What is #f(x) = int xe^(2-x) + 3x^2 dx# if #f(0 ) = 1 #?

Question

What is #f(x) = int xe^(2-x) + 3x^2  dx# if #f(0  ) = 1  #?

Ezra Adams · Accepted Answer

#-xe^(2-x)-e^(2-x)+x^3+1+e^2# Begin by using the sum rule for integrals and splitting these into two separate integrals: #intxe^(2-x)dx+int3x^2dx# The first of these mini-integrals is solved using integration by parts: Let #u=x->(du)/dx=1->du=dx# #dv=e^(2-x)dx->intdv=inte^(2-x)dx->v=-e^(2-x)# Now using the integration by parts formula #intudv=uv-intvdu#, we have: #intxe^(2-x)dx=(x)(-e^(2-x))-int(-e^(2-x))dx# #=-xe^(2-x)+inte^(2-x)dx# #=-xe^(2-x)-e^(2-x)# The second of these is a case of the reverse power rule, which states: #intx^ndx=(x^(n+1))/(n+1)# So #int3x^2dx=3((x^(2+1))/(2+1))=3(x^3/3)=x^3# Therefore, #intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+C# (remember to add the constant of integration!) We are given the initial condition #f(0)=1#, so: #1=-(0)e^(2-(0))-e^(2-(0))+(0)^3+C# #1=-e^2+C# #C=1+e^2# Making this final substitution, we obtain our final solution: #intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+1+e^2#

Paisley Johnson · Answer

undefined The function f(x) = ∫xe^(2-x) + 3x^2 dx with f(0) = 1 evaluates to f(x) = -(x + 2)e^(2-x) + x^3 + C, where C is the constant of integration.

Cameron Alexander · Answer

undefined To find $ f(x) $, we first need to integrate the given function with respect to $ x $, then apply the given condition $ f(0) = 1 $. Integrating $ xe^{2-x} + 3x^2 $ with respect to $ x $ yields:

$$ \int xe^{2-x} + 3x^2 \, dx = \int xe^{2-x} \, dx + \int 3x^2 \, dx $$

$$ = -xe^{2-x} + \int e^{2-x} \, dx + x^3 + C $$

$$ = -xe^{2-x} - e^{2-x} + x^3 + C $$

Now, to find $ f(x) $, we'll substitute $ f(0) = 1 $:

$$ f(0) = -0e^{2-0} - e^{2-0} + 0^3 + C = -e^2 + C = 1 $$

$$ C = 1 + e^2 $$

So, $ f(x) = -xe^{2-x} - e^{2-x} + x^3 + 1 + e^2 $.