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For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5]
subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as
n--> infinity to calculate the area under the curve over [2,5]
please show all of your work as be as descriptive as you can I appreciate your help thank you!
Solution
a=2, b =5, x =(b-a)/n =(5-2)/n =3/n
f(x)=2x
xi=a+ ix =2+i(3/n)
f(xi)=2(2+3(i/n))
the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x
=limn->[i =1 to n]2(2+3(i/n))(3/n)
=limn->[i =1 to n]6((2/n)+3(i/n2))
=limn-> 6((2n/n)+3(n(n+1)/2n2))
=limn-> 6(2 +3((1+(1/n))/2))
= 6(2 +3((1+0)/2))
=6(2 +3(1/2))
=6(2 +(3/2))
=6(7/2)
=3*7
=21
the area under the curve over [2,5] is 21

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  • 1. For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5] subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as n--> infinity to calculate the area under the curve over [2,5] please show all of your work as be as descriptive as you can I appreciate your help thank you! Solution a=2, b =5, x =(b-a)/n =(5-2)/n =3/n f(x)=2x xi=a+ ix =2+i(3/n) f(xi)=2(2+3(i/n)) the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x =limn->[i =1 to n]2(2+3(i/n))(3/n) =limn->[i =1 to n]6((2/n)+3(i/n2)) =limn-> 6((2n/n)+3(n(n+1)/2n2)) =limn-> 6(2 +3((1+(1/n))/2)) = 6(2 +3((1+0)/2)) =6(2 +3(1/2)) =6(2 +(3/2)) =6(7/2) =3*7 =21 the area under the curve over [2,5] is 21