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#### dangerousHobo

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Well I have the following definite intergal:<br><br>
integral from 0 to 4(-20/3)(sin(2x))dx<br><br>
I got -3.818<br>
I'm I able to get a negative number?? Also had to solve it using Simpsons rule and got -2.27<br><br>
Also took the derivative to the forth so I could do error estimating for Simpsons rule and got f^4(x)=(-320/3)sin(2x)<br>
and pluged 4 in and got f^4(4)=-105.5315<br>
I was using Simpsons error estimate to find an n value that would be accurate within .12, so<br><br>
.12=(-105.5315)(4)^5/(180n^4)<br>
and got n to equal 8.41 then rounding up to the neatest even number for Simpson so 10.<br><br>
What I want to know though is, is the -150.53 and -3.818 and -2.27 correct? A negative -3.818 and -2.27 area doesn't make sense.

#### ouroboros1827

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Ok. In the calculator, (-20/3)*fnint(sin(2x),x,0,4) = -1.85, no?<br><br>
It's been a while since I did that stuff...that's all I can offer atm lol

#### dangerousHobo

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<div style="margin:20px;margin-top:5px;">
<div class="smallfont" style="margin-bottom:2px;">Quote:</div>
<table border="0" cellpadding="6" cellspacing="0" width="99%"><tr><td class="alt2" style="border:1px inset;">
<div>Originally Posted by <strong>ouroboros1827</strong></div>
<div style="font-style:italic;">Ok. In the calculator, (-20/3)*fnint(sin(2x),x,0,4) = -1.85, no?<br><br>
It's been a while since I did that stuff...that's all I can offer atm lol</div>
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forgot about the calculator way, just did it and got -3.818333<br><br>
fnint((-20/3)sin(2x),X,0,4)

#### rabidgnome229

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A negative result for an integral simply means that most of the area lies below the x-axis. For example - the graph below has a negative integral from 0 to whatever the edge is.<br><a href="http://assets.overclock.net.s3.amazonaws.com/8/8f/8f11238d_vbattach32739.jpeg"><img src="http://www.overclock.net/image/id/617432/width/525/height/525/flags/LL"></a>

#### dangerousHobo

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Thanks rabidgnome. <img src="/images/smilies/smile.gif" border="0" alt="" title="Smile" class="inlineimg" /><br />
<br />
I have another question.<br />
<br />
I needed to find the second derivitive of the following function:<br />
<br />
10ln(4x+1)<br />
<br />
and got it to be: 160ln(4x+1)<br />
<br />
f'(x)=10ln(4x+1)(4)<br />
f''(x)=40ln(4x+1)(4) or 160ln(4x+1)<br />
<br />
Is that correct?

#### rabidgnome229

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no<br><br>
ln() is a function, so you have to apply the chain rule to it<br>
d/dx ln(f(x)) = 1/f(x) f'(x)<br><br>
so<br><br>
d/dx 10ln(4x+1) = 10*1/(4x+1)*4 = 40/(4x+1)<br><br>
d/dx 40/(4x+1) = (40(4) - (4x+1)(0))/(4x+1)^2 = f''(x)=-160/(4x+1)^2

#### dangerousHobo

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I'm glade your here <img src="/images/smilies/smile.gif" border="0" alt="" title="Smile" class="inlineimg" /><br />
<br />
Thanks for all the constant help <img src="/images/smilies/thumb.gif" border="0" alt="" title="Thumb" class="inlineimg" />

#### ouroboros1827

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<div style="margin:20px;margin-top:5px;">
<div class="smallfont" style="margin-bottom:2px;">Quote:</div>
<table border="0" cellpadding="6" cellspacing="0" width="99%"><tr><td class="alt2" style="border:1px inset;">
<div>Originally Posted by <strong>rabidgnome229</strong></div>
<div style="font-style:italic;">no<br><br>
ln() is a function, so you have to apply the chain rule to it<br>
d/dx ln(f(x)) = 1/f(x) f'(x)<br><br>
so<br><br>
d/dx 10ln(4x+1) = 10*1/(4x+1)*4 = 40/(4x+1)<br><br>
d/dx 40/(4x+1) = (40(4) + (4x+1)(0))/(4x+1)^2 = f''(x)=<b><i><span style="color:#FF0000;">-</span></i></b>160/(4x+1)^2</div>
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Other than being negative, that'd be it...<img alt="" class="inlineimg" src="/images/smilies/cheers.gif" style="border:0px solid;" title="Cheers">

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#### dangerousHobo

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what about to the forth?<br />
<br />
f^4(x)= -15360/(4x+1)^4

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