Area under the graph using Right and Left end-points

**Steelers72** · February 8th 2013, 09:51 PM

Problem 2: Estimate the area under the graph of f(x) = 4 cos(x) from x = 0 to x = π/2. (Round the answer to four decimal places.)(a) Use four approximating rectangles and right endpoints.

My work and answer (which is wrong according to the site)
Delta x= Pi/2 / 4 = pi/8
R4= pi/8 (3.8+3.4+2.5+1.6)
R4= 4.4375

(b) b) Use four approximating rectangles and left endpoints.
L4= pi/8 f(pi/8) + pi/8 f(pi/4) +pi/8 f(3pi/4)
L4= pi/8 (4 + 3.9 +3.3 + 2.7)
L4=5.4585

I made a graph so these values are my estimates

**abender** · February 9th 2013, 08:29 AM

Originally Posted by Steelers72

Problem 2: Estimate the area under the graph of f(x) = 4 cos(x) from x = 0 to x = π/2. (Round the answer to four decimal places.)(a) Use four approximating rectangles and right endpoints.

My work and answer (which is wrong according to the site)
Delta x= Pi/2 / 4 = pi/8
R4= pi/8 (3.8+3.4+2.5+1.6)
R4= 4.4375

(b) b) Use four approximating rectangles and left endpoints.
L4= pi/8 f(pi/8) + pi/8 f(pi/4) +pi/8 f(3pi/4)
L4= pi/8 (4 + 3.9 +3.3 + 2.7)
L4=5.4585

I made a graph so these values are my estimates

I only checked the value I bolded, and it is wrong.

**abender** · February 9th 2013, 08:49 AM

Your $\Delta x$ is correct.

Remember, part (a) instructs you to use right end points.

As such,

$f(x)\approx \Delta x \left(f(\tfrac{\pi}{8})+f(\tfrac{\pi}{4})\right +f(\tfrac{3\pi}{8})+f(\tfrac{\pi}{2}))$

is the right Riemann approximation with four (evenly spaced) rectangles.

EDIT: GO RAVENS!!

**Steelers72** · February 9th 2013, 06:22 PM

Thanks for your replies. So which values do I multiply the values you gave me by? so it's Pi/8 ( f(pi/8)+f(pi/4)+...)and so on. I'm lost on how to find the values needed (as you saw from 3.8 you bolded. Do I make graphs? Really confused on that.

6. Enough said

**abender** · February 9th 2013, 06:59 PM

Originally Posted by Steelers72

Thanks for your replies. So which values do I multiply the values you gave me by? so it's Pi/8 ( f(pi/8)+f(pi/4)+...)and so on. I'm lost on how to find the values needed (as you saw from 3.8 you bolded. Do I make graphs? Really confused on that.

$f(x)=4\cos(x)$

$f(\tfrac{\pi}{8})=4\cos(\tfrac{\pi}{8})\approx 3.69551813004515$

$f(\tfrac{\pi}{4})=4\cos(\tfrac{\pi}{4})\approx 2.82842712474619$

$f(\tfrac{3\pi}{8})=4\cos(\tfrac{3\pi}{8})\approx 1.53073372946037$

$f(\tfrac{\pi}{2})=4\cos(\tfrac{\pi}{2}) = 0$

Right Riemann Approx: $f(x)\approx \Delta x \left(f(\tfrac{\pi}{8})+f(\tfrac{\pi}{4})\right +f(\tfrac{3\pi}{8})+f(\tfrac{\pi}{2}))$

**Steelers72** · February 9th 2013, 07:41 PM

Thank you! I got 3.1631 as the R4 value.

From the left, how would the values change?
It would still be pi/8 (f(pi/8) + f (pi/4) + f(p(3pi/8) + f(pi/2)) and I know from the left side sum it is an over-estimation as compared to the under-estimation from the right.

thanks again for all of your help.

**abender** · February 9th 2013, 07:57 PM

Originally Posted by Steelers72

Thank you! I got 3.1631 as the R4 value.

From the left, how would the values change?
It would still be pi/8 (f(pi/8) + f (pi/4) + f(p(3pi/8) + f(pi/2)) and I know from the left side sum it is an over-estimation as compared to the under-estimation from the right.

thanks again for all of your help.

A LEFT Riemann sum uses LEFT endpoints.

Left Riemann Approximation: $f(x)\approx \Delta x \left(f(0)+ f(\tfrac{\pi}{8})+f(\tfrac{\pi}{4}) +f(\tfrac{3\pi}{8})\right)$

Literally draw the rectangles.

**Steelers72** · February 9th 2013, 08:02 PM

Will do. Thanks again!

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