Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4 I keep getting .4853 but is incorrect Thanks
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Originally Posted by waite3 Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4 I keep getting .4853 but is incorrect Thanks
Originally Posted by waite3 Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4 I keep getting .4853 but is incorrect Thanks See the image below for the intersection region. (This entire area fits within the required bounds, so there are no cut-offs. The intersection points of the two curves may be found by Keeping in mind that cosine is an even function of x: We obviously integrate over x.
y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4 doh!! I did not That was just plain lazy
I tried the 1.3833 answer but was incorrect and the teacher said im missing the sqrt 2 so any help would be greatly appreciated. thanks
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