stuck with ii)

[toxic charm]

Think about what (x-1) does in terms of graph transformations

[zealous merlin]

Is it bad that Idek how do do bi💀

[polar kettle]

surely if you can do a) you can do both prts for b)

just make a new table of values

[toxic charm]

Think about trapezium rule and what wouldve been done in a

[zealous merlin]

Is it 4* answer to a?

[zealous merlin]

So every y value + 4?

[toxic charm]

What is 4* ?

[zealous merlin]

4 multiply

Because isn’t the curve being translated up 4

[toxic charm]

ohh no think of it shifting the graph up by 4 units at everpoint

so theres almost an extra rectangle at the bottom of height 4 and width of the difference of bounds

so its not a multiplier of 4

[zealous merlin]

So how would you work it out?

[toxic charm]

Use trapezium rule for part a and add an area of 4*2

[zealous merlin]

Ohhh

[zealous merlin]

Then for bii that’s a translation 1 to the right

So can we just add on to the tables of values x=2.5 and 3 and find the y value

And do another trap rule eqn

[toxic charm]

I dont think we need to do that at all consider the u sub of u = x-1 and what that would make the new integral

[zealous merlin]

$e^\frac{1}{5}u^2$

[old panther]

Puzzle

[zealous merlin]

You get what I mean

Wait

So it’s the same?

[toxic charm]

Yh

[zealous merlin]

Interesting

[zealous merlin]

Would that method work too

[toxic charm]

should do

everything i did in the image is legit

Changed the limit and changed it to du

[zealous merlin]

Noo the method I said

[toxic charm]

ohh idk maybe i have a feeling it should but im not too sure

[zealous merlin]

Cos no chance I’m thinking of what you did in the exam

[toxic charm]

Fair but the examiners are trying to point you this way i think as they purposely chose 2 bounds 1 higher then the initial and then they made x become x-1

[fair sable]

Ahh that makes a lot of sense idk how I didn’t see that thanks a lot!