Question

lim n-> Infinity of summation i=0 to n of
(1+i3/n)^(1/2) * 3/n

Answer 1

looks like reimann to integration

Answer 2

yes my final answer was 2/3*X^(3/2) from 1to 4

Answer 3

is that right

Answer 4

b-a = 3, a = 1, so b = 4 is good

Answer 5

\[f(a+i\frac{b-a}{n})=(1+x)^{.5}\]
the change in x represents x right?

Answer 6

sqrt(x) is prolly more like it

Answer 7

\[\int_{1}^{4}x^{.5}dx=\frac{4^{1.5}}{1.5}-\frac{1^{1.5}}{1.5}\]

Answer 8

okay yes that's what i got i just haven't plugged in the limits to x yet

Answer 9

thanks for your help would you mind helping me with one more

Answer 10

if i can remember that far back .. sure :)

Answer 11

\[d/dx \int\limits_{2x}^{x^2}\sqrt[3]{1+x^4}\]

Answer 12

in general:\[let:~F(x)=\int_{\alpha(x)}^{\beta(x)}f(t)~dt=F[\beta(x)]-F[\alpha(x)]\]
\[\frac d{dx}F(x)=\frac d{dx}\int_{\alpha(x)}^{\beta(x)}f(t)~dt\]
\[\frac d{dx}F(x)=\frac d{dx}F[\beta(x)]-F[\alpha(x)]\]
\[\frac d{dx}F(x)=f[\beta(x)]\beta'-f[\alpha(x)]\alpha'\]

Answer 13

\[d/dx \int\limits_{2x}^{x^2}\sqrt[3]{1+t^4}~dt=\sqrt[3]{1+(2x)^4}(2x)'-\sqrt[3]{1+(x^2)^4}(x^2)'\]

Answer 14

thanks a lot I really appreciate it

Answer 15

youre welcome