What is #f(x) = int xe^(2-x) -2 x^2 dx# if #f(0 ) = 1 #?

Answer 1

#f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+1+e^2#

#f(x)=int(xe^(2-x)-2x^2)dx#
and #f(0)=1#
#f(x)=int(xe^(2-x)-2x^x2)dx=color(red)(int(xe^(2-x))dx)-int2x^2dx#

the second integral is integrated using the power rule; the first integral is done by integrating by parts#

#I_1=int2x^2dx=2/3x^3#
#I_2=int(xe^(2-x))dx#

#IBP formula

#I_2=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#
#u=x=>du=dx#
#(dv)/(dx)=e^(2-x)=>v=-e^(2-x)#
#:. I_2=-xe^(2-x)-int(-e^(2-x))dx#
#I_2=-xe^(2-x)-e^(2-x)#
#:. f(x)=I_2-I_1#
#f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+c#
#f(0)=1#
#=>1=0-e^2-0+c#
#:. c=1+e^2#
#f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+1+e^2#

which can be simplified as required

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Answer 2

To find ( f(x) = \int xe^{2-x} - 2x^2 , dx ) given ( f(0) = 1 ), integrate the function and then use the given initial condition to solve for the constant of integration. The integral of ( xe^{2-x} - 2x^2 ) with respect to ( x ) is ( -xe^{2-x} + 2x^3/3 ). Then, plug in ( x = 0 ) and ( f(0) = 1 ) to solve for the constant of integration. Solving for the constant yields ( C = 1 + 2/3 ), so ( f(x) = -xe^{2-x} + 2x^3/3 + 5/3 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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