Answer
For $\int_{a}^{b} (1-x^2)dx$, the values of $a$ and $b$ that maximize this integral are $a = -1$ and $b = 1$ because they completely enclose only the positive area under the curve of the function $1-x^2$.
Work Step by Step
For $\int_{a}^{b} (1-x^2)dx$ you are finding the maximum area under the curve of the function $(1-x^2)$. Since this function is an upside down parabola because it has a negative coefficient in the $x^2$, then the limits of integration must be the roots of the parabola because they enclose the positive area of the parabola. To find the zeros, simply equate the function to 0:
$1-x^2 = 0$
$1 = x^2$
$\sqrt 1 = x$
$x = -1$ and $x = 1$
Therefore, the answer is:
$\int_{-1}^{1} (1-x^2)dx$
