Sketch the region enclosed by y= 6|x| and y = x^2 -7

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To find the limits of integration for the area enclosed by the curves y = 6|x| and y = x^2 - 7, set the equations equal: 6|x| = x^2 - 7. This leads to two cases based on the absolute value: for x ≥ 0, solve 6x = x^2 - 7, resulting in the quadratic x^2 - 6x - 7 = 0. For x < 0, solve -6x = x^2 - 7, leading to x^2 + 6x - 7 = 0. Both cases yield intersection points that can be used to evaluate the integral for the area between the curves.
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i know that the way to solve this is by evaluating the integral from a to b of the first function minus the second one but how would i solve for x to find out what the limits of integration should be?

if you set them equal to each other you get 6|x| = x^2 - 7
but I am not exactly sure what to do with the |x|
 
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like rootX said, when dealing with absolute value functions, it's usually best to define it peace-wise.

f(x) = 6x for x >= 0
-6x for x < 0

Or you could notice that the two functions are even, and you could thus solve for x using 6x, and keep in mind that there is another intersection point opposite the y-axis.
 
apiwowar said:
i know that the way to solve this is by evaluating the integral from a to b of the first function minus the second one but how would i solve for x to find out what the limits of integration should be?

if you set them equal to each other you get 6|x| = x^2 - 7
but I am not exactly sure what to do with the |x|
If x\ge 0, |x|= x so this is 6x= x^2- 7 which is the same as x^2- 6x- 7= 0.

If x< 0, |x|= -x so this is -6x= x^2- 7 which is the same as x^2+ 6x- 7= 0.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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