Evaluate `int_0^2(x^2+3) dx` as limit of sums.

Answer

N/a

Explanation

Let `I = int_(0)^(2)(x^(2)+3)dx`
Here, `a = 0, b = 2` an d ` h = (b-a)/(n) = (2-0)/(n)`
`rArr h = 2/n rArr nh = 2 rArr f(x) = (x^(2)+3)`
Now, `int_(0)^(2) (x^(2)+3)dx= underset(hrarr0)("lim")h[f(0)+f(0+h)+f(0+2h)+"....."+f{0+(n-1)h}]`
`f(0) = 3`
`rArr f(0+h)=h^(2)+3, f(0+2h)=4h^(2)+3=2^(2)h^(2)+3`
`f[0+(n-1)h]=(n^(2)-2n+1)h+3=(n-1)^(2)h+3`
From Eq. (i),
`int_(0)^(3)(x^(2)+3)dx=underset(hrarr0)("lim")h[3+h^(2)+3+2^(2)h^(2)+3+3^(2)h^(2)+3+"......"+(n-1)^(2)h^(2)+3]`
` = underset(hrarr0)("lim")h[3n+h^(2){1^(2)+2^(2)+"....."+(n-1)^(2)}]`
`= underset(hrarr0)("lim")h[3n+h^(2)(((n-1)(2n-2+1)(n-1+))/(6))]` , `[:' sumn^(2)=(n(n+1)(2n+1))/(6)]`
`= underset(hrarr0)limh[3n+h^(2)(((n^(2)-n)(2n-1))/(6))]`
`=underset(hrarr0)("lim")h[3n(h^(2))/(6)(2n^(2)-n^(2)-2n^(2)+n)]`
` =underset(hrarr0)("lim")[3nh+(2n^(3)h^(3)-3n^(2)h^(2).h+nh.h^(2))/(6)]`
`= underset(hrarr0)("lim")[3.2+(2.8-3.2^(2).h+2.h^(2))/(6)]=underset(hrarr0)("lim")[6+(16-12h+2h^(2))/(6)]`
`= 6+ (16)/(6)=6+8/3=(26)/(3)`