What is #f(x) = int xe^(x-1)-x^2e^-x dx# if #f(2) = 7 #?

Question

What is #f(x) = int xe^(x-1)-x^2e^-x   dx# if #f(2) = 7   #?

Christopher Allers · Accepted Answer

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+7-e-10/e^2#

#intxe^(x-1)dx-intx^2e^-xdx#

Let us now deal with only the first integral

#intxe^(x-1)dx# Let #u=x-1# then #du=dx#

#int(u+1)e^udu#

Let #w=u+1#, #dw=du # , # dv=e^u du#, #v=e^u#

#(u+1)e^u-inte^udu#

#(u+1)e^u-e^u+c_1#

Substitute back in

#xe^(x-1)-e^(x-1)+c_1#

Now move onto the second integral

#-intx^2e^-xdx#

Let #u=-x# , #u^2=x^2# , #-du=dx#

#intu^2e^udu#

Let #w=u^2 , dw=2udu , dv=e^udu , v=e^u#

#u^2e^u-2intue^udu#

Let #w=u , dw=du , dv=e^udu , v=e^u#

#u^2e^u-2(ue^u-inte^udu)#

#u^2e^u-2ue^u+2e^u+c_2#

Substitute

#x^2e^-x+2xe^-x+2e^-x+c_2#

Combine both Integrals

#xe^(x-1)-e^(x-1)+c_1+ x^2e^-x+2xe^-x+2e^-x+c_2#

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+C#

Solve for C

#7=e+e^-2(2^2+2(2)+2)+C#

#7=e+10e^-2+C#

#C=7-e-10/e^2#

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+7-e-10/e^2#

Ezra Adams · Answer

undefined To find $ f(x) = \int xe^{x-1} - x^2e^{-x} \, dx $ given $ f(2) = 7 $, integrate the function to get the expression for $ f(x) $, and then solve for the constant of integration using the given condition $ f(2) = 7 $.

First, integrate $ xe^{x-1} - x^2e^{-x} $ with respect to $ x $ to get:

$$ f(x) = \int xe^{x-1} - x^2e^{-x} \, dx = \frac{x^2e^{x-1}}{2} + xe^{-x} + C $$

Now, apply the given condition $ f(2) = 7 $ to find $ C $:

$$ f(2) = \frac{2^2e^{2-1}}{2} + 2e^{-2} + C = 4e + \frac{2}{e^2} + C = 7 $$
$$ C = 7 - 4e - \frac{2}{e^2} $$

Therefore, the function $ f(x) $ is:

$$ f(x) = \frac{x^2e^{x-1}}{2} + xe^{-x} + (7 - 4e - \frac{2}{e^2}) $$