# Math Help - [SOLVED] easy limits area under the function

1. ## [SOLVED] easy limits area under the function

The value of the limit lim n--> inf E (n above and i=1 below) pi/8n tan(ipi/24n) is equal to the area below the graph of a function f(x) on an interval [A,B]. Find f, A, and B. (Do not evaluate the limit.) for f i get tan(x) for a i get -pi/12 for b i get .1308996939 i used delta x = (b-a)/n i also used xi = a + (delta x)(i) so i know a + pi/8n = ipi/24n solved for a then (b-a)/n = pi/8n solved for b my webwork is saying its wrong???? help please. i dont know what else to do

2. Originally Posted by mybrohshi5
The value of the limit lim n--> inf E (n above and i=1 below) pi/8n tan(ipi/24n) is equal to the area below the graph of a function f(x) on an interval [A,B]. Find f, A, and B. (Do not evaluate the limit.) for f i get tan(x) for a i get -pi/12 for b i get .1308996939 i used delta x = (b-a)/n i also used xi = a + (delta x)(i) so i know a + pi/8n = ipi/24n solved for a then (b-a)/n = pi/8n solved for b my webwork is saying its wrong???? help please. i dont know what else to do
$\lim_{n\to\infty}\sum_{i=1}^n\frac{\pi}{8n}\tan\le ft(\frac{i\pi}{24n}\right)=\lim_{n\to\infty}\sum_{ i=1}^n\frac{\pi}{8n}\tan\left(\frac{i\frac{\pi}{8n }}{3}\right)$

From this you can see that the function is $\tan\left(\frac{x}{3}\right)$, and your interval length is $\frac{\pi}{8}$.

However, since you have no constant being added to the argument of tangent, your interval is from $0$ to something. Since the length of the interval is $\frac{\pi}{8}$, this implies that the upper bound is $0+\frac{\pi}{8}=\frac{\pi}{8}$.

$\lim_{n\to\infty}\sum_{i=1}^n\frac{\pi}{8n}\tan\le ft(\frac{i\pi}{24n}\right)=\int_0^{\pi/8}\tan\left(\frac{x}{3}\right)\,dx=-3\ln\left(\cos\left(\frac{\pi}{24}\right)\right)\a pprox0.258$

PS:

Originally Posted by mybrohshi5
for b i get .1308996939
This value is $\frac{\pi}{24}$, in case you were wondering.

3. Great explanation. That makes way more sense then how i figured it out a little bit ago. thank you very much =)