NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.8

Exercise 7.8

Direction (Q.1 to 6) : Evaluate the following definite integrals as limit of sums.

$$\textbf{1.\space}\int^{b}_{a}\space\textbf{x dx}\\\textbf{Sol.\space}\text{We know that}\\\space\int^{b}_{a}\text{f(x)dx}=\lim_{h\to0}h[f(a) + f(a+h)+\\f(a+2h)+...+ f\lbrace a+(n-1)h\rbrace]$$

where, nh = b – a

Here, a = a, b = b and f(x) = x

$$\therefore\space \int^{b}_{a}\space\text{x dx}=\lim_{h\to0}h[a + (a+h)+(a+2h)+...\\+a+(n-1)h]\\=\lim_{h\to 0}h[(a+a+...+ n\space \text{times})+\\h(1+2+...(n-1))]\\\lbrack\because\space\Sigma a=ma\rbrack\\=\lim_{h\to0} h[na+h(1+2+3+...+(n-1))]\\=\lim_{h\to0}\bigg[hna+h^2\frac{n(n-1)}{2}\bigg]\\\bigg(\because\space\Sigma n=\frac{n(n+1)}{2}\therefore\space \Sigma(n-1)=\frac{(n-1)n}{2}\bigg)\\$$

$$=\lim_{h\to0}\bigg[hna + \frac{h^2n^2}{2}-\frac{h^2n}{2}\bigg]\\=\lim_{h\to0}\bigg[(b-a) a +\frac{1}{2}(b-a)^2-\frac{(b-a)^2}{n^2}.\frac{n}{2}\bigg]\\\lbrack\because\space nh=b-a\rbrack\\=\bigg[(b-a)a+\frac{(b-a)^2}{2}\bigg]\\\begin{bmatrix}\because\space\text{when}\space h\to0, \text{then n}\to \infty\space\\\therefore\text{lim}\space\frac{(b-a)^2}{2}=0\end{bmatrix}\\=(b-a)\bigg[a +\frac{b-a}{2}\bigg]\\=(b-a)\bigg[\frac{2a+b-a}{2}\bigg]$$

$$=(b-a)\bigg[\frac{a+b}{2}\bigg]\\=\frac{b^2-a^2}{2}$$

$$\textbf{2.}\space\int^{\textbf{5}}_{\textbf{0}}\space\textbf{(x+1)}\textbf{dx}\\\textbf{Sol.}\space\text{we know that}\\\int^{b}_{a}\space\text{f(x)}dx=\lim_{h\to0}\space h[f(a)+f(a+h)+\\f(a+2h)+......+f\lbrace(a+(n-1))h\rbrace],$$

where nh = b – a

Here, a = 0, b = 5 and nh = 5 – 0 = 5 and

f(x) = (x + 1)

$$\Rarr\space f(0)=1\\\therefore\space\int^{5}_{0}\space(x+1)dx=\lim_{h\to0}h[1+(1+h)+\\(1+2h)+.....+(1+(n-1))h]\\=\lim_{h\to0}\space h\bigg[n+h\frac{n(n-1)}{2}\bigg]\\\begin{pmatrix}\because\Sigma n=\frac{n(n+1)}{2}\text{and}\space\Sigma1=n\\\therefore\space\Sigma(n-1)=\frac{(n-1)n}{2}\end{pmatrix}$$

$$=\lim_{h\to0}\space\bigg[nh+h^2\frac{n(n-1)}{2}\bigg]\\=\lim_{h\to0}\bigg[nh+\frac{(h^2n^2-h^2n)}{2}\bigg]\\=\lim_{h\to0}\bigg[hn+\frac{(hn-h)(hn)}{2}\bigg]\\=\lim_{h\to0}\bigg[5+\frac{(5-h)(5)}{2}\bigg]\\\lbrack\text{Here, nh=5}\rbrack\\=5+\frac{5×5}{2}=\frac{35}{2}$$

$$\textbf{3.\space}\int^{\textbf{3}}_{\textbf{2}}\space \textbf{x}^\textbf{2}\space \textbf{dx}$$

Sol. We know that

$$\int^{b}_{a}\space f(x)=\lim_{h\to0}h[f(a)+f(a+h)+\\f(a+2h)+......+f\lbrace a + (n-1)h\rbrace]$$

where, nh = b – a

Here a = 2, b = 3 and nh = b – a = 3 – 2 = 1

f(x) = x², f(2) = 2², f(2 + h) = (2 + h)²f(2 + 2h) = (2 + 2h²),....., f(2 + (n – 1)h)

