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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 : 171 of 271
Marks: +1, -0
∫05(x+1)dx\int\limits_{0}^{5}(x+1)dx
Solution:  
Let f(x)=x+1f(x)=x+1 then ∫05f(x)dx=∫05(x+1)dx\int\limits_{0}^{5}f(x)dx=\int\limits_{0}^{5}(x+1)dx
where a=0,b=5a = 0, b = 5 , nh=b−a=5−0=5nh = b - a = 5 - 0 = 5
Now, f(a)=f(0)=0+1=1f(a) = f(0) = 0 + 1 = 1
f(a+h)=f(h)f(a+h) = f(h) =h+1=1+h= h + 1 = 1 + h
f(a+2h)=f(2h)f(a+2h) = f(2h) =2h+1=1+2h= 2h + 1 = 1 + 2h
.........\begin{array}{lllllllllllll} &&.&&&&.&&&&.\\ &&.&&&&.&&&&.\\ &&.&&&&.&&&&.\\ \end{array}
f(a+(n−1)hf(a+(n-1)h =f((n−1)h)=f((n-1)h) =1+(n−1)h=1+(n-1)h
By definition
∫abf(x)dx=lim⁡h→0[f(a)+f(a+h)+f(a+2h)+⋯+f(a+(n−1)h)]\int\limits_{a}^{b}f(x)dx=\lim\limits_{h\rightarrow 0}\left[f(a)+f(a+h)+f(a+2h)+\dots+f(a+(n-1)h)\right]
∴∫05(x+1)dx=lim⁡h→0h[1+(1+h)+(1+2h)+⋯+(1+(n−1)h)]\therefore\int\limits_{0}^{5}(x+1)dx=\lim\limits_{h\rightarrow 0}h\left[1+(1+h)+(1+2h)+\dots+(1+(n-1)h)\right]
=lim⁡h→0h[1+1+1+⋯→n terms)+h(1+2+3+⋯+(n−1)]=\lim\limits_{h\rightarrow 0}h\left[1+1+1+\dots\to n\text{ terms})+h(1+2+3+\dots+(n-1)\right]
=lim⁡h→0[nh+(nh+h)nh2]=\lim\limits_{h\rightarrow 0}\left[nh+\frac{(nh+h)nh}{2}\right]
=lim⁡h→0(5+(5−h)52)=\lim\limits_{h\rightarrow 0}\left(5+\frac{(5-h)5}{2}\right)
=5+252=352=5+\frac{25}{2}=\frac{35}{2}
=5+25/2=35/2=5+25/2=35/2
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