Find $a$ and $b$ that maximize the value of the integral $\int ^{b}_{a}\left( 1-x^{2}\right) dx$.

Question

Original Title: Range that maximizes the value of the integral

Can you help me with this question? $\int ^{b}_{a}\left( 1-x^{2}\right) dx$ dFind the constants and that make the maximum of the curve. (By explaining and explaining the reasons find).

Answer 1

The integrand is positive on $(-1,1)$ and negative outside $[-1,1]$, so the integral is maximized if it is performed over $[-1,1]$, that is, if $a=-1$ and $b=1$.

Technically speaking, the integral can be made arbitrarily large by choosing $a=-1$ and $b$ a large negative number, but I doubt that this is the intention of the question.

Answer 2

For $y>x$,

$\int_{x}^{y} (1-t²) dt=(y-x)-\frac{(y³-x³)}{3}=A(x,y)$

Now, for some $x<1$, $A(x,y)$ will have maximum value for y=1. As, $\frac{\partial A(x,y)}{\partial y}=1-y²=0 \Rightarrow y=1$. Because for $y=-1, \frac{\partial^2 A(x,y)}{\partial y^2}=3>0$.

So, for fixed $x<1$, $A$ will be maximum for $y=$ and that value will be $A(x,1)=\frac{2}{3}+(\frac{x³}{3}-x)$.

And $A(x,1)$ will be extremum when $ \frac{\partial A(x,1)}{\partial x}=0 \Rightarrow 1-x²=0 \Rightarrow x=±1$.

And $\frac{\partial^2 A(x,1)}{\partial x^2}<0$ for $x=-1$.

So, $A(x,y)$ will be maximum when $x=a=-1$ and $y=b=1$.