Begin by using the sum rule for integrals and splitting these into two separate integrals:
intxe^(2-x)dx+int3x^2dx∫xe2−xdx+∫3x2dx
The first of these mini-integrals is solved using integration by parts:
Let u=x->(du)/dx=1->du=dxu=x→dudx=1→du=dx
dv=e^(2-x)dx->intdv=inte^(2-x)dx->v=-e^(2-x)dv=e2−xdx→∫dv=∫e2−xdx→v=−e2−x
Now using the integration by parts formula intudv=uv-intvdu∫udv=uv−∫vdu, we have:
intxe^(2-x)dx=(x)(-e^(2-x))-int(-e^(2-x))dx∫xe2−xdx=(x)(−e2−x)−∫(−e2−x)dx
=-xe^(2-x)+inte^(2-x)dx=−xe2−x+∫e2−xdx
=-xe^(2-x)-e^(2-x)=−xe2−x−e2−x
The second of these is a case of the reverse power rule, which states:
intx^ndx=(x^(n+1))/(n+1)∫xndx=xn+1n+1
So int3x^2dx=3((x^(2+1))/(2+1))=3(x^3/3)=x^3∫3x2dx=3(x2+12+1)=3(x33)=x3
Therefore, intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+C∫xe2−x+3x2dx=−xe2−x−e2−x+x3+C (remember to add the constant of integration!)
We are given the initial condition f(0)=1f(0)=1, so:
1=-(0)e^(2-(0))-e^(2-(0))+(0)^3+C1=−(0)e2−(0)−e2−(0)+(0)3+C
1=-e^2+C1=−e2+C
C=1+e^2C=1+e2
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∫xe2−x+3x2dx=−xe2−x−e2−x+x3+1+e2
