frac{5(5+h)}{2}+5\right)$
$=\frac{5(5+0)}{2}+5=\frac{25}{2}+5=\frac{35}{2}$
(iii) $\int_{0}^{2}(x+4) d x$
Sol :
a=0, b=2, nh=2
f(x)=x+4
$\int_{0}^{2}(x+4) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(0+r h)$
$=\lim _{h \rightarrow 0} h\sum_{r=1}^{n} f(rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(rh+4)$
$=\lim _{h \rightarrow 0} h\left[\sum_{r=1}^{n} rh+\sum_{r=1}^{n} 4\right]$
$=\lim _{h \rightarrow 0} h\left[h\sum_{r=1}^{n} r+4n\right]$
$=\lim _{h \rightarrow 0} h\left(\frac{hn(n+1)}{2}+4 n\right]$
$=\lim_{h\rightarrow 0}\left[h^{2} \frac{n(n+1)}{2}+4 nh\right]$
$=\lim_{h\rightarrow 0} \left[ \frac{nh(nh+h)}{2}\right]+4nh$
$=\lim _{h \rightarrow 0}\left[\frac{2(2+h)}{2}+4 \times 2\right]$
=2+0+8
=10
(v) $\int_{-1}^{1}(x+3) dx$
Sol :
a=-1, b=1
nh=1-(-1)
f(x)=x+3
$\int_{-1}^{1}(x+3) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(-1+rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(-1+rh+3)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(rh+2)$
$=\lim _{h \rightarrow 0} h\left[\sum_{r=1}^{n} rh+\sum_{r=1}^{n} 2\right]$
$=\lim _{h \rightarrow 0} h\left[h \frac{n(n+1)}{2}+2 n\right]$
$=\lim _{h \rightarrow 0}\left[\frac{nh(nh+h)}{2}+2 nh\right]$
$=\lim _{h \rightarrow 0}\left[\frac{2(2+h)}{2}+2 \times 2\right]$
=2+0+4
=6
(ix) $\int_{a}^{b} x dx$
Sol :
f(x)=x ; nh=b-a
$\int_{a}^{b} f(x) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(a+rh)$
$=\lim _{h \rightarrow 0} h\left[\sum_{r=1}^{n} a+\sum_{r=1}^{n} rh\right]$
$=\lim _{h \rightarrow 0} h\left[\operatorname{an}+h \frac{n(n-1)}{2}\right]$
$=\lim_{h\rightarrow 0}\left[a nh+\frac{ nh(nh+h)}{2}\right)$
$=\lim_{h \rightarrow 0}\left[a(b-a)+\frac{(b-a)(b-a+h)}{2}\right]$
$=\left[a(b-a)+\frac{b-a)(b-a)}{2}\right]$
$=(b-a)\left[a+\frac{b-a}{2}\right]$
$=(b-a)\left[\frac{2a+b-a}{2}\right]$
$=\frac{(b-a)(b+a)}{2}$
$=\frac{b^{2}-a^{2}}{2}$
Question 2
$\int_{1}^{2} x^{2} d x$
Sol :
a=1 ,b=2
nh=1,
f(x)=x2
$\int_{1}^{2} f(x) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h\rightarrow 0} h \sum_{r=1}^{n}f(1+x h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(1+r h)^{2}$
$=\lim_{h\rightarrow 0} h \sum_{r=1}^{n}\left(1+r^{2}h^2+2 rh\right)$
$=\lim _{h \rightarrow 0} h\sum_{r=1}^{n}1+h\sum_{r=1}^{n} r^{2}+2 h \sum_{r=1}^{n} r$
$=\lim _{h \rightarrow 0} h\left[n+h^{2} \frac{n(n+1)(2 n+1)}{6}+2h\frac{n(n+1)}{2}\right]$
$=\lim _{h \rightarrow 0}\left[n h+\frac{nh(nh+h)(2nh+h)}{6}+n h(nh+h)\right]$
$=\lim _{h \rightarrow 0}\left[1+\frac{1(1+h)(2\times 1+h)}{6}+1(1+b)\right]$
$=1+\frac{2}{6}+1$
$=2+\frac{1}{3}=\frac{7}{3}$
(v) $\int_{1}^{3}\left(2 x^{2}+5\right) d x$
Sol :
a=1, b=3,
nh=2
f(x)=2x2+5
$\int_{1}^{3}\left(2 x^{2}+5\right) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}f(1+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2(1+rh)^{2}+5\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2\left(1+r^{2} h^{2}+2 r h\right)+5\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2+2 r^2 h^{2}+4rh+5\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2 r^{2} h^{2}+4rh+7\right]$
