Mathematics
Kiritokita
2016-04-10 19:04:36
How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between x=1 and x=b is equal to 240?
ANSWERS
abcd2
2016-04-11 01:24:15

The Fundamental Theorem of Calculus regarding geometry states that  [latex] intlimits^b_a g{(x)} , dx = F(b)-F(a)[/latex] Where F is the indefinite integral of [latex]g(x)[/latex] The first step is to integrate [latex]g(x)[/latex] [latex] int {4x} , dx = frac{4x^{1+1} }{1+1} = frac{4x^{2} }{2} =2 x^{2} [/latex] Then substitute the value of [latex]b[/latex] and [latex]a=1[/latex] into [latex] 2x^{2} [/latex] [latex][2 (b)^{2}]-[2 (1)^{2}] = 240 [/latex] [latex] 2b^{2} -2=240[/latex] [latex] 2b^{2}=240+2 [/latex] [latex] 2b^{2}=242 [/latex] [latex] b^{2}= frac{242}{2} [/latex] [latex] b^{2}=121 [/latex] [latex]b=11[/latex] Hence the limit of the area under [latex]g(x)[/latex] is between [latex]a=1[/latex] and [latex]b=11[/latex]

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