Probably because my original plan was to base it off an area of a rectangle and a triangle, but when I realised what the definition did I forgot to swap it back.
This isn't really a thing; more the concept of visualising the question.
From the graph of y=e^(2x), it appears that the region bound by the curve and the lines y=1 and x=2, is almost the value of the integral, however not quite.
The integral of e^(2x) from 0 to 1 also includes a rectangle beneath it. This rectangle, that does belong to the value of the integral, does not belong in the area of the question required. All he did was subtract off the area to match the question.
(I tried to use GeoGebra to create a graph, but e^4 is too large (about 60) and if I change the scale, the rectangle will appear far too narrow.)
Next question:
Hint for (ii):
Consider the two points in part (i), i.e. (0,0) and (r,h). What must be the boundaries of the integral?
Probably because my original plan was to base it off an area of a rectangle and a triangle, but when I realised what the definition did I forgot to swap it back.
This isn't really a thing; more the concept of visualising the question.
From the graph of y=e^(2x), it appears that the region bound by the curve and the lines y=1 and x=2, is almost the value of the integral, however not quite.
The integral of e^(2x) from 0 to 1 also includes a rectangle beneath it. This rectangle, that does belong to the value of the integral, does not belong in the area of the question required. All he did was subtract off the area to match the question.
(I tried to use GeoGebra to create a graph, but e^4 is too large (about 60) and if I change the scale, the rectangle will appear far too narrow.)
Next question:
Hint for (ii):
Consider the two points in part (i), i.e. (0,0) and (r,h). What must be the boundaries of the integral?
(A line through the origin is of the form , where is the slope, so once we've found the slope, we can quickly write down its equation without having to use point-slope form.)
In the following answer, please pretend that the negative square root of two is actually positive on the second to last line of working. For some reason, if I remove the negative, LaTeX screws up. If anyone knows why, let me know.
In the following answer, please pretend that the negative square root of two is actually positive on the second to last line of working. For some reason, if I remove the negative, LaTeX screws up. If anyone knows why, let me know.