NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.8
Exercise 7.1 Solutions 22 Questions
Exercise 7.2 Solutions 39 Questions
Exercise 7.3 Solutions 24 Questions
Exercise 7.4 Solutions 25 Questions
Exercise 7.5 Solutions 23 Questions
Exercise 7.6 Solutions 24 Questions
Exercise 7.7 Solutions 11 Questions
Exercise 7.8 Solutions 6 Questions
Exercise 7.9 Solutions 22 Questions
Exercise 7.10 Solutions 10 Questions
Exercise 7.11 Solutions 21 Questions
Miscellaneous Exercise on Chapter 7 Solutions 44 Questions
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)2 f(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) = ex, 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 + e2x)
∴ f(0) = (0 + e2(0)) = 1,
f(0 + h) = h + e2h, f(0 + 2h) = 2h + e4h,
and f(0 + (n – 1)h) = (n – 1)h + e2(n – 1)h
$$\therefore\space\int^{4}_{0}(x+e^{2x})dx=\lim_{h\to0}\space h[1+(h+e^{2h}) + \\(2h+e^{4k})+......+(n-1)\space h + e^{2(n-1)h}]\\=\lim_{h\to0}\space h[(h+2h+3h+.....+(n-1))h+\\\lbrace 1+ e^{2h}+e^{4h}+.....+e^{2(n-1)h}\rbrace]\\=\lim_{h\to0}\space h[h\lbrace 1+2+3+..+(n-1)\rbrace+\\\lbrace1+ e^{2k} + e^{4k} + .....+e^{2nh-2h}\rbrace]\\=\lim_{h\to0}\space h\begin{Bmatrix}h\frac{(n-1)n}{2}+1\bigg(\frac{(e^{2h})^n-1}{e^{2h}-1}\bigg)\end{Bmatrix}\\\bigg(\because\space\Sigma n=\frac{n(n+1)}{2}\bigg)$$
$$\because\space 1+ e^{2k}+e^{4k}+....+e^{2(n-1)h}\\\text{is a GP series whose first term is 1}\\\text{and common ratio r =}\frac{e^{2h}}{1}=e^{2h}$$
$$\therefore\space\text{Sum}=\frac{a(r^n-1)}{r-1}$$
$$=\lim_{h\to0}\begin{Bmatrix}h^2\frac{(n-1)n}{2}+h\frac{(e^{2nh}-1)}{e^{2h}-1}\end{Bmatrix}\\=\lim_{h\to0}\space\frac{(nh-h)(nh)}{2}+\lim_{h\to0}\frac{(e^2)^{nh}-1}{\frac{e^{2h-1}}{2h}×2}\\=\lim_{h\to0}\begin{Bmatrix}\frac{(4-h)4}{2}+\frac{e^8-1}{2}\end{Bmatrix}\\\bigg(\because\space\lim_{h\to0}\frac{e^x-1}{x}=1\bigg)\\\lbrack\because\space nh=4\rbrack\\=\frac{4×4}{2}+\frac{e^8-1}{2}\\=\bigg(8-\frac{1}{2}\bigg)+\frac{e^8}{2}=\frac{15+e^8}{2}$$
