What is #f(x) = int (3-x)e^x dx# if #f(0)=-2 #?

Answer 1

#f(x)=(3-x)e^x+e^x-6#

Given #f(x)=int(3-x)e^xdx# and that #f(0)=-2#.
Now, in the given function, we can solve using the formula #int(uv)dx=uintvdx-int(frac{d}{dx}(u)*intvdx)dx# Takin #u=3-x# and #v=e^x# you'll end up with #int(3-x)e^xdx=(3-x)inte^xdx-int(d/dx(3-x)*inte^xdx)dx# So the answer if you solve the above is #f(x)=(3-x)e^x+e^x+c#
Now, they have also said that #f(0)=-2# which means we're going to have to find out what that #c# is. So #f(0)=(3-0)e^0+e^0+cimplies-2=3+1+cimpliesc=-6#

So, taking it as such, we get the answer as given above.

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Answer 2

The function ( f(x) = \int (3 - x)e^x , dx ) evaluates to ( f(x) = (2 - x)e^x + C ), where ( C ) is the constant of integration. Given that ( f(0) = -2 ), we can find ( C ) by substituting ( x = 0 ) into the function. This yields ( -2 = (2 - 0)e^0 + C ), which simplifies to ( -2 = 2 + C ). Solving for ( C ) gives ( C = -4 ). Therefore, the function ( f(x) ) is ( f(x) = (2 - x)e^x - 4 ).

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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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