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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 : 172 of 271
Marks: +1, -0
∫23x2dx\int\limits_{2}^{3}x^2dx
Solution:  
Let f(x)=x2f(x)=x^2
⇒∫23f(x)dx=∫23x2dx\Rightarrow \int\limits_{2}^{3}f(x)dx=\int\limits_{2}^{3}x^2dx
a=2,b=3,nh=1a = 2, b = 3, nh = 1
Now, f(a)=f(2)=22f(a)=f(2)=2^2
f(a+h)=f(2+h)f(a+h)=f(2+h) =(2+h)2=4+h2+4h=(2+h)^2=4+h^2+4h
f(a+2h)=f(2+2h)f(a+2h)=f(2+2h) =(2+2h)2=(2+2h)^2 =4+4h2+8h=4+4h^2+8h
.........\begin{array}{c c c c c c c c c c c c} &&&.&&&.&&&. \\ &&&.&&&.&&&. \\ &&&.&&&.&&&. \end{array}
f(a+(n−1)h)f(a+(n-1)h) =f[2+(n−1)h]=f[2+(n-1)h] =[2+(n−1)h]2=[2+(n-1)h]^2 =4+(n−1)2h2+4(n−1)=4+(n-1)^2h^2+4(n-1)
By definition
∫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]
∴∫23x2dx=lim⁡h→0h[4n+h2(1+4+9+...+(n−1)2+4h[1+2+...+(n−1)]]\therefore \int\limits_{2}^{3}x^2dx=\lim\limits_{h\rightarrow0}h[4n+h^2(1+4+9+...+(n-1)^2+4h[1+2+...+(n-1)]]
=lim⁡h→0h[4n+h2(n−1)n(2n−1)6+4h(n−1)n2]=\lim\limits_{h\rightarrow0}h\left[4n+\frac{h^2(n-1)n(2n-1)}{6}+\frac{4h(n-1)n}{2}\right]
=lim⁡h→0[4nh+(nh−h)(nh)(2nh−h)6+2(nh−h)nh]=\lim\limits_{h\rightarrow0}\left[4nh+\frac{(nh-h)(nh)(2nh-h)}{6}+2(nh-h)nh\right]
=4×1+1×26+2×1×1=4\times 1+\frac{1\times 2}{6}+2\times 1\times 1
=4+13+2=4+\frac{1}{3}+2
=6+13=193=6+\frac{1}{3}=\frac{19}{3}
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