Recycle Bin Posted June 22, 2017 Report Share Posted June 22, 2017 Reply Link to post Share on other sites More sharing options...
SC2Player Posted June 22, 2017 Report Share Posted June 22, 2017 I'm not too sure what do you mean by solve; that should be the answer right there. If you're asking why that's the answer, then first consider the integral of 1 with respect to y. 1 Reply Link to post Share on other sites More sharing options...
kw0573 Posted June 22, 2017 Report Share Posted June 22, 2017 10 hours ago, Recycle Bin said: How far are you into integeation? What integrations problems can you solve? Reply Link to post Share on other sites More sharing options...
Recycle Bin Posted June 25, 2017 Author Report Share Posted June 25, 2017 On 6/22/2017 at 6:47 PM, SC2Player said: I'm not too sure what do you mean by solve; that should be the answer right there. If you're asking why that's the answer, then first consider the integral of 1 with respect to y. On 6/22/2017 at 10:31 PM, kw0573 said: How far are you into integeation? What integrations problems can you solve? I'm so sorry, my IA involves this particular chapter so I'm trying to teach myself but I realised this question made no sense LMAO 1 Reply Link to post Share on other sites More sharing options...
Nomenclature Posted June 25, 2017 Report Share Posted June 25, 2017 If you're interested in the process without using a calculator: First, recognize that log10(2) is a real number—you don't have to put it in a calculator—, but knowing that will allow you to not worry about the chain rule. Then, if you've done any basic integration you should know that any lone constant c simply becomes cy where y is the variable with respect to which we are integrating. So now, we just plug the upper limit into the equation as y and then subtract the quantity of the lower limit into the equation as y. So log10(2) - 0 = log10(2). And voilá, there you have it. Reply Link to post Share on other sites More sharing options...
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