hard exponential integration area under curve a2

Please can someone who is really good at maths/ explaining stuff please help. It looks really messy the way I've set it out but trust once you start reading its easy to follow.

Appreciate it so much to anyone that helps. :smile:

Please see attatched (oh please don't make it really complicated as my head is already hurts lol)

The very first attatchment outlines the exact place I'm stuggling with however its quite busy! to say the least lol so I've uploaded the question and mark scheme so you can get your head around the situation and then refer back to the very first one to see what I was stuggling with :biggrin:

(edited 11 years ago)

oh my.png140.1KB

question 1111111111111111.png55.4KB

mark scheme.png74.7KB

Reply 1

boner in jeans

Original post by CasualSoul

Please can someone who is really good at maths/ explaining stuff please help. It looks really messy the way I've set it out but trust once you start reading its easy to follow.

Appreciate it so much to anyone that helps. :smile:

Please see attatched (oh please don't make it really complicated as my head is already hurts lol)

dat text o.o

Reply 2

CasualSoul

OP

16

Original post by boner in jeans

dat text o.o

Original post by Haesights

It might be my eyes but the text is too small to read- can you just type out the question?

Wow I was gonna say that was fast reply LOl. If you click on the attatchment it comes up and appears sall but if you click again a little magnifying glass sshould come up and then if you click again it magnified it to the size you want. Does that work?

Reply 3

CasualSoul

OP

16

Original post by boner in jeans

dat text o.o

Original post by Haesights

It might be my eyes but the text is too small to read- can you just type out the question?

Wow I was gonna say that was fast reply LOL. If you click on the attachment it comes up. If you click again it opens in a new window and then if you click again a little magnifying glass should come up and then if you click again it magnifies it to the size you want. Does that work?

(edited 11 years ago)

Reply 4

boner in jeans

Original post by CasualSoul

Wow I was gonna say that was fast reply LOL. If you click on the attachment it comes up. If you click again it opens in a new window and then if you click again a little magnifying glass should come up and then if you click again it magnifies it to the size you want. Does that work?

wish you told me that before I did the impossible and read that tiny text :smug:

OP, sorry, but could you post the link to the full question? in the midst of your working and the sketch and the random bits and bobs, i'm getting kind of confused

Reply 5

CasualSoul

OP

16

Original post by boner in jeans

wish you told me that before I did the impossible and read that tiny text :smug:

OP, sorry, but could you post the link to the full question? in the midst of your working and the sketch and the random bits and bobs, i'm getting kind of confused

Original post by Haesights

So am I.. what part of the question are you stuck on exactly? You've made so many annotations I'm not quite sure where to start!

I was trying to make it easier lol :colondollar:

but yeah you're right. Gimme a sec and I'll upload the whole question and the answer

Reply 6

Dark Lord of Mordor

1

Original post by CasualSoul

I was trying to make it easier lol :colondollar:

but yeah you're right. Gimme a sec and I'll upload the whole question and the answer

If you look at the graph of f and g sideways, i.e. with the y-axis as the new x-axis and the x-axis as the new y-axis, then can you see anything similar about the two functions, in terms of the area below the graphs? EDIT: And their shapes.

I mean like this:

(edited 11 years ago)

Reply 7

F1Addict

12

Original post by CasualSoul

Please can someone who is really good at maths/ explaining stuff please help. It looks really messy the way I've set it out but trust once you start reading its easy to follow.

Appreciate it so much to anyone that helps. :smile:

Please see attatched (oh please don't make it really complicated as my head is already hurts lol)

I think the key thing you need to understand/remember here is that the inverse of the function is a reflection about the line y=x, hence you have symmetry. So the area bounded by f(x) and y=0 from 0 to ln2, is equal to the area bounded by g(x), x=0 and y=ln2. The image below might help reinforce what I've said. The green and blue areas are identical, the red is what the question asks you to calculate.

(Also, it may help if you look at the curve g(x), and rotate the axes by 90º anti-clockwise, then flip horizontally. The curve should look like f(x). You can then deduce the relevant areas from there.)

(edited 11 years ago)

Reply 8

boner in jeans

Original post by CasualSoul

I was trying to make it easier lol :colondollar:

but yeah you're right. Gimme a sec and I'll upload the whole question and the answer

get what you're asking now and dark lord's right :smug:

OP, what do you know about functions and their inverses? the inverse is the reflection of the function on the line y = x

now as the integral of f(x) from 0 to ln2 (the area between f(x) and the x-axis and the line x = ln2) is the same as the area between g(x) and the y-axis from 0 to ln2 on the y-axis (and the corresponding x-coordinate of ln2 on g(x) is x = 1, the same thing you're finding the area of)

and since you have the area of the rectangle, the area between g(x) and the y-axis from 0 to ln2, you have the area between g(x) and the x-axis from 0 to 1 by subtracting them, which is what the m.s. did

Reply 9

CasualSoul

OP

16

Original post by F1Addict

I think the key thing you need to understand/remember here is that the inverse of the function is a reflection about the line y=x, hence you have symmetry. So the area bounded by f(x) and y=0 from 0 to ln2, is equal to the area bounded by g(x), x=0 and y=ln2. The image below might help reinforce what I've said. The green and blue areas are identical, the red is what the question asks you to calculate.

(Also, it may help if you look at the curve g(x), and rotate the axes by 90º anti-clockwise, then flip horizontally. The curve should look like f(x). You can then deduce the relevant areas from there.)

Original post by boner in jeans

get what you're asking now and dark lord's right :smug:

OP, what do you know about functions and their inverses? the inverse is the reflection of the function on the line y = x

now as the integral of f(x) from 0 to ln2 (the area between f(x) and the x-axis and the line x = ln2) is the same as the area between g(x) and the y-axis from 0 to ln2 on the y-axis (and the corresponding x-coordinate of ln2 on g(x) is x = 1, the same thing you're finding the area of)

and since you have the area of the rectangle, the area between g(x) and the y-axis from 0 to ln2, you have the area between g(x) and the x-axis from 0 to 1 by subtracting them, which is what the m.s. did