$=\lim _{h \rightarrow 0} h\left[2 h\sum_{r=1}^{n} r^{2}+4h\sum_{r=1}^{n} r+\sum_{r=1}^{n} 7\right]$
$=\lim _{h \rightarrow 0} h\left[2h^{2} \frac{n(n+1)(2 n+1)}{6}+4 h \frac{n(n+1)}{2}+7 n\right]$
$=\lim _{h \rightarrow 0}\left[\frac{n h(n h+h)(2nh+h)}{3}+2 n h(n h+h)+7 n\right]$
$=\lim _{h \rightarrow 0}\left[\frac{2(2+h)(2\times 2+4)}{3}+2 \times 2(2+h)+7 \times 2\right]$
$=\frac{2(2+0)(4+0)}{3}+4(2+0)+14$
$=\frac{16}{3}+8+14$
$=\frac{16}{3}+22$
$=\frac{16+66}{3}=\frac{82}{3}$
(vi) $\int_{1}^{3}\left(x^{2}+x\right) d x$
Sol :
a=1, b=3,
nh=2
f(x)=x2+x
$\int_{1}^{3} \left(x^{2}+x\right) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(1+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[(1+r h)^{2}+1+r h\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[1+r^2h^{2}+2rh+1+x h\right]$
$=\lim _{x \rightarrow 0} h \sum_{r=1}^{n}\left[r^{2} h^{2}+3rh+1\right]$
$=\lim _{h \rightarrow 0} h\left[h^{2} \sum_{r=1}^{n} r^{2}+3 h \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1\right]$
$=\lim _{h \rightarrow 0} h\left[h^{2} \frac{n(n+1)(2n+1)}{6}+3 h \frac{n(n+1)+n}{2}\right]$
$=\lim _{h \rightarrow 0}\left[\frac{n h\left(nh+h\right)(2nh+h)}{6}+3 n \frac{h(nh+h)}{2}+nh\right]$
$=\lim _{h \rightarrow 0}\left[\frac{2(2+h)(2 \cdot 2+h)}{6}+\frac{3 \cdot 2(2+h)}{2}+2\right]$
$=\frac{(2+0)(4+0)}{3}+3(2+0)+2$
$=\frac{8}{3}+6+2$
$=\frac{8}{3}+8$
$=\frac{8+24}{3}=\frac{32}{3}$
(xiii) $\int_{3}^{1}\left(2 x^{2}+5 x\right) d x$
Sol :
a=3, b=1,
nh=-2
f(x)=2x2+5x
$\int_{3}^{1}\left(2 x^{2}+5 x\right) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(3+r h)$
$=\lim _{h \rightarrow 0}\sum_{r=1}^{n}\left[2(3+r h)^{2}+5(3+r h)\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2\left(9+r^{2}h^2+6rh)+15+5 rh\right]\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[18+2 r^{2}h^2+12rh+15+5rh\right]$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left[2 \pi^{2} h^{2}+17 r h+33\right]$
$=\lim _{h \rightarrow 0} h\left[2 h^{2} \sum_{r=1}^{n} r^{2}+\operatorname{17h} \sum_{r=1}^{n} r+\sum_{r=1}^{n} 33\right]$
$=\lim_{h \rightarrow 0} h\left[2 h^{2} \frac{n(n+1)(2 n+1)}{6}+17 h \frac{n(n+1)}{2}+33 n\right]$
$=\lim _{h \rightarrow 0}\left[\frac{nh(nh+h)(2 n h+h)}{3}+\frac{17}{2} n h(nh+h)+33 n h\right]$
$=\lim _{h \rightarrow 0}\left[\frac{-2(-2+h)(-4+h)}{3}+\frac{17}{2}(-2)(-2+0)+33(-2)\right]$
$=\frac{-2(-2+0)(-4+0)}{3}+\frac{17}{2} \times 4-66)$
$=\frac{-16}{3}+34-66$
$=-\frac{16}{3}-32$
$=-\frac{16}{3}-32$
$=\frac{-16-96}{3}=\frac{-112}{3}$
Question 3
$\int_{0}^{2} e^{x} d x$
Sol :
a=0, b=2
nh=2
f(x)=ex
$\int_{0}^{2} e^{x} d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(x h)$
$=\lim _{h \rightarrow 0} \operatorname{h} \sum_{r=1}^{n} e^{r h}$
