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\(\int_1^a(3x^2-6x+3)dx=27\)

Guest Jun 6, 2018

Best Answer 

 #1
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\(\text{Find } a: \int_1^a(3x^2-6x+3)dx=27\)

 

After factoring the integrand:\(\int_1^a3(x-1)^2dx=(x-1)^3|_1^a=(a-1)^3=27\)
Taking the cube root:\(a-1=3 \Rightarrow a=\boxed4\)

 

\(\text{The solution you seek is } a=\boxed4\\ \)

 

laugh

GYanggg  Jun 6, 2018
 #1
avatar+945 
+2
Best Answer

\(\text{Find } a: \int_1^a(3x^2-6x+3)dx=27\)

 

After factoring the integrand:\(\int_1^a3(x-1)^2dx=(x-1)^3|_1^a=(a-1)^3=27\)
Taking the cube root:\(a-1=3 \Rightarrow a=\boxed4\)

 

\(\text{The solution you seek is } a=\boxed4\\ \)

 

laugh

GYanggg  Jun 6, 2018

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