= (2 + (n – 1) h)²

$$\therefore\space\int^{3}_{2}x^2\space dx=\lim_{h\to0} h[f(2)+f(2+h)+\\f(2+2h)+...+f(2+(n-1)h)]\\=\lim_{x\to0}\space h[2^2+(2+h)^2+(2+2h)^2+...+\\(2+(n-1)h)^2]\\=\lim_{h\to0}\space h[2^2+(2^2+h^2+4h)+(2^2+2^2h^2+8h)\\+....+(2^2+(n-1)^2)h^2+4(n-1)h]\\\lbrack\because (a+b)^2=a^2+b^2+2ab\rbrack\\=\lim_{h\to0}\space h[2^2+ 2^2+......+ n\space\text{times}]+\\h^2(1^2+2^2+3^2+...+(n-1)^2)+\\4h(1+2+....+(n-1))]$$

$$=\lim_{h\to0}\begin{Bmatrix}4nh + h^3\frac{(n-1)n(2n-1)}{6}+4h^2\frac{(n-1)n}{2}\end{Bmatrix}\\\begin{bmatrix}\because\space\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\\\therefore\space\Sigma(n-1)^2=\frac{1}{6}(n-1)n(2n-1)\space\\\text{and}\space\Sigma(n-1)=\frac{n(n-1)}{2}\end{bmatrix}$$

$$=\lim_{h\to0}\begin{Bmatrix}4\pi h+\frac{(hn-h)hn(2nh-h)}{6}+\\2(nh-h)(nh)\end{Bmatrix}\\=\lim_{h\to0}\begin{Bmatrix}4×1+\frac{(1-h)1(2×1-h)}{6}+2(1-h)1\end{Bmatrix}\\\lbrack\because\space nh=1\rbrack\\=\lim_{h\to0}\bigg[4+\frac{(1-h)(2-h)}{6}+2(1-h)\bigg]\\=4+\frac{2}{6}+2=6+\frac{1}{3}=\frac{19}{3}$$

$$\textbf{4.\space}\int^{\textbf{4}}_{\textbf{1}}\space\textbf{(x}^\textbf{2}\textbf{-x)dx.}$$

Sol. We know that

$$\int^{b}_{a}\space\text{f(x)dx}=\lim_{h\to0}\space h[f(a)+f(a+h)+\\f(a+2h)+.....+f\lbrace a+(n-1)\rbrace h]$$

where nh = b – a

$$\text{Given,\space}\int^{4}_{1}(x^2-x)dx,\space$$

here a = 1, b = 4 and nh = 3

and f(x) = x² – x = x (x – 1)

$$\therefore\space\int^{4}_{1}(x^2-x)dx=\lim_{h\to0}h[f(1)+f(1+h)+\\f(1+2h)+...+f(1+(n-1)h)]\\=\lim_{h\to0}\space h[1(1-1)+(1+h)h+(1+2h)(2h)\\+.....+\lbrace1+(n-1)h\rbrace \lbrace(n-1)h\rbrace]\\\lbrack\because\space f(x)=x(x-1)f(1)=1(1-1);f(1+h)\rbrack\\=(1+h)(1+h-1)=(1+h)h\space\text{and so on}\rbrack$$

$$=\lim_{h\to0}\space h^2[(1+h)+2(1+2h)+3(1+3h)\\+...+(n-1)(1+(n-1))h]\\=\lim_{h\to0}\space h^2[\lbrace(1+2+3+.. +(n-1))\rbrace +\\\lbrace h+2^2h+3^2h +......+ (n-1)^2h\rbrace]\\=\lim_{h\to0}\space h^2\begin{Bmatrix}\frac{(n-1)n}{2}+h(1^2+2^2+......+(n-1)^2)\end{Bmatrix}\\\begin{bmatrix}\because\space 1+2+3+.....(n-1)\\=\Sigma(n-1)=\frac{n(n-1)}{2}\end{bmatrix}$$

$$=\lim_{h\to0}\begin{Bmatrix}\frac{(n-1)nh^2}{2}+\frac{h^3(n-1)n(2n-1)}{6}\end{Bmatrix}\\\begin{pmatrix}\because\space \Sigma n^2=\frac{n(n+1)(2n+1)}{6},\\\therefore\space \Sigma(n-1)^2=\frac{(n-1)n(2n-1)}{6}\end{pmatrix}\\=\lim_{h\to0}\begin{Bmatrix}\frac{(nh-h)(hn)}{2}+\frac{(nh-h)(hn)(2nh-h)}{6}\end{Bmatrix}\\=\lim_{h\to0}\begin{Bmatrix}\frac{(3-h)3}{2}+\frac{(3-h)3(2×3-h)}{6}\end{Bmatrix}\\\lbrack\because\space nh=3\rbrack\\=\frac{9}{2}+\frac{3×3×6}{6}\\=\frac{9}{2}+9=\frac{27}{2}$$

Note : While determining the value of definite integrals by the method of sum of a limits, the value can be checked by simple integration method to avoid the any chance of error.

$$\textbf{5.}\space\int^{\textbf{1}}_{\normalsize\textbf{-1}}\space \textbf{e}^\textbf{x} \textbf{dx}.$$

