Your numerical integral

(10 pts)

In this problem, you’re going to work out an integral numerically from start to finish. To get your specific problem, choose your name from the following menu:

My integral is

Here’s a graph of the integrand f, together with f''(x):

Over the interval [1,3], it sure looks like

So I can take M_2 = 2. Now, our error bound says

Solving that equation for n, we find that we need

Thus, we’ll take n = 26 and break the interval from 1 to 3 up into 26 parts to get

This yields the approximation

Note that the final approximation was computed with this code.

My integral: \int_0^2 \sqrt{1+x^4}\ dx
Here is the graph of f and f^"(x):

Over the interval [0, 2]:
|f^"(x)| < 3
,so M_2=3.f
n\geq \sqrt{27000/24} \approx33.54
I’ll take n=34, so \Delta x= \frac{2}{34}= \frac{1}{17}, \text{and } x_i= \frac{i}{17}
The result is:
\int_0^2 \sqrt{1+x^4} \approx \sum_{i=1}^{34} \sqrt{1+(\frac{\frac{i}{17} + \frac{i-1}{17}}{2})^4} \approx \sum_{i=1}^{34} \sqrt{1+ (\frac{2i-1}{34})^4} \approx 3.6529

This is the code.

My integral is

Here is the graph of f and f"(x)

Over the interval [0,2], it appears to be |f''(x)| < 1 So M1 = 1. The error bound says error < or = M1 (b-a)^2/n = 4/n < or = 1/1000 Solving for n, we need n > or = 4000. We'll take n=4000 and break the interval from 0 to 2 up into 4000 parts to get delta(x) = 2/4000=1/2000 and xi= 0 + i/2000 This yields the approximation ~~ 7.6527. This was computed using this code: f(x) = (16-x^2)^(1/2) a = 0; b = 2; n = 4000;

xi = a;
dx = (b-a)/n;
sum = 0;
for i in range(n):
xi = xi+dx;
sum = sum + f(xi)
N(sum*dx)

Apologies this is not formatted at the moment, I will improve on it. I struggled with getting any formatting to work successfully.

My integral is

The graph of this (red) and f''(x) (blue) is shown below:

From this it seems that over the interval [-1,1],

So we can take M_2 = 3. The error bound for a midpoint sum states

When solving the equation for n we find that

well round up and make n = 32. We’ll then break up the interval from -1 to 1 into 32 parts to get

This yields the approximation

This code I used to calculate this was here: Sage Cell Server.

My integral is

My graph of f and f''(x):

I can set my M_{2}=10

Using n=21 to find \Delta x and x_i, yields:

Midpoint approximation:

https://sagecell.sagemath.org/?z=eJwljUsKhEAMRPcNfYcsE0X8LG28wtyhpacli8mIIuT4U45ZpF4VFFXZhRY61XjXxiWGDDsmWiFTisFuHQExuIIzsDiA1y5Lb7Dn9YEfQPV7kJIaHdm2N5vMMRDu33Rti6cneCr3b6mya1e8nzD-YmRNcfkByGMljQ==&lang=sage&interacts=eJyLjgUAARUAuQ==

An image of the function (in red) and it’s first and second derivative
|f"(x)|<3...f"(x)=\dfrac{6x^2}{\sqrt{x^4+1}}-\dfrac{4x^6}{\left(x^4+1\right)^\frac{3}{2}}
Error Calculation: M2 =3, n=>\sqrt(\frac{27000}{34})=33.5

This gives us the equation \int_0^2 \sqrt{1+x^4} \, dx \approx \sum_{i=1}^{34} \sqrt{1+(i/17+(\frac{2i-1}{34})^4}=3.6848
Code: Sage Cell Server
f(x) = sqrt(1+x^4)
a = 0; b = 2;
n = 34;

xi = a;
dx = (b-a)/n;
sum = 0;
for i in range(n):
xi = xi+dx;
sum = sum + f(xi)
N(sum*dx)
I have confused myself.

