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NCERT solutions Class 12 math’s Chapter 7 Exercise 7.8 consists of 6 questions that impart a clear understanding of the definite integrals, their intervals, and limits. Learn about the concept used and the solution to Chapter 7– Integrals Exercise 7.8 in this article.
Question 1: [Tex]\int_{a}^{b} x \,dx [/Tex]Solution:
We known that,
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex][Tex]where \,, h = \frac{b-a}{n}[/Tex]
Here, a = a, b = b and f(x) = x
[Tex]\int_{a}^{b} x \,dx [/Tex][Tex]=(b-a) \lim_{n\to\infty}\frac{1}{n} [(a+a+…+a\,(n\,times))+(h+2h+..+(n-1)h)][/Tex]
[Tex]=(b-a) \lim_{n\to\infty}\frac{1}{n} [na+h(1+2+3..(n-1))][/Tex]
[Tex]=(b-a) \lim_{n\to\infty}\frac{1}{n} [na+h{\frac{(n-1)(n)}{2}}][/Tex]
[Tex]=(b-a) \lim_{n\to\infty}\frac{n}{n} [a+{\frac{(n-1)(h)}{2}}][/Tex]
[Tex]=(b-a) \lim_{n\to\infty} [a+{\frac{(n-1)(h)}{2}}][/Tex]
[Tex]=(b-a) \lim_{n\to\infty} [a+{\frac{(n-1)(b-a)}{2n}}][/Tex]
[Tex]=(b-a) \lim_{n\to\infty} [a+{\frac{(1-\frac{1}{n})(b-a)}{2}}][/Tex]
[Tex]=(b-a) [a+{\frac{(b-a)}{2}}][/Tex]
[Tex]=(b-a) [{\frac{2a+b-a}{2}}][/Tex]
[Tex]={\frac{(b-a)(b+a)}{2}}[/Tex]
[Tex]={\frac{(b^2-a^2)}{2}}[/Tex]
Question 2:[Tex]\int_{0}^{5} (x+1) \,dx [/Tex]Solution:
Let I = [Tex]\int_{0}^{b} (x+1) \,dx [/Tex]
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex] [Tex]where \,, h = \frac{b-a}{n}[/Tex]
Here, a = 0, b = b and f(x) = (x+1)
[Tex]=> h = \frac{5-0}{n} = \frac{5}{n}[/Tex]
[Tex]\int_{0}^{5} (x+1) \,dx [/Tex]= [Tex](5-0) \lim_{n\to\infty}\frac{1}{n} [f(0)+f(\frac{5}{n})+f(\frac{(n-1)5}{n})][/Tex]
[Tex]=5 \lim_{n\to\infty}\frac{1}{n} [1+(\frac{5}{n}+1)+…+[1+(\frac{(n-1)5}{n})][/Tex]
[Tex]=5 \lim_{n\to\infty}\frac{1}{n} [(1+1..+1(n\, times)\,)+[\frac{5}{n}+2.\frac{5}{n}+3.\frac{5}{n}+…+(\frac{(n-1)5}{n})][/Tex]
[Tex]=5 \lim_{n\to\infty}\frac{1}{n} [n+{\frac{5(n-1).n}{n.2}}][/Tex]
[Tex]=5 \lim_{n\to\infty}\frac{1}{n} [n+{\frac{5(n-1)}{2}}][/Tex]
[Tex]=5 \lim_{n\to\infty} [1+{\frac{5(1-\frac{1}{n})}{2}}][/Tex]
[Tex]= 5[1+\frac{5}{2}][/Tex]
[Tex]= 5[\frac{7}{2}][/Tex]
[Tex]= [\frac{35}{2}][/Tex]
Question 3:[Tex]\int_{2}^{3} x^2 \,dx [/Tex]Solution:
We known that,
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex] [Tex]where \,, h = \frac{b-a}{n}[/Tex]
Here, a = 2, b = 3 and f(x) = x2 => h = (3-2)/n = 1/n
