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what is a if area of region between x =1,x =2,y=2x^2-1 is a^2-16/3?
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You need to evaluate the area of the region located between the curves `y = 2x^2 - 1, x = 1, x = 2` , hence, you need to evaluate the following definite integral, such that:
`int_1^2(2x^2 - 1)dx`
You need to use the property of linearity of integral, such that:
`int_1^2(2x^2 - 1)dx = int_1^2(2x^2) dx - int_1^2 dx`
`int_1^2(2x^2 - 1)dx = 2x^3/3|_1^2 - x|_1^2`
Using the fundamental theorem of calculus, yields:
`int_1^2(2x^2 - 1)dx = (16/3 - 2/3) - (2 - 1)`
`int_1^2(2x^2 - 1)dx = 14/3 - 1`
`int_1^2(2x^2 - 1)dx = 11/3`
The problem provides the information that the area of the given region is equal to` a^2 - 16/3` , hence, you need to evaluate a, equating the the values of the area, such that:
`11/3 = a^2 - 16/3 => a^2 = 11/3 + 16/3`
`a^2 = (11+16)/3 => a^2 = 27/3 => a^2 = 9 => a = +-sqrt9 => a = +-3`
Hence, evaluating the area of the given region, under the given conditions, yields `a = +-3` .
Posted by sciencesolve on June 30, 2013 at 4:56 PM (Answer #1)
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