Matematické Fórum

stuart clark

Consider the polynomial $g(x)=ax^2+bx+c$ and $g(0)=0$ and $g(2)=2.$

Then minimum value of $\displaystyle \int^{2}_{0}|g'(x)|dx$ is

stuart clark · 05. 05. 2019 09:55

From $g(0),$ we have $c=0$ and from $g(2)=4a+2b=2\Rightarrow b=1-2a$

So $g'(x)=2ax+b=2ax+(1-2a)=2a(1-x)+1$ and $\int^{2}_{0}|2a(1-x)+1|dx$

Put $2a(1-x)=t$ and $dx=-\frac{1}{2a}dt$

we have $-\frac{1}{2a}\int^{1}_{-1}|2at+1|dt$

How i solve it please explain. Thanks

byk7 · 05. 05. 2019 12:59

Let's start from the integral
$\int^{2}_{0}|2a(1-x)+1|\d x=\int^{2}_{0}|-2a|\cdot\left|x-1-\frac{1}{2a}\right|\d x$
apparently,
$\int\pm\left(x-1-\frac{1}{2a}\right)\d x=\pm\frac{x^2}{2}\mp x\left(1+\frac{1}{2a}\right)+C$
now you should discuss three cases
- 1+1/(2a) < 0,
- 0 ≤ 1+1/(2a) ≤ 2,
- 2 < 1+1/(2a),
compute the definite integrals, which give you some functions depending on 'a'.
Maximize them and choose the largest value.

krakonoš · 05. 05. 2019 17:00

↑ stuart clark:
Hi Stuart Clark
g(0)=0 $\Rightarrow$ c=0
g(2)=0 $\Rightarrow$ $a=\frac{1}{2}-\frac{1}{2}\cdot b$
$I=\int_{0}^{2}|(1-b)x+b|dx$

a)If b<1 then the function is growing.
If b<1 and b>0 then we calculate the area of trapezoid $((2-b)+b)\cdot \frac{2}{2}=2$
If b<1 and b<0 then we calculate the area of two triangles
$(1-b)x+b=0 \Leftrightarrow x=\frac{b}{b-1}$
If $\frac{b}{b-1}<2$ then the area is $\frac{b^{2}}{2\cdot (b-1)}+\frac{b-2}{b-1}\cdot \frac{2-b}{2}=2$
The situation $\frac{b}{b-1}>2$ is not possible (b<1 & b>2)

b)If b>1 then the function is declining
There is the similar situation as in a).
If $\frac{b}{b-1}>2$ you will get the same trapezoid.
If $\frac{b}{b-1}<2$ you will get the same triangles.

c) If b=1 the function is constant I=2

stuart clark

Thanks ↑ byk7: and ↑ krakonoš:.

Can we solve this way

$\int^{2}_{0}|g'(x)|dx \geq \bigg|\int^{2}_{0}g'(x)dx\bigg|=|g(x)|\bigg|^{2}_{0}=|g(2)-g(0)|=2.$

Equality hold when $g(x)=x$

Matematické Fórum

#1 04. 05. 2019 09:51 — Editoval stuart clark (04. 05. 2019 09:51)

function

#2 05. 05. 2019 09:55

Re: function

#3 05. 05. 2019 12:59

Re: function

#4 05. 05. 2019 17:00

Re: function

#5 06. 05. 2019 08:37 — Editoval stuart clark (06. 05. 2019 08:38)

Re: function

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