[Tex]\int_{2}^{3} x^2 \,dx [/Tex] = [Tex](3-2) \lim_{n\to\infty}\frac{1}{n} [f(2)+f(2+\frac{2}{n})+…+f(2+\frac{(n-1)}{n})][/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [(2)^2+(2+\frac{1}{n}))^2+…+[(2+(\frac{(n-1)}{n}))^2][/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [(2)^2+[2^2+[\frac{1}{n}]^2+2.2.\frac{1}{n}]+…+[2^2+(\frac{(n-1)}{n})^2+2.2.\frac{n-1}{n}]\,][/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [(2^2+2^2..+2^2(n\, times)\,)+[(\frac{1}{n})^2+(\frac{2}{n})^2+….+(\frac{n-1}{n})^2]+2.2\{\frac{1}{n}+\frac{2}{n}+….+\frac{(n-1)}{n}\}][/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\frac{1}{n^2}\{1^2+2^2+…+(n-1)^2\}+\frac{4}{n}\{1+2+…+(n-1)\}] \,[/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\frac{1}{n^2}\{\frac{n(n-1)(2n-1)}{6}\}+\frac{4}{n}\{\frac{n(n-1)}{2}\}] \,[/Tex]
[Tex]=\lim_{n\to\infty}\frac{1}{n} [ \,4n+\{\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}\}+\{\frac{4(n-1)}{2}\}] \,[/Tex]
[Tex]=\lim_{n\to\infty} [ \,4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}] \,[/Tex]
[Tex]= 4+\frac{2}{6}+2
= \frac{19}{3}[/Tex]
Question 4: [Tex]\int_{1}^{4} (x^2-x) \,dx [/Tex]Solution:
Let I [Tex]=\int_{1}^{4} (x^2-x) \,dx [/Tex]
[Tex]=\int_{1}^{4} x^2 \,dx -\int_{1}^{4} x \,dx [/Tex]
[Tex]Let \, I = I_1-I_2,where [/Tex][Tex]I1=\int_{1}^{4} x^2 \,dx \, and \, I2=\int_{1}^{4} x \,dx [/Tex] ……….(1)
We known that,
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex] [Tex]where \,, h = \frac{b-a}{n}[/Tex]
For [Tex]I_1=\int_{1}^{4} x^2 \,dx \,[/Tex]
a = 1, b = 4, and f(x) = x2
h = (4-1)/n = 3/n
[Tex]I_1=\int_{1}^{4} x^2 \,dx \,[/Tex][Tex]=(4-1) \lim_{n\to\infty}\frac{1}{n} [f(1)+f(1+h)+…+f(1+(n-1)h)][/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [(1)^2+(1+\frac{3}{n}))^2+…+[(1+(\frac{(n-1)3}{n}))^2][/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [(1)^2+[1^2+[\frac{3}{n}]^2+2.3.\frac{1}{n}]+…+[1^2+(\frac{(n-1)3}{n})^2+2.3.\frac{n-1}{n}]\,][/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [1^2+1^2+..+1^2(n\, times)\,)+(\frac{3}{n})^2[ \,1^2+2^2+…+(n-1)^2] \,+2.\frac{3}{n}\{1+2++…+(n-1)\}] \,[/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [ \,n+\frac{9}{n^2}\{\frac{n(n-1)(2n-1)}{6}\}+\frac{6}{n}\{\frac{n(n-1)}{2}\}] \,[/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [ \,n+\{\frac{9n(1-\frac{1}{n})(2-\frac{1}{n})}{6}\}+\{\frac{6(n-1)}{2}\}] \,[/Tex]
[Tex]=3\lim_{n\to\infty} [ \,1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3-\frac{3}{n}] \,[/Tex]
= 3[1+3+3]
I1 = 3[7] = 21 ………….(2)
For I2 [Tex]=\int_{1}^{4} x \,dx [/Tex]
a = 1, b = 4 and f(x) = x
=> h = (4-1)/n = 3/n
[Tex]I_2=(4-1) \lim_{n\to\infty}\frac{1}{n} [f(1)+f(1+h)+…+f(1+(n-1)h)][/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [(1)+(1+\frac{3}{n})+…+[(1+(\frac{(n-1)3}{n})][/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [1+1+…+1(n\, times)\,)+\frac{3}{n}[ \,1+2+..+(n-1)] \,[/Tex]
