How to calculate Norm for a Polynomial in P(R) of (1+x)

Question

Editing my question to add more details.

I am doing a course for data science where the first module is Linear Algebra. In one of the chapter about Norm, they taught about vectors.

In practice quiz question is "Question 5 Find the norm of the following polynomial in P(R) of (1+x)."

For this i know that ∥1+x∥= ∫ (1+x)^2 ​ but the answer is √7/3 which i am not able to understand how.

I found another question in this site at LINK but again there is no explanation for how he got the answer.

I am looking reference where i can read and study about norm for a polynomial so that I can calculate the norm for a polynomial P(R) of (1+x).

Answer 1

Finally I am able to figure out how to solve this. I have to apply the calculus which I was missing.

For my question ||1+x||=square root of ∫(1+x)^2 dx

1. I have to solve power of 2 which will give me x^2 +2x+1
2. Then I have to apply anti derivative on this which will give me (x^3/3)(x:1->0)+2(x^2/2)(x:1->0)+1
3. This will give me ((1^3/3)-0^3/3)+ 2((1^2/2)-0^2/2) +1)
4. Then i will have (1/3+ 2*1/2 +1)
5. This will result in 1/3 + 1 + 1
6. 1/3 +2
7. (1+6)/3
8. 7/3
9. Last we need to take the square root of 7/3 to reach the ans

I am sorry as i am new to this site and was not able to explain the answer in more paper symbols. If some can help me and explain the ans in paper symbols, that will be great.

Answer 2

If the inner product $\langle .,. \rangle$ is defined as exactly as in the link, you mentioned in your comment, then:

$$ \|1+x\| = \sqrt{\int_0^1 (1+x)^2 dx} $$ Convince yourself that $$ \int_0^1 (1+x)^2 dx=\frac{7}{3}. $$ Then insert this result in above to get: $$ \|1+x\|= \sqrt{\frac{7}{3}} $$ Maybe this is what you wanted? If not, please let me know, then I can delete this answer.