Correct, but there shouldn't have been a need to split PV into 2 parts. It seems you split it then recombined without a purpose.
What is this "area of a rectangle" thing? Where can I find info how to do this in my cambridge textbook?First sketch the graph
Since its that little triangle thingy:
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The trapezoidal rule is more accurate than The Simpson's Rule when more subintervals are used.
When is trapezoidal rule for accurate than the Simpson's Rule? Use a diagram to justify your answer.
d/dx(xe^[2x])
Quick question.
no you're all gee if you leave it unfactorised version but you Have to simplify fully
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.
Part i. FInd gradientProbably 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?
Or we could use the point (h, r) instead, so we can find the volume by rotation about the x-axis (instead of y-axis).Next question:
 A straight line passes through the origin and the point (r, h). State it's equation. \\ (ii) By considering the graph of this line, use integration to verify the volume of a cone:$ V=\frac { \pi { r }^{ 2 }h }{ 3 } \\ $where r is the radius of the cone, and h is the height.$)
The line for (i) is simply
done. I've got my working here http://imgur.com/VatVnPF
http://community.boredofstudies.org/12/mathematics/234262/short-guide-latex.html
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.
.}\\f(x) = x+3x+\frac{8}{x}\\f'(x)=4-\frac{8}{x^2}\\0=4-\frac{8}{x^2}\\0=4x^2-8\\0=x^2-2\\0=(x-\sqrt{2})(x+\sqrt{2})\\x=\sqrt{2}\text{ OR }x=-\sqrt{2}\\f''(x)=\frac{16}{x^3}\\f''(-\sqrt{2})=\frac{16}{-2\sqrt{2}}\text{ which is negative.}\\f''(-\sqrt{2})=\frac{16}{2\sqrt{2}}\text{ which is positive.}\\\therefore\text{The only minimum in the function is at }x=\sqrt{2})