2015 HSC maths ext question on integration by substitution (1 Viewer)

narges-ts · Jul 15, 2017

This question is 11 e) from the 2015 paper. Im confused as too how to get rid of the x on the top of the function after differentiating the u and subbing in for dx because when you differentiate 2x-1 it turns into just 2 and i cant cancel out the x on the numerator. Someone please help thanks!!

Use the substitution u = 2x – 1 to evaluate integration between 1 and 2 of x/(2x-1)^2 dx

1729 · Jul 15, 2017

narges-ts said:
This question is 11 e) from the 2015 paper. Im confused as too how to get rid of the x on the top of the function after differentiating the u and subbing in for dx because when you differentiate 2x-1 it turns into just 2 and i cant cancel out the x on the numerator. Someone please help thanks!!

Use the substitution u = 2x – 1 to evaluate integration between 1 and 2 of x/(2x-1)^2 dx

$$\noindent $x = \frac{1}{2}(2x-1) + \frac{1}{2}$ so $\int_1^2\frac{x}{(2x-1)^2}dx = \int_1^2\left(\frac{1}{2(2x-1)}+\frac{1}{2(2x-1)^2}\right)dx$

$$\noindent Or just write $x = \frac{1}{2}(u+1)$

narges-ts · Jul 15, 2017

how did you figure out that x=1/2 (2x-1)+1/2?

InteGrand · Jul 15, 2017

narges-ts said:
how did you figure out that x=1/2 (2x-1)+1/2?

$\noindent This (finding constants $a$ and $b$ such that $x \equiv \underbrace{a(2x-1) + b}_{= 2ax + (b-a)}$) can be done by inspection or by equating coefficients in the identity $x \equiv 2a x + (b-a)$.$

InteGrand · Jul 15, 2017

narges-ts · Jul 15, 2017

Thanks guys!

2015 HSC maths ext question on integration by substitution (1 Viewer)

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