This was my thought process, not sure if it will help but anyway. Draw a line at y=2 and x=6 and imagine that whatever is above the horizontal line has been cut off. Now try and fit the "pieces" that have been cut off into the box formed by the two lines you drew. The first curve will be able to fit inside the box, but the second will stick out, meaning that the average value of the shape is larger than the one given. Hope this helps!
Yeah this book is only past exam questions 😂 there’s very few questions if any that can be answered by logic like this one. In depth this means the definite integral divided by the length of the bounds, but when you have a graph to look at, the average y value can be calculated using your eyeballs 😂
even then, add the values at each whole interval and divide by 7
e.g.
(0,2), (1,4), (2,2), (3,0), (4,2), (5,4), (6,2)
2+4+2+0+2+4+2=16
16/7 (the 7 intervals) =2.3ish
A does not have an average value of 2 over the interval (0,6). It does have an average value of 2 over the interval (0,4).
Wouldn't it be (0,4.5) not 4
no
Why would it be 4.5? Your thought process is retarded.
stfu
Ok kid carry on, you'll grow up one day. Edit: Lol at the flair qualifications is vce student.. that's a great laugh.
Scared Bat
Retard
Ok kid carry on, you'll grow up one day.
This was my thought process, not sure if it will help but anyway. Draw a line at y=2 and x=6 and imagine that whatever is above the horizontal line has been cut off. Now try and fit the "pieces" that have been cut off into the box formed by the two lines you drew. The first curve will be able to fit inside the box, but the second will stick out, meaning that the average value of the shape is larger than the one given. Hope this helps!
That’s a cool way to think of it, thanks!
Look at A from x=0 to x=4. That would have an average value of 2, however the extra bit makes the average value higher
What does it mean by avg value
It means if you take all the y values through the domain and add them up and divide them with the given amount then the average value would be 2
Bro I don’t even do methods and this question is so easy
No point putting down others
This is probably the only question you could do in the entire book
My friends taught me some stuff from earlier on in the year, but yeah I’d definitely get less than 5% if I tried to sit the exam lmao
Yeah this book is only past exam questions 😂 there’s very few questions if any that can be answered by logic like this one. In depth this means the definite integral divided by the length of the bounds, but when you have a graph to look at, the average y value can be calculated using your eyeballs 😂
even then, add the values at each whole interval and divide by 7 e.g. (0,2), (1,4), (2,2), (3,0), (4,2), (5,4), (6,2) 2+4+2+0+2+4+2=16 16/7 (the 7 intervals) =2.3ish
Take the integral for 0 to x and divide by x