Evaluate the following integrals.

aortiH
2021-01-13
Answered

Evaluate the following integrals.

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dieseisB

Answered 2021-01-14
Author has **85** answers

Step 1

Consider the integrals,

Suppose that,

Differentiating with respect to "x"

Step 2

Substitute all value in given integrals,

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I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^{n}$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:

$(0.1,\varphi (1)-{y}_{10}),$

$(.05,\varphi (1)-{y}_{20}),$

$(.025,\varphi (1)-{y}_{40}),$

$(.0125,\varphi (1)-{y}_{80}),$

$(.00625,\varphi (1)-{y}_{160})$

And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is ${y}^{\prime}=x+2y,y(0)=1$ and the exact solution I found was $\varphi (x)=\frac{1}{4}(-2x+5{e}^{2x}-1)$.

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Formula is F = 9/5 C + 32

= {F/F less than or equal to ?}

Formula is F = 9/5 C + 32

= {F/F less than or equal to ?}