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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