$=\lim _{h \rightarrow 0} h\left[\left(e^{h}\right)+e^{2 h}+e^{3 h}+e^{4 h}+\ldots+e^{nh}\right]$
$=\lim _{h \rightarrow 0} h e^{h}\left(\frac{\left(e^{h}\right)^{n}-1}{e^{h}-1}\right)$
$=\lim _{h \rightarrow 0} h e^{h}\left(\frac{e^{n h}-1}{e^{h}-1}\right)$
$=\lim _{h \rightarrow 0} {h}e^h \frac{\left(e^{2}-1\right)}{\left(\frac{e^{h}-1}{h}\right)\times h}$
=e0(e2-1)=e2-1
(ii) $\int_{1}^{3} e^{-x} d x$
Sol :
a=1, b=3
nh=2
f(x)=e-x
$\int_{1}^{3} e^{-\pi} d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(1+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{-(1+r h)}$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{-1-r h}$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n}\left({e}^{-1}\right) e^{-rh}$
$=\lim _{h \rightarrow 0} \frac{h}{e} \sum_{r=1}^{n} e^{-rh}$
$=\lim _{h \rightarrow 0} \frac{h}{e}\left[e^{-h}+e^{-2 h}+e^{-3 h}+\cdots+e^{-n h}\right]$
$=\lim _{h \rightarrow 0} \frac{h}{e} e^{-h \left(\frac{(e^{-h})^{n}-1}{e^{-h}-1}\right)}$
$=\lim _{h \rightarrow 0} \frac{h}{e} e^{-h}\left(\frac{e^{-n h}-1}{e^{-h}-1}\right)$
$=\lim _{h \rightarrow 0} \frac{h}{e} e^{-h}\left(\frac{e^{-2}-1}{\left(\frac{e^{-1}-1}{-h}\right)(-h)}\right)$
$=-\lim _{h \rightarrow 0} \frac{1}{e} e^{-h}\left(e^{-2}-1\right)$
$=-\left(e^{-3}-e^{-1}\right)$
=e-1-e-2
$=\frac{1}{e}-\frac{1}{e^{3}}$
$=\frac{e^{2}-1}{e^{3}}$
(iii) $\int_{a}^{b} e^{x} dx$
Sol :
nh=b-a
f(x)=ex
$\int_{a}^{b} f(x) d x=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{a+r h}$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{a} \cdot e^{rh}$
$=\lim _{h \rightarrow 0} h e^{a} \sum_{r=1}^{n} e^{r h}$
$=\lim _{h \rightarrow 0} h e^{a}\left[e^{h}+e^{2 h}+e^{3 h}+\cdots+e^{n h}\right]$
$=\lim _{h \rightarrow 0} he^{a}e^h{\left(\frac{e^{\operatorname{nh}}-1}{\frac{e^h-1}{h}\times h}\right)}$
=ea(=eb-a-1)
=ea+b-a-ea
=eb-ea
(iv) $\int_{0}^{1} e^{2-3 x} d x$
Sol :
a=0, b=1
nh=1
f(x)=e2-3x
$\int_{0}^{1} e^{2-3x} d x=\lim_{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{2-3r h}$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} e^{2} \cdot e^{-3r h}$
$=\lim _{h \rightarrow 0} h e^{2} \sum_{r=1}^{n} e^{-3rh}$
$=\lim _{h \rightarrow 0} he^{2}\left[e^{-3(h)}+e^{-3(2 h)}+e^{-3(3h)}+\cdots \cdot e^{-3nh}\right]$
$=\lim _{h \rightarrow 0} h e^{x} e^{-3h}\left(\frac{e^{-3 n h}-1}{\frac{e^{-3h}-1}{-3 h}\times (-3 h)} \right)$
$=-\frac{1}{3} \lim _{h \rightarrow 0} e^{x} e^{-3 h}\left(e^{-3}-1\right)$
$=-\frac{1}{3} e^{2\left(e^{-3}-1\right)}$
$=-\frac{1}{3}\left(e^{-1}-e^{2}\right)$
$=\frac{e^{2}-e^{-1}}{3}$
(v) $\int_{2}^{4} 2^{x} d x$
Sol :
a=2,b=4
nh=2
f(x)=2x
$\int_{2}^{4} 2^{x} d n=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(2+rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} 2^{2+rh}$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} 2^{2} \cdot 2^{2 h}$
$=\lim _{h \rightarrow 0} 4 h \sum_{r=1}^{n} 2^{rh}$
