solve inequality integral (1-->2) e^x*f0(x)dx=<e(e-1) 1=<x=<2 limits in integration 1 to 2fa(x)=(a^2-a+1)x^2+2(a-1)x+1 -infinity<a<+infinity

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to check the inequality and not to solve it, since computing the definite integral yields a value that does not depend on x.

Hence, you need to verify if `int_1^2 e^x*f_0(x)dx =lt e(e-1)`

Notice that the problem provides the equation of the function `f_a(x),`  hence, substituting 0 for a in equation yields:

`f_0(x) = (0^2-0+1)x^2+2(0-1)x+1`

`f_0(x) = x^2 - 2x+ 1`

`f_0(x) = (x-1)^2`

You need to substitute `(x-1)^2`  for `f_0(x)`  in integrand such that:

`int_1^2 e^x*(x-1)^2 dx`

Solving by parts the definite integral yields:

`u = (x-1)^2 =gt du = 2(x-1)dx`

`dv = e^xdx =gt v = e^x`

`int_1^2 e^x*(x-1)^2 dx = e^x*(x-1)^2|_1^2 - 2int_1^2 e^x*(x-1)dx`

You need to solve `int_1^2 e^x*(x-1)dx`  using parts such that:

`u = x - 1 =gt du = dx`

`dv = e^xdx =gt v = e^x`

`int_1^2 e^x*(x-1)dx = e^x*(x-1) -int_1^2 e^x dx`

`2int_1^2 e^x*(x-1)dx = 2e^x(x - 1 - 1)|_1^2`

`2int_1^2 e^x*(x-1)dx = 2e^x(x - 2)|_1^2`

`2int_1^2 e^x*(x-1)dx = 2e^2(2 - 2) - 2(1-2)`

`2int_1^2 e^x*(x-1)dx = 2e`

`int_1^2 e^x*(x-1)^2 dx = e^2*(2-1)^2 - e^1*(1-1)^2 - 2e`

`int_1^2 e^x*(x-1)^2 dx = e^2 - 2e`

`int_1^2 e^x*(x-1)^2 dx = e(e-2)`

Notice that `e(e-2) lt e(e-1), ` hence, the last line checks the inequality `int_1^2 e^x*f_0(x)dx =lt e(e-1).`


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