MHB Approximate the area of the region

[shamieh]

Consider the area between the curve $$y = x^2 + 2x$$ from $$x = 1$$ to $$x = 5.$$

View attachment 1677
Approximate the area of the region by using a regular partition of 4 sub intervals.

a) using L4 i,e, left hand endpoints
b) using R4 i,e, right hand endpoints

So for the left hand endpoints would I just plug into the function? like for example;

$$(1^2 + 2(1) + (2^2 + 2(2) + (3^2 + 2(3) + (4^2 + 2(4) = L4?$$
and$$
(2^2 + 2(2) + (3^2 + 2(3) + (4^2 + 2(4) + (5^2 + 2(5) = R4?$$

Or am I on the wrong track here?

 

[MarkFL]

With the exception of missing closing parentheses, it looks good. For clarity, I would write:

$$f(x)=x^2+2x$$

$$\text{L}4=\frac{5-1}{4}\sum_{k=1}^4f(k)=\sum_{k=1}^4\left(k^2+2k \right)$$

$$\text{R}4=\frac{5-1}{4}\sum_{k=2}^5f(k)=\sum_{k=2}^5\left(k^2+2k \right)$$

 

[shamieh]

Thank you Mark,

I got $$L4 = 50$$ &$$ R4 = 82$$...

The next question it says is

c) using T4 i.e. Trapezoid Rule
d)using simpson rule i,e, S4

(I would have posted a new topic BUT, these are all questions for one question, or relating to this problem, they are like subquestions of this question. Let me know if this is a problem)

basically just need someone to check my answers

I got

$$T4 = 66$$

and $$s4 = \frac{196}{3}$$

and finally

d)find the exact value by using the fundament theorem.

and I wasn't sure about this one but i got

$$\int^5_1 x^2 + 2x dx = \frac{1}{3}x^3 + x^2$$

so then

upper limit minus lower limit to get $$\frac{196}{5}$$

 

[MarkFL]

Thank you Mark,

I got $$L4 = 50$$ &$$ R4 = 82$$...

I get the same results. :D

The next question it says is:

(I would have posted a new topic BUT, these are all questions for one question, or relating to this problem, they are like subquestions of this question. Let me know if this is a problem)

basically just need someone to check my answers.

Yes, since these are all related to the same problem, it is better for you to post them in the same thread. We consider this to be one question, having multiple parts.

c) using T4, i.e. Trapezoid Rule

I got

$$T4 = 66$$

$$T_4=\frac{5-1}{2\cdot4}\sum_{k=0}^{4-1}\left(f(1+k)+f(1+(k+1)) \right)=\frac{1}{2}\sum_{k=0}^3\left(2k^2+10k+11 \right)=66$$

d)using Simpson's Rule, i.e. S4

and $$s4 = \frac{196}{3}$$

$$S_4=\frac{5-1}{3\cdot4}\sum_{k=0}^{4-3}\left(f(1+2k)+4f(1+(2k+1))+f(1+(2k+2)) \right)=\frac{2}{3}\sum_{k=0}^1\left(12k^2+36k+25 \right)=\frac{196}{3}$$

and finally

d)find the exact value by using the fundament theorem.

and I wasn't sure about this one but i got

$$\int^5_1 x^2 + 2x dx = \frac{1}{3}x^3 + x^2$$

so then

upper limit minus lower limit to get $$\frac{196}{5}$$

$$\int_1^5 x^2+2x\,dx=\left[\frac{1}{3}x^3+x^2 \right]_1^5=\left(\frac{1}{3}5^3+5^2 \right)-\left(\frac{1}{3}1^3+1^2 \right)=\frac{200}{3}-\frac{4}{3}=\frac{196}{3}$$

I am assuming you just had a typo in your final result.

Can you explain why Simpson's Rule gave the exact result?

 

[shamieh]

$$\int_1^5 x^2+2x\,dx=\left[\frac{1}{3}x^3+x^2 \right]_1^5=\left(\frac{1}{3}5^3+5^2 \right)-\left(\frac{1}{3}1^3+1^2 \right)=\frac{200}{3}-\frac{4}{3}=\frac{196}{3}$$

I am assuming you just had a typo in your final result.

Can you explain why Simpson's Rule gave the exact result?

Yep was just a typo, that should be $$\frac{196}{3}$$

I'm not sure why Simpsons gives us the same result. I'm guessing it has something to do with $$\frac{1}{3}$$ and $$\frac{deltaX}{3}$$ being in both problems?

 

[MarkFL]

Simpson's Rule uses parabolic arcs to approximate a definite integral, and since the given function was parabolic, this is why Simpson's Rule returned the exact value. In the same way, the Trapezoidal Method would return the exact value for a linear function. :D