$=\lim _{h \rightarrow 0} 4 h\left[2^{h}+2^{2 h}+2^{2 h}+\cdots-+2^{nh}\right]$
$=\lim _{h \rightarrow 0} 4h2^{h}\left(\frac{2^{nh}-1}{\frac{2^{nh}-1}{h}\times h}\right)$
$=\lim _{h \rightarrow 0} 4 \cdot 2^h{\left(\frac{2^{2}-1}{\log 2}\right)}$
$=\frac{4 \times 3}{\log 2}$
$=\frac{12}{\log 2}$
(vii) $\int_{0}^{\pi /2} \cos x d x$
Sol :
$\sin (a+h)+\sin (a+2 h)+\sin (a+3 h)+\cdots+\sin \left(a+n^{h}\right)=\frac{\sin \left(\frac{n h}{2}\right)}{\sin \left(\frac{h}{2}\right)} \sin \left(\frac{(a+h)+(a+n h)}{2}\right)$
$\cos (a+h)+\cos (a+2 h)+\cos (a+3h)+-\cdots+\cos \left(a+n^{h}\right)=\frac{\sin \left(\frac{n h}{2}\right)}{\sin \left(\frac{h}{2}\right)} \cos \left(\frac{(a+h)+\left(a+n^{h}\right)}{2}\right)$
a=0, $b=\frac{\pi}{2},n h=\frac{\pi}{2}$
f(x)=cosx
$\int_{0}^{\pi/2} \cos d n=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(a+r h)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} f(rh)$
$=\lim _{h \rightarrow 0} h \sum_{r=1}^{n} \cos rh$
$=\lim _{h \rightarrow 0} h\left[\cos h+\cos 2h+\cos 3h+\cos 4h+\dots+\cos rh\right]$
$=\lim _{h \rightarrow 0} h \frac{\sin \left(\frac{n h}{2}\right)}{\sin \left(\frac{h}{2}\right)} \cos \left(\frac{h+nh}{2}\right)$
$=\lim _{h \rightarrow 0} h\left[\frac{\sin \pi /4}{\frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{h}{2}} \sin \left(\frac{h+\pi/2}{2}\right)\right]$
$=\lim _{h \rightarrow 0} 2\left[\frac{1}{\sqrt2} \sin \left(\frac{h+\pi /2}{2}\right)\right]$
$=\quad 2\left[\frac{1}{\sqrt{2}} \sin \frac{\pi}{4}\right]$
$=2 \frac{1}{\sqrt2} \cdot \frac{1}{\sqrt2}$
$=2 \times \frac{1}{2}=1$
Question 4
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right)$
Sol :
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{r}{n^{2}}$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n-1}\left(\frac{r}{n}\right) \cdot \frac{1}{n}$
$=\int_{0}^{1} xdx$
$=\frac{\left[x^{2}\right]_{0}^{1}}{2}$
$=\frac{1-0}{2}=\frac{1}{2}$
Question 5
$\lim _{n \rightarrow \infty}\left[\frac{1}{n+0}+\frac{1}{n+1}+\cdots+\frac{1}{n+5 n}\right]$
Sol :
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{5n} \frac{1}{n+r}$
$=\lim _{n \rightarrow \infty} \sum_{n=1}^{5 n} \frac{1}{n+\left(\frac{r}{h}\right)}$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{5n}\left(\frac{1}{1+\frac{r}{n}}\right)\frac{1}{n}$
$=\int_{0}^{5} \frac{1}{1+x} d x$
$=[\log |1+x|]_{0}^{5}$
=log6-log1
=log6
Question 6
$\lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{4 n^{2}-r^{2}}}$
Sol :
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{n \sqrt{4-(\frac{r}{n})^{2}}}$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{n} \frac{1}{\sqrt{4-(\frac{r}{n}})^{2}}$
$=\int_{0}^{1} \frac{1}{\sqrt{4-x^{2}}} d x$
$=\left[\sin \frac{x}{2}\right]_{0}^{1}$
$=\sin^{-1} \frac{1}{2}-\sin ^{-1} \frac{0 }{2}$
$=\frac{\pi}{6}-0$
$=\frac{\pi}{6}$
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