Sol. We know that

$$\int^{b}_{a}\space f(x)dx=\lim_{h\to0}\space h[f(a)+f(a+h)+\\f(a+2h)+.....+f\lbrace a+(n-1)h\rbrace]$$

where, nh = b – a

$$\text{Given,}\space \int^{1}_{\normalsize-1}\space e^x\space dx$$

Here, a = – 1, b = 1 and nh = b – a = 2

and f (x) = e^x, then f(a) = f (1) = e^–1,

f(– 1) = f(– 1 + h) = e^{(– 1 + h)} .....,

f [(– 1 + (n – 1) h] = e^{(– 1 + (n – 1) h)}

$$\therefore\space\int^{1}_{\normalsize-1}\space e^xdx=\lim_{h\to0} h[e^{\normalsize-1}+e^{(-1+h)}+\\e^{(\normalsize-1+2h)}+...+e^{(-1+(n-1)k)}]$$

$$=\lim_{h\to0}\space he^{\normalsize-1}[1+e^k+e^{2h}+...+e^{((n-1)h)}]$$

$$\begin{bmatrix}\because\space a+ar+ar^2+......+ar^n\\=\frac{a(r^n-1)}{(r-1)}\end{bmatrix}\\=\lim_{h\to0}\space\frac{h}{e}\frac{\lbrace((e^h)^n-1)\rbrace}{e^h-1}\\=\frac{1}{e}\lim_{x\to0}\frac{e^{nh}-1}{\frac{e^h-1}{h}}\\=\frac{1}{e}\bigg(\frac{e^2-1}{1}\bigg)\\\bigg(\because\space\lim_{x\to0}\frac{e^x-1}{x}=1\space\text{and nh=2}\bigg)\\=\frac{e^2-1}{e}= e-\frac{1}{e}$$

$$\textbf{6.\space}\int^{\textbf{4}}_{\textbf{0}}\textbf{(x+e}^{\textbf{2x}})\textbf{dx.}$$

Sol. We know that

$$\int^{b}_{a}\space\text{f(x)}dx=\lim_{h\to0}h[f(a)+f(a+h)\\+f(a+2h)+......+f\lbrace a+(n-1)h\rbrace]$$

where, nh = b – a

$$\text{Given,}\int^{4}_{0}(x+e^{2x})dx$$

Here, a = 0, b = 4 and nh = 4

and f(x) = (x + e^2x)

∴ f(0) = (0 + e²⁽⁰⁾) = 1,

f(0 + h) = h + e^2h, f(0 + 2h) = 2h + e^4h,

and f(0 + (n – 1)h) = (n – 1)h + e^{2(n – 1)h}

$$\therefore\space\int^{4}_{0}(x+e^{2x})dx=\lim_{h\to0}\space h[1+(h+e^{2h}) + \\(2h+e^{4k})+......+(n-1)\space h + e^{2(n-1)h}]\\=\lim_{h\to0}\space h[(h+2h+3h+.....+(n-1))h+\\\lbrace 1+ e^{2h}+e^{4h}+.....+e^{2(n-1)h}\rbrace]\\=\lim_{h\to0}\space h[h\lbrace 1+2+3+..+(n-1)\rbrace+\\\lbrace1+ e^{2k} + e^{4k} + .....+e^{2nh-2h}\rbrace]\\=\lim_{h\to0}\space h\begin{Bmatrix}h\frac{(n-1)n}{2}+1\bigg(\frac{(e^{2h})^n-1}{e^{2h}-1}\bigg)\end{Bmatrix}\\\bigg(\because\space\Sigma n=\frac{n(n+1)}{2}\bigg)$$

$$\because\space 1+ e^{2k}+e^{4k}+....+e^{2(n-1)h}\\\text{is a GP series whose first term is 1}\\\text{and common ratio r =}\frac{e^{2h}}{1}=e^{2h}$$

$$\therefore\space\text{Sum}=\frac{a(r^n-1)}{r-1}$$

$$=\lim_{h\to0}\begin{Bmatrix}h^2\frac{(n-1)n}{2}+h\frac{(e^{2nh}-1)}{e^{2h}-1}\end{Bmatrix}\\=\lim_{h\to0}\space\frac{(nh-h)(nh)}{2}+\lim_{h\to0}\frac{(e^2)^{nh}-1}{\frac{e^{2h-1}}{2h}×2}\\=\lim_{h\to0}\begin{Bmatrix}\frac{(4-h)4}{2}+\frac{e^8-1}{2}\end{Bmatrix}\\\bigg(\because\space\lim_{h\to0}\frac{e^x-1}{x}=1\bigg)\\\lbrack\because\space nh=4\rbrack\\=\frac{4×4}{2}+\frac{e^8-1}{2}\\=\bigg(8-\frac{1}{2}\bigg)+\frac{e^8}{2}=\frac{15+e^8}{2}$$

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