Question

Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

Answer 1

Hello,

Explanation:

We divide the interval [a b] in n equal parts.

`\Delta x=(b-a)/(n) \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^(th) \ rectangle=R(x_i)=\Delta x * y_i\\`

`\displaystyle \sum_(i=1)^(n) R(x_i)=(b-a)/(n)*\sum_(i=1)^(n)\ (a^2 +((b-a)/(n))^2*i^2+2*a*(b-a)/(n)*i)\\`

`=(b-a)^2*a^2+((b-a)/(n))^3*(n(n+1)(2n+1))/(6) +2*a*((b-a)/(n))^2*\frac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_(n \to \infty) \sum_(i=1)^(n) R(x_i)\\\\=(b-a)*a^2+((b-a)^3 )/(3) +a(b-a)^2\\\\=a^2b-a^3+(1)/(3) (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=(b^3)/(3)-ab^2+ab^2+a^2b+a^2b-2a^2b-(a^3)/(3) \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =(b^3)/(3) -(a^3)/(3)}\\`