Question

Question #b3951

Answer 1

`int_0^1 (x-1)^2*e^x*dx=2e-5`

Explanation:

`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`

Answer 2

`2e-5`

Explanation:

.
`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`

Answer 3

See explanation. `2e-5`

Explanation:

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`