We use Excel Solver to find optimal values for the limits `a` and `b` which
maximize an integral objective function
$$\underset{a}{\overset{b}{\int}}f\left(x\right)\cdot dx$$
subject to the constraint `a < b`.

For illustration we will use the parabola `f(x) = 1 - x ^{2}`
for which we easily see from the graph that the maximum positive integral of

Working with named variables as shown in Table 1, we define our decision variables `a` and `b` in `B1:B2`
with initial values 0 and 2, our objective integral in `B5`, and a formula in `B6` which we
will supply to Excel solver to impose the constraint `a < b`

A | B | |

1 | a | 0 |

2 | b | 2 |

3 | x | |

4 | f | =1-x^2 |

5 | objective | =QUADF(f,x,a,b) |

6 | constraint | =a-b |

Next, we configure and run Excel Solver as shown in Figure 2. We select to maximize `B5` by varying the limits `B1` and `B2` subject to the constraint `B6 ≤ 0 `.

The Solver iterates until it finds the optimal solution `a = -1, b = -1`, then displays a Results Dialog as shown in Figure 3.

A | B | |

1 | a | -1 |

2 | b | -1 |

3 | x | |

4 | f | =1 |

5 | objective | 1.3333333 |

6 | constraint | -2 |

We change the objective function to
$$f\left(x,b\right)=\underset{a}{\overset{b}{\int}}1-{x}^{2}+b\cdot dx$$
by modifying cell `B4` of Table 1 to `=1-x^2+b`

The solution is no longer obvious visually by graphing.

We run Excel Solver again. It remembers the values form last run so we do not need to make any changes.
Now it reports the new solution `a = -2` and `b = 3`.
It is interesting to note that the integral includes a negative portion between 2 and 3 yet the net value is maximized over
any other values for `a` or `b`. The achieved maximum in this case is 8.3333333.