o.O ... more like *bu-sy* these days ðŸ˜›

Gosh feels like in a time machine right now ...

]]>Welcome to JÏ†ss Sticks Gavin!

Must say, your approach certainly radiates elegance and is rather brilliant! The gradient of the line being e, together with P(1, e), would *straightaway* point to the fact that the newly-shifted line passes through the origin - something Miss Loi missed way back in 2007.

Oh why didn't she think of this? WHY DIDN'T SHE THINK OF THIS??!!! WHYYYYY????!!!! *stops her hand from slapping herself*

Alright, the least she could do is present your solution in a proper diagram for the benefit of all:

Thanks for reviving and bringing this old blog post from those carefree days in 2007 back to life ðŸ˜‰

]]>The required area is therefore minus the area of a simple triangle caused by the tangent.

Easier to see with a graph, obviously.

]]>Please see Miss Loi's corrections on your workings! Thanks for your great efforts once more! ðŸ˜‰

]]>m = e

You need to sub in `x` = 1 *before* you find the equation of the tangent (straight line) at P. Hence the gradient at P, `m` = e

equation of line y -(1 + e) = e~~^x~~ (x - 1)

Hence straight line equation is:

`y` = e`x` + 1

at R, x = 0

with the curve y =e

intergrate curve with x cords with point 1 and 0

and we'll get 2.718

2.718 -( e x 1) - (1/2 x(1 e - 1) x1)

= 2.718 -e - 1/2(e)

= 2.718 - 3/2e

er tats de most i can simply up to-.- whats de ans anyway

Shaded area = - (or you can use trapezium area formula as you've done)

= [`x` + e^{x}]^{1}_{0} - [ (e`x`^{2}/2) + `x`]^{1}_{0}

= (e/2 - 1) unit^{2} (after simplification)

]]>= [

= (e/2 - 1) unit

You are sweet and cute in an 'unworldly' way *lol*

]]>not all students are unworldly lar ... some are sweet and cute like me, keke

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