• Go back-and-forth until satisfied Ray Atkinson, Bachelor's Degree
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(a) (i) Explain why the graph of the function f(x) = x^2 ln(1/3x)

(a) (i) Explain why the graph of the function f(x) = x^2 ln(1/3x) lies below
the x-axis for 2 < x < 3 and above the x-axis for 3 < x < 5.

(ii) Use this fact to ﬁnd the area enclosed by the graph and the x-axis
between x = 2 and x = 5, giving your answer to four decimal
places.

(b) Find the volume of the solid of revolution obtained when the graph of
f(x) = cos x + sin x, from x = −1/6pi to x =2/3pi,
a1. For a graph to be negative to the left of one point, and positive to the right, it must be 0 at that point and the derivative must be positive at that point.
3² * ln(3/3) = 9 * ln(1) = 0. That checks.
d/dx of x² ln(x/3) is x+2x ln(x/3) and at x=3 this is 3 + 6 ln(3/3) = 3 + 6*0 = 3. That checks, too.

a2. (5+10 ln(5/3)) - (2 + 4 ln(2/3)) = 10.10826 - 0.37814 = 9.7301

b. The volume of a solid of revolution is represented by π*integral[(f(x))², a, b], so here it is
π * Integral[sin²(x) + cos²(x) + 2*sin(x)*cos(x), -π/6, 2π/3]
π * (x - cos(2x)/2) evaluated from -π/6 to 2π/3
π * 3.11799
9.7955

I can go into a lot more detail in the derivatives and rather messy integrals in a Word file if you need me to.
Customer: replied 8 years ago.

I would appreciate a more detailed answer in a word file, if you could do that as soon as possible that would be great (question a2 is a 12 mark question so I could do with more detail than the line provided). Thanks for your help so far though. :)

I am so glad you said that. It made me check it, and I did it backwards. a2 is completely wrong. Here is what I did. I proved a1 using the derivative and used "what I found" to find a2. What that means is to find the areas separately using the integral. I will go and do it right in the word file.

Here is your corrected file. If you have any questions about it, let me know. It tried to make it as clear as possible, but that last one takes a lot of substitutions, so it may be a little confusing.

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