Let $f(x)=x^2+2x-t^2$ and $f(x)=0$ has two roots $\alpha(t)$ and $\beta(t)(\alpha<\beta)$

Question

Let $f(x)=x^2+2x-t^2$ and $f(x)=0$ has two roots $\alpha(t)$ and $\beta(t)(\alpha<\beta)$ where $t$ is a real parameter.Let $I(t)=\int_{\alpha}^{\beta}f(x)dx$.If the maximum value of $I(t)$ be $\lambda.$Find $\lambda.$

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I found $I'(t)=f(\beta)\frac{d\beta}{dt}-f(\alpha)\frac{d\alpha}{dt}$ by Leibnitz theorem.But $f(\beta)$ and $f(\alpha)$ are both zero.I am confused now,what to do?

Answer 1

hint

$$\alpha (t)=-1-\sqrt {1+t^2} $$ $$\beta (t)=-1+\sqrt {1+t^2}$$

$$f (x)=(x+1)^2-(t^2+1)$$

$$I (t)=\int_\alpha^\beta f (x)dx=$$

$$(\frac {2}{3}-2)(1+t^2)\sqrt {1+t^2}$$

$$\max I=I (0)=-4/3$$ $I $ is an even function.

Answer 2

$\alpha(t)=\sqrt{1+t^2}-1$ and $\beta(t)=-(\sqrt{1+t^2}+1)$.

$I(t)=\left(\frac{x^3}{3}+x^2-t^2x\right)|_{\beta(t)}^{\alpha(t)}=\frac{(\sqrt{1+t^2}-1)^3}{3}+(\sqrt{1+t^2}-1)^2-t^2(\sqrt{1+t^2}-1)+\frac{(\sqrt{1+t^2}+1)^3}{3}-(\sqrt{1+t^2}-1)^2-t^2(\sqrt{1+t^2}-1)$

$I'(t)=0 \Rightarrow t(t^2+1)=0\Rightarrow t=0$.

$I(0)=-\frac{4}{3}=\lambda$ is max.

Answer 3

A more geometric approach. Note that $$f(x)=(x+1)^2-(t^2+1).$$

is a concave-up parabola with an axial symmetry through $x=-1$. A change in $t$ corresponds to a vertical shifting of the graph.

Therefore, because between the roots $f(x)$ is negative, to maximise the integral we need to shift the graph upwards as much as possible (to make the negative area as small as possible). We do this by choosing $t=0$:

Plots of $f(x)$ for $t=0,1,\dots,5$.

Hence the $f$ that maximises the area is $x^2+2x$.

The roots are $-2$ and $0$ and

$$\int_{-2}^0(x^2+2x)\,dx=-\frac43.$$