Integrating a Taylor Expansion with Limits: Finding the Exact Value

[chemphys1]

Homework Statement

expand
f(x) = x^4 - 3x^3 + 9x^2 +22x +6 in powers of (x-2)

Hence evaluate integral,
(limits 2.2 - 2) f(x) dx

Homework Equations

Taylor expansion for the first part
integral f(x) dx with limits 2.2-2

The Attempt at a Solution

Expansion of the function I've done comes to
78 +46(x-2) +18(x-2)^2 +9/2(x-2)^3 +3/4(x-2)^4

But then I don't know how the x-2 relates to the limits 2.2 - 2,
do I integrate the original integral or the expanded one? And then how do I use the integral to get the exact value

any help appreciated

 

[pasmith]

Homework Statement

expand
f(x) = x^4 - 3x^3 + 9x^2 +22x +6 in powers of (x-2)

Hence evaluate integral,
(limits 2.2 - 2) f(x) dx

Homework Equations

Taylor expansion for the first part

The function is a fourth-order polynomial. Expansion in powers of u = x - 2 is nothing more than substituting x = u + 2 and collecting powers of u.

integral f(x) dx with limits 2.2-2

The Attempt at a Solution

Expansion of the function I've done comes to
78 +46(x-2) +18(x-2)^2 +9/2(x-2)^3 +3/4(x-2)^4

But then I don't know how the x-2 relates to the limits 2.2 - 2,
do I integrate the original integral or the expanded one? And then how do I use the integral to get the exact value

any help appreciated

Start with \int_2^{2.2} f(x)\,dx and consider the substitution x = u + 2.

 

[HallsofIvy]

The function is a fourth-order polynomial. Expansion in powers of u = x - 2 is nothing more than substituting x = u + 2 and collecting powers of u.

Yes, but finding the Taylor series expansion about x= 2, as chemphys1 does, is a good way of doing that.

Start with \int_2^{2.2} f(x)\,dx and consider the substitution x = u + 2.

Exactly right!

 

[chemphys1]

Have I got this right,

I integrate f(x) but with x = u-2

i.e integral of (u-2)^4 - 3(u-2)^3 etc

with the new limits being 0.2 - 0?

Not really sure what the point of the expansion I did was?