My integral is

Here’s a graph of f, together with f'(x)

Over the interval [-1,1], it looks like

So I can take M_1=1.5. Now my error bound says

solving for n, I find that I need

Thus, I’ll take n=6000 and break the interval from -1 to 1 up into 6000 parts to get

This yields the approximation

The final approximation was computed with this code: Sage Cell Server

My integral is:

Here’s a graph of the integrated f, together with f''(x):

Over the interval [0,2], it looks like:

So, we can take M_2=3. Now, our error bound says:

Solving that equation for n, we will find that we need:

Thus, we’ll take n=32 and break the interval from 0 to 2 up into 32 parts to get:

This yields the approximation:

Note that the final approximation was computed with this code.

my integral is

Here’s a graph of the integrand f , together with f"(x)

Over the interval [-1,1],

So I can take M_2 = 3 . Now, our error bounds says

Solving that equation for n , we find that we need

Thus, we’ll take n=32 and break the interval from -1 to 1 up into 32 parts to get

this yields the approximation

https://sagecell.sagemath.org/?z=eJwljcEKAjEMRO-F_kOOiWVZq54s_sJ-gtClW8nBKCtCPt9xN4fMm4FhOrvQjRZ_cx45J79fRGKoyIZcaIbmEoNBzydADK7gCmwO4HmoMhrs5_uEP4L6ayUlNVqrPRY2ucZAuK3pmpqXPdgr_5-osyuGJ4Y9NJcfkwwk8A==&lang=sage&interacts=eJyLjgUAARUAuQ==

my integral is

Here is the graph of f(x) and f’(x):

Over the interval [1,2] it looks like

So, M1 = 4. Now, our error bound says

Solving that equation for n, we find that

Thus, we’ll take n=4000 and break the interval from 1 to 2 up into 4000 parts to get

This yields the approximation

Note that the final approximation was computed with Sage Cell Server

My integral is

Here is a graph of the integrand f, together with f''(x):

Over the interval [-1,1], it looks like

So I can take M_2 = 20. Now, our error bounds says

Solving for n, we find that we need

Now we will take n=82 and break the interval from -1 to 1 up into 82 parts to get

This yields the approximation

The final approximation was computed using this code.

My integral is:

Here’s a graph of the integrand f, together with f, together with f''(x).

Over the interval [0,2], it sure looks like |f''(x)|<1

So I can take M_2 =1. Now, our error bound says,

Solving that equation for n, we find that we need,

Thus we’ll take n =18 and break the interval from 0 to 2 up into 18 parts to get,

This yields the approximation,

Note that the final approximation was computed with this code.

My integral is

Here is a graph of the f(x) and f''(x):

Over the interval [2,4] it looks like

So, M1=1. Now, our error bound says

Solving that equation form, we find that

Thus, we’ll take n=4000 and break the interval from 2 to 4 into

This yields the approximation

The final approximation was computed with this code

My intergral is

Here’s a graph of the integrand f, together with f''(x)

Over the interal [0,2], it looks like

So I can take M_2 = 1. Now, our error bound says

Solving that equation for n, we find that we need

We take n=19 and break the interval from 0 to 2 up into 19 parts to get

This yields the approximation

The final approximation was computed with this code.

My Integral is:

The graph is f(Orange) and f′′(x)(Green) is shown below:

The interval is over [0,1] it looks like
|f''(x)|<10
So I set M_2=10
error \leq M_2 \frac{(b-a)^3}{24n^2}=\frac{10}{24n^2} \stackrel{\color{red}\text{want}}{\leq} \frac{1}{1000}
n \geq \sqrt{\frac{1250}{3}}\approx20.412
Using n= 20 and break the interval from 0 to 1 up into 20 parts to get
\Delta x = \frac{1}{20}\text{ and } x_i = 1 + \frac{i}{20}.
This yields the approximation
\int_{0}^{1}sin({\pi}x)dx\approx\displaystyle\sum_{i=1} ^{20} sin(\pi\frac{(1+\frac{i}{20})+(1+\frac{i+1}{20})}{2})\frac{1}{20}\approx0.6372
Note that the final approximation was computed with Sage Cell Server

My integral is

Over the interval [0,2] it looks like

So I will set

The error bound for a left sum states

So,

So I’ll take n = 16000 and break the interval 0 to 2 into 16000 parts

This yields the approximation

Sage code: Sage Cell Server