Given
[tex]f(x)=2x,[2,4][/tex]Find
Area under the curve.
Explanation
first , we split interval [2 , 4] into n subintervals of length
[tex]\frac{4-2}{n}=\frac{2}{n}[/tex]so ,
[tex][2,4]=[2,2+\frac{2}{n}]\cup[2+\frac{2}{n},2+\frac{4}{n}]\cup[2+\frac{4}{n},2+\frac{6}{n}]\cup.....\cup[2+\frac{2(n-1)}{n},4][/tex]so that the right endpoints are given by the sequence
[tex]x_i=2+\frac{2i}{n}=\frac{2(n+i)}{n},for\text{ }1\leq i\leq n[/tex]then Riemann sum approximating ,
[tex]\int_2^42xdx=\sum_{i\mathop{=}1}^nf(x_i)\frac{4-2}{n}=\frac{12n+4}{n}[/tex]the integral is given exactly as n tends to infinity , for which we get ,
[tex]\int_2^42xdx=\lim_{n\to\infty}(\frac{12n+4}{n})=12[/tex]Final Answer
Hence ,
[tex]\int_2^42xdx=12[/tex]