How to use the limit definition (Riemann sum) to evaluate the following integral ?

Question

2961 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-05T12:57:05

#I=int_0^2(x^3+x)dx=6#

We know that ,

#(1)int_a^b f(x)dx=lim_(nto oo ) hsum_(i=1)^n f(a+ih) ,where, h=(b-a)/n#

We have ,

#I=int_0^2(x^3+x)dx=>f(x)=x^3+x ,and a=0 ,b=2#

#f(a+ih)=f(0+ih)=f(ih)=(ih)^3+(ih)=i^3h^3+ih#

#:.h=(b-a)/n=(2-0)/n=2/n#

#I=lim_(nto oo)hsum_(i=1)^n{i^3h^3+ih}#

#=lim_(nto oo)h{h^3sum_(i=1)^n i^3 +hsum_(i=1)^n i}#

#=lim_(nto oo)2/n{8/n^3sum_(i=1)^n i^3+2/nsum_(i=1)^ni}....to[because h=2/n]#

#=lim_(nto oo)16/n^4sum_(i=1)^n i^3+lim_(nto oo)4/n^2sum_(i=1)^ni#

#=lim_(nto oo)16/n^4*n^2/4(n+1)^2+lim_(nto oo) 4/n^2*n/2(n+1)#

#=lim_(ntooo)4/n^2(n+1)^2+lim_(nto oo)2/n(n+1)#

#=4lim_(nto oo)((n+1)/n)^2+2lim_(nto oo)((n+1)/n)#

#=4lim_(nto oo)(1+1/n)^2+2lim_(nto oo)(1+1/n)#

#=4(1+0)^2+2(1+0)...to[becauselim_(nto oo)1/n=0]#

#=4+2#

#=6#

Note :
#diamondsum_(i=1)^ni=n/2(n+1)#

#diamondsum_(i=1)^ni^3=n^2/4(n+1)^2#