Area Under the Curve

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car0le_la

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Let R be the region in the fourth quadrant enclosed by the x-axis and the curve y=x²-2kx, where k is a constant. If the area of the region R is 36, then the value of k is
a)-3
b)3
c)4
d) 6

I know that in order to solve this problem, area under the curve is found by using an intergral under which case we would need to know the bottom and top x values. The bottom value would be 0 because it is only in the fourth quadrant whose area is being measured. We know that the total area is 36 but I dont know how to attain the top value (b) in order to then obtain the value of k.
b
∫ (x²-2kx) dx = 36
a
The answer is supposed to be B. But I still dont know how to get to the answer.
 
As x goes from being slightly less than b to slightly greater than b the curve leaves the fourth quadrant and heads into the first quadrant.

So, what will be the value of y when x=b? (what line does the curve cross at x=b)?

Also, will the result of the definite integral be positive or negative?
 
The value of y is supposed to be 36 when x=b. The result of this definite integral, the area under the curve is supposed to be positive.
 
I think you might be a bit confused between the integral, and the curve itself.

The definite integral's value will be -36.

But when x=b, then the curve y=x²-2kx will cross the x axis, therefore y will be 0 at this point. Can you use this to find the value of b?
 
The area lies entirely in the fourth quadrant. Since it lies below the x-axis it is a signed area.
The area is negative. To make it come out positive you must account for the upper function which in this case is the x-axis or y = 0
To find the other zero you just factor........ x(x-2k)=0 and find out where it is equal to zero
So x = 0 and x = 2k

You Integral becomes

[MATH]\int_{0}^{2k}0-(x^2-2kx)dx=36[/MATH]
[MATH]\int_{0}^{2k}2kx-x^2dx=36 [/MATH]
Evaluate and solve for k

Hope this helps.
 
Last edited:
[MATH]\int_{0}^{2k}0-(x^2-2kx)dx=36[/MATH]
[MATH]\int_{0}^{2k}2kx-x^2dx=36 [/MATH]
Evaluate and solve for k
....

________2k
kx²-1/3x³ | = 36
________0
(k(2k)²-(1/3)(8k³))-(k(0)²-(1/3)(0³))=36
(k(2k)²-(1/3)(8k³))=36
4k³-(8/3)k³=36
4/3k³=36
k³=27
k=3

Thanks for the help! ?
 
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