[Tex]=3\lim_{n\to\infty}\frac{1}{n} [ \,n+\frac{3}{n}\{\frac{n(n-1)}{2}\}] \,[/Tex]
[Tex]=3\lim_{n\to\infty} \frac{1}{n}[ \,1+\frac{3}{2}(1-\frac{1}{n})[/Tex]
[Tex]=3[ \,1+\frac{3}{2}] \,[/Tex]
[Tex]I_2 = \frac{15}{2} \,\,\,……….(3)[/Tex]
From eq (2) and (3) we obtain,
I = I1 – I2 = 21 – [Tex]\frac{15}{2}[/Tex] = [Tex]\frac{27}{2}[/Tex]
Question 5:[Tex]\int_{-1}^{1} {\rm e^x} \,dx [/Tex]Solution:
Let I = [Tex]\int_{-1}^{1} {\rm e^x} \,dx [/Tex]
We known that,
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex] [Tex]where \,, h = \frac{b-a}{n}[/Tex]
Here a = -1, b = 1 and f(x) = ex
=> h = (1+1)/n = 2/n
[Tex]=(1+1)\lim_{n\to\infty}\frac{1}{n} [f(-1)+f(-1+\frac{2}{n})+..+f(-1+\frac{2(n-1)}{2})][/Tex]
[Tex]=2\lim_{n\to\infty}\frac{1}{n} [ e^{(-1)}+e^{(-1+\frac{2}{n})}+..+e^{(-1+\frac{2(n-1)}{n})}][/Tex]
[Tex]=2\lim_{n\to\infty}\frac{1}{n} [ e^{-1}\{1+e^{\frac{2}{n}}+e^{\frac{4}{n}}+e^{\frac{6}{n}}+…+e^{\frac{2(n-1)}{n}}\}][/Tex]
[Tex]=2\lim_{n\to\infty}\frac{e^{-1}}{n} [ \,\frac{e^{\frac{2n}{n}-1}}{e^{\frac{2}{n}-1}}] \,[/Tex]
[Tex]=e^{-1}*2\lim_{n\to\infty}\frac{1}{n} [ \,\frac{e^{2-1}}{e^{\frac{2}{n}-1}}] \,[/Tex]
[Tex]= \frac{e^{-1}*2(e^{2}-1)}{\lim_{\frac{2}{n}\to0}( \,\frac{e^{\frac{2}{n}}-1}{\frac{2}{n}}) \,*2 }[/Tex]
[Tex]= e^{-1}[ \,\frac{2(e^{2}-1)}{2}] \,[/Tex]
[Tex]=\frac{e^{2}-1}{e}[/Tex]
[Tex]=\{e-\frac{1}{e}\}[/Tex]
Question 6:[Tex]\int_{0}^{4} (x+{\rm e^{2x}}) \,dx [/Tex]Solution:
Let I = [Tex]\int_{0}^{4} (x+{\rm e^{2x}}) \,dx [/Tex]
We known that,
[Tex]\int_{a}^{b} f(x) \,dx =(b-a) \lim_{n\to\infty}\frac{1}{n} [f(a)+f(a+h)+…..f(a+(n-1)h)][/Tex] where [Tex]h = \frac{b-a}{n}[/Tex]
Here a = 0, b = 4, and f(x) = x+e2x
h = [Tex]\frac{4-0}{n} = \frac{4}{n}[/Tex]
=> [Tex]\int_{0}^{4} (x+{\rm e^{2x}}) \,dx [/Tex][Tex]=(4-0)\lim_{n\to\infty}\frac{1}{n} [(f(0)+f(h)+f(2h)+….+f((n-1)h)][/Tex]
[Tex]=4\lim_{n\to\infty}\frac{1}{n} [(0+{\rm e}^0)+(h+{\rm e}^{2h})+(2h+e^{2.2h})+..+\{(n-1)h+e^{2(n-1)h}\}][/Tex]
[Tex]=4\lim_{n\to\infty}\frac{1}{n} [1+(h+{\rm e}^{2h})+(2h+e^{4h})+..+\{(n-1)h+e^{2(n-1)h}\}][/Tex]
[Tex]=4\lim_{n\to\infty}\frac{1}{n} [\{h+2h+3h+…+(n-1)h\}+(1+e^{2h}+e^{4h}+..+e^{2(n-1)h}][/Tex]
[Tex]=4\lim_{n\to\infty}\frac{1}{n} [\{h\{1+2+..(n-1)\}+ (\frac{e^{2hn}-1}{e^{2h}-1})][/Tex]
[Tex]=4\lim_{n\to\infty}\frac{1}{n} [\frac{4}{n}.\frac{n(n-1)}{2}+\{\frac{e^8-1}{e\frac{8}{n}-1}\}[/Tex]
[Tex]=4(2)+4\lim_{n\to\infty} \frac{e^8-1}{(\frac{e\frac{8}{n}-1}{\frac{8}{n}})8}[/Tex]
[Tex]=8+\frac{4(e^8-1)}{8}[/Tex]
[Tex]=8+\frac{e^8-1}{2} = \frac{15+e^8}{2}[/Tex]
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