Friday, July 7, 2017

Calculus: Early Transcendentals, Chapter 6, 6.2, Section 6.2, Problem 36

We can see the two graphs intersect at about x = .75 and about x = 1.5.
Now we want to set up an integral where we rotate this area around the x axis and make washers.
The formula for washers is:

where r1 is the outside function and r2 is the inside function.
y = e^(x/2) + e^(-2x) is the inside function and y = 3sin(x^2) is the outside function.
our integral now becomes
pi int_.75^1.5 (3sin(x^2))^2 - (e^(x/2) + e^(-2x))^2 dx
Which evaluates to about 7.5 cubic units.

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