Question #b3951

#int_0^1 (x-1)^2*e^x*dx#

After using tabular integration,

#int_0^1 (x-1)^2*e^x*dx#

=#([(x-1)^2-2*(x-1)+2]*e^x)_0^1#

=#[(x^2-4x+5)*e^x]_0^1#

=#2e-5#

.
#I=inte^x(x-1)^2dx#

Since it is a product of two functions we use integration by parts:

#u=(x-1)^2#

#du=2(x-1)dx#

#dv=e^xdx#

#v=e^x#

#intudv=uv-intvdu#

#I=e^x(x-1)^2-2inte^x(x-1)dx=e^x(x-1)^2-2II#

We use integration by parts again for the remaining integral #II#:

#w=x-1#

#dw=dx#

#dz=e^xdx#

#z=e^x#

#intwdz=wz-intzdw#

#II=e^x(x-1)-inte^xdx=e^x(x-1)+e^x#

#I=e^x(x-1)^2-2e^x(x-1)+2e^x=e^x(x^2-2x+1-2x+2+2)=e^x(x^2-4x+5)#

Now, we can evaluate this from #0# to #1#

#I=e(1-4+5)-5=2e-5#

We will solve this using integration by parts, which is essentially an inversion of the product rule.

#(uv)' = u'v + uv' -> int(uv)' = uv = int(u'v)+ int(uv') -> int(uv') = uv - int(u'v)#

It appears that we will have to perform this operation two times.

For our first operation:

#u(x) = (x-1)^2, u'(x) = 2(x-1), v(x) = v'(x)=e^x#

Then...

#int_0^1 (x-1)^2e^xdx =[(x-1)^2e^x]_0^1 - int_0^1(2(x-1)e^x)dx# (A)

One more iteration must be performed, upon the second part of this. Now we have:

#u_2(x) = 2(x-1), u_2'(x) = 2, v_2(x) = v_2'(x) = e^x#

Giving us:

#int_0^1 (2(x-1)e^x)dx = [2(x-1)e^x]_0^1 - int_0^1 2e^xdx= (2)(0)(e) - (2)(-1)(1) - [2e^x]_0^1 = 0 +2 -(2e-2) = -2e +4#

Returning back to our initial integration by parts...

#int_0^1 (x-1)^2e^xdx =[(x-1)^2e^x]_0^1 - int_0^1(2(x-1)e^x)d x= -1 - (-2e+4) = 2e-5#