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KC Sinha Solution Class 12 Chapter 21 (निश्चित समकलनों के गुणधर्म) Property of Definite Integral Exercise 21.2

 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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