Question #b3951

3 Answers
Dec 7, 2017

int_0^1 (x-1)^2*e^x*dx=2e-510(x1)2exdx=2e5

Explanation:

int_0^1 (x-1)^2*e^x*dx10(x1)2exdx

After using tabular integration,

int_0^1 (x-1)^2*e^x*dx10(x1)2exdx

=([(x-1)^2-2*(x-1)+2]*e^x)_0^1([(x1)22(x1)+2]ex)10

=[(x^2-4x+5)*e^x]_0^1[(x24x+5)ex]10

=2e-52e5

Dec 7, 2017

2e-52e5

Explanation:

.
I=inte^x(x-1)^2dxI=ex(x1)2dx

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

u=(x-1)^2u=(x1)2

du=2(x-1)dxdu=2(x1)dx

dv=e^xdxdv=exdx

v=e^xv=ex

intudv=uv-intvduudv=uvvdu

I=e^x(x-1)^2-2inte^x(x-1)dx=e^x(x-1)^2-2III=ex(x1)22ex(x1)dx=ex(x1)22II

We use integration by parts again for the remaining integral IIII:

w=x-1w=x1

dw=dxdw=dx

dz=e^xdxdz=exdx

z=e^xz=ex

intwdz=wz-intzdwwdz=wzzdw

II=e^x(x-1)-inte^xdx=e^x(x-1)+e^xII=ex(x1)exdx=ex(x1)+ex

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)I=ex(x1)22ex(x1)+2ex=ex(x22x+12x+2+2)=ex(x24x+5)

Now, we can evaluate this from 00 to 11

I=e(1-4+5)-5=2e-5I=e(14+5)5=2e5

Dec 7, 2017

See explanation. 2e-52e5

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