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Let $P_2$ be the space of polynomials of degree $\leq 2$.

For $T: P_2 \to \mathbb R,$ $T(p)= \int_0^1 p(x)\,dx $

I need to find all values of $p(x)$ such that $T(p)=1$. However, I am unsure how to solve the resulting equation:

$$1=\int_0^1 p(x) \, dx $$

I know that in the case of the indefinite integral, you would simply take the derivative of both sides but that does not seem like it will work (given that you get $0=p(x)$ which does not work).

What is the method for finding all polynomials that give 1 as the value of this integral?

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  • $\begingroup$ Is $P_2$ the space of polynomials of order $2$? $\endgroup$ Jul 23, 2017 at 18:59
  • $\begingroup$ I'm assuming $P_2$ is all polynomials of degree $\leq 2$? $\endgroup$ Jul 23, 2017 at 18:59
  • $\begingroup$ Does the notation $P_2$ Mean anything particular? (For example the space of quadratic polynomials?) $\endgroup$
    – Student
    Jul 23, 2017 at 18:59
  • $\begingroup$ If $p=ax^2+bx+c$ then $T(p)=\frac{a}{3}+\frac{b}{2}+c$. $\endgroup$ Jul 23, 2017 at 19:01
  • $\begingroup$ $P_2$ is the space of polynomials of degree $\le 2$ $\endgroup$
    – wzbillings
    Jul 23, 2017 at 19:02

3 Answers 3

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Let $p(x)=ax^2+bx+c$. Then we have:

$$\int_0^1p(x)dx = \left[\frac{a}{3}x^3+\frac{b}{2}x^2+cx\right]_0^1 = \frac{a}{3}+\frac{b}{2}+c$$

Thus, your solution should be all $p(x)$ with $\frac{a}{3}+\frac{b}{2}+c=1$, or if you prefer integer coefficients, $2a+3b+6c=6$. That's one linear equation, so its solution set has two free variables in it. Do you know how to find it?

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  • $\begingroup$ I believe so--you set the answer such that $c=\frac{-a}{3}+\frac{-b}{2}+1$ and then any polynomial in $P_2$ where this is true would be a solution, correct? $\endgroup$
    – wzbillings
    Jul 23, 2017 at 19:14
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    $\begingroup$ Yes, that works. You can pick any values at all for $a$ and $b$, and then find $c$ like you said. $\endgroup$ Jul 23, 2017 at 19:32
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if $P_2$ are polynomials of degree at most $2$, then you can write them as $ax^2+bx+c$.

$$\int_0^1 (ax^2+bx+c) dx = 1$$

$$\frac{a}{3}+\frac{b}{2}+c=1$$

Hence these polynomials are of the form of $ax^2+bx+\left(1-\frac{a}3-\frac{b}2\right)$

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I think if we try taking polynomials $ax^2+bx+c$ then integrate $\frac{ax^3}3+\frac{bx^2}2+cx+d$ then putting limits will give $\frac{a}3+\frac{b}2+c$ And we can also take polynomials of degree $1$ and constants ...:)

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  • $\begingroup$ Hi, I think the $d$ should not be there after you substitute the limit and take the difference. $\endgroup$ Jul 23, 2017 at 19:06

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