Integration Help: Pls Explain Attached Picture

[jderulo]

Hi

Pls can anyone explain how the attached picture was worked out?

Thanks

 

[SteamKing]

Hi

Pls can anyone explain how the attached picture was worked out?

Thanks

You should be able to work this integral out by using what you learned in Calc I.

Treat T_C as a constant and take T_B to be the variable of integration.

After you find the antiderivative, substitute the limits and use the rules of logarithms to obtain the final result. That's all there is to it.

BTW, if you haven't learned this, ##\int \frac{dx}{x}= ln\,x + C##

 

[HallsofIvy]

Perhaps you have seen [itex]\int_a^b \frac{dx}{c- x}[/itex]. To integrate that let u= c- x so du= -dx.

 

[jderulo]

You should be able to work this integral out by using what you learned in Calc I.

Treat T_C as a constant and take T_B to be the variable of integration.

After you find the antiderivative, substitute the limits and use the rules of logarithms to obtain the final result. That's all there is to it.

BTW, if you haven't learned this, ##\int \frac{dx}{x}= ln\,x + C##

Bit confused as the equation isn;t in the format ##\int \frac{dx}{x}= ln\,x + C## ??

 

[geoffrey159]

Let's look at what happens on [itex]\int_a^b \frac{dx}{c- x}[/itex], and you'll do the necessary substitutions later.

1. If ## c \notin [a,b]##, then you are integrating a continuous function on ##[a,b]##. In this case, there is no discussion needed, and the theory says your integral is equal to ##F(b) - F(a)##, where ##F## is an antiderivative of ##\frac{1}{c-x}##.
   If ## b < c ## then you can choose ##F(x) = - \ln(c-x) ##
   If ## c < a ## then you can choose ##F(x) = - \ln(x - c) ##
   In any case, you can choose ##F(x) = -\ln |c-x| ## and [itex]\int_a^b \frac{dx}{c- x} = F(b) - F(a) = \ln |\frac{ c-a }{c-b}| = \ln \frac{ c-a }{c-b}[/itex]
2. If ##c\in[a,b]##, then there is a discontinuity at ##x=c##, and it can be shown that ##\frac{1}{c-x}## is not integrable. That's why I think you are in case (1) given the answer.

 

[SteamKing]

Bit confused as the equation isn;t in the format ##\int \frac{dx}{x}= ln\,x + C## ??

Just how much integral calculus have you studied?

 

[Ssnow]

A primitive for the inverse is the natural logaritm, it is important understand how the module work in the argument of logaritm in relation of your physical quantities...