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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.8
Evaluate the following definite integrals as limit of sums.
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Question : 170 of 271
Marks: +1, -0
∫abxdx\int\limits_{a}^{b}xdx
Solution:  
f(x)=x,f(x)=x, then ∫abf(x)dx\int\limits_{a}^{b}f(x)dx =∫abxdx=\int\limits_{a}^{b}xdx
Now, f(a)=af (a) = a
f(a+h)=a+hf (a + h) = a + h
f(a+2h)=a+2hf (a + 2h) = a + 2h
...\begin{array}{c c c} & & . \\ & & . \\ & & . \end{array}
f(a+(n−1)h)=a+(n−1)hf (a + (n - 1)h) = a + (n - 1)h
since
∫abf(x)dx=lim⁡h→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h]\int\limits_{a}^{b}f(x)dx=\lim\limits_{h\rightarrow0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h]
where nh=b−anh = b - a
∴∫abxdx=lim⁡h→0h[a+(a+h)+(a+2h)+...+(a+(n−1)h]\therefore \int\limits_{a}^{b} x dx=\lim\limits_{h\rightarrow0}h[ a + (a+h)+ (a+2h)+...+ (a+(n-1)h]
=lim⁡h→0h[na+h(1+2+...+(n−1))]=\lim\limits_{h\rightarrow0}h[na+h(1+2+...+(n-1))]
=lim⁡h→0h[na+h(n−1)(n)2]=\lim\limits_{h\rightarrow0}h\left[na+h\frac{(n-1)(n)}{2}\right]
=lim⁡h→0[2nah+nh(nh−h)2]=\lim\limits_{h\rightarrow0}\left[\frac{2nah+nh(nh-h)}{2}\right]
=2a(b−a)+(b−a)(b−a)2=\frac{2a(b-a)+(b-a)(b-a)}{2}
=2{a+b−a)(b−a)}2=\frac{2\{a+b-a)(b-a)\}}{2}
=(b+a)(b−a)2=\frac{(b+a)(b-a)}{2}
=b2−a22=\frac{b^2-a^2}{2}
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