From i) you know that :

d/dx (x ln x)

= x(1/x) + (1)ln x

= 1 + ln x


ii) asks you to integrate region bounded by the y-axis and lines y = 2 and y = 5.

So we have to integrate with respect to y instead of with respect to x instead.


We'll have to rewrite the equation of the curve as :

y = e^x
x = ln y

Since x = ln y, integrate with respect to y, from y = 2 to y = 5

∫ x dy = ∫ ln y dy

How to proceed from here and use i)'s result?

Since we already know from i) that d/dx (x ln x) = 1 + ln x, we know that integrating it will give us back x ln x.

But the term x is just a placeholder letter for an unknown and so we can replace it with any other letter/symbol accordingly and the result should read the same way.

For example : (x + 1)² = x² + 2x + 1

We can replace x with any letter eg. y and say:

(y + 1)² = y² + 2y + 1
Likewise,
(a + 1)² = a² + 2a + 1
(λ + 1)² = λ² + 2λ + 1


So, i)'s result can be changed to :

d/dy (y ln y) = 1 + ln y

Integrating it will give us back y ln y.


Now we need to rewrite the following :

∫ ln y dy

= ∫ (1 + ln y - 1) dy
= ∫ (1 + ln y) dy - ∫ 1 dy
= y ln y - y + c


Since we need a definite integral then we sub y = 5 and y = 2

⁵₂∫ (1 + ln y - 1) dy

= [y ln y - y]⁵₂
= [5 ln5 - 5] - [2 ln2 - 2]
= 5 ln5 - 5 - 2 ln2 + 2
= 5 ln5 - 2 ln2 - 3

(Shown)


In summary,

①Make x the subject of the equation.
②Replace the 'x' s of your result from i) with 'y' s
③ Rewrite the term to be integrated, such that one part of it is directly integrable back to the original expression. Integrate the other parts accordingly.