Question

integration of x² as the limit of the sum.

Answer 1

x^3/3+c is the answer of question.

Answer 2

Let the upper limit be b and the lower limit be a.
No of divisions of area = n
Width= of the area = h

b = a + nh
⇒ nh = b - a

                           r = n     
Area = $\lim_{n \to \infty}$   Σ     h f( a + nh)
                           r = 1
      

                          r = n     
        = $\lim_{n \to \infty}$   Σ     h ( a + nh)²
                          r = 1


                       r = n     
  = $\lim_{n \to \infty}$   Σ     h [(a+h)²+(a+2h)²+(a+3h)²+.....+(a+nh)²]
                       r = 1
     
                        r = n     
        = $\lim_{n \to \infty}$  h     Σ    a² + 2arh + r²h²
                        r = 1
 
                        r = n     
        = $\lim_{n \to \infty}$  h     Σ    a² + 2arh + r²h²
                        r = 1

Hence as we solve further,
 
        = $\lim_{n \to \infty}$ [ a²(b-a) + a(b-a)(b-a-h) +  
                                                       $\frac{(b-a-h)(b-a)(2(b-a))}{6}$
As we solve this, we get
 
Area = a²(b-a) + a(b-a)² + (b-a)³/3
         = (b-a)/3  [ b² + ab + a²]
         = (b³-a³)/3
         
Now we put upper limit b = 3 and lower limit a = 2, 

Integration = (b³-a³)/3
                  = (3³-2³)/3                                               
                  = (27-8)/3
                  = 19/3

Integration of x² as the limit of the sum with upper limit 3 and lower limit 2 is 19/3.

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