Well, you can split the integral to f(x-3) and 2. We know that the integral from 3 to 7 of f(x-3) equals the integral of f(x) from 0 to which equals 6. The integral of 2 from 3 to 7 is 2(7)-2(3)=8. Sum the two and you get 14.How?
yeah but the 2 is included in the function - it's not outside itWell, you can split the integral to f(x-3) and 2. We know that the integral from 3 to 7 of f(x-3) equals the integral of f(x) from 0 to which equals 6. The integral of 2 from 3 to 7 is 2(7)-2(3)=8. Sum the two and you get 14.
This might be hard to visualise- I can latex it later if no one does it soon
Oh I didn't see that, that's a bit silly then because (x-3)+2 is just (x-1). Would be much more interesting if 2 was outside because you have to realise you're just adding a 2x4 rectangle.yeah but the 2 is included in the function - it's not outside it
Yea that's exactly what I thought, so I thought it'd just become and I have no idea how to do thatOh I didn't see that, that's a bit silly then because (x-3)+2 is just (x-1). Would be much more interesting if 2 was outside because you have to realise you're just adding a 2x4 rectangle.
Also, I'm not entirely happy with the question and I have a feeling it is invalid due to assumptions. Will post up a pic in a few minutes.
Not enough detail to determine if there exists a numerical solution. It would have worked if they had specified that in the interval 0 to 4, f(x) is greater than or equal to 0.Does the question even work?