Consider $f$ a real integrable function, usually we want to evaluate the integral $\int_a^bf(x)dx$ for some $a<b$ given. Now, suppose we know $a$ but we don't know $b$, further, we know the value of this integral, let $\int_a^bf(x)dx=\lambda$. My question is, in what conditions we can find $b$, and how?
A simple case is when $f$ has a known primitive $F$ and $F$ has inverse $F^{-1}$. $$ \int_a^bf(x)dx=\lambda\implies F(b)-F(a)=\lambda\implies F(b)=F(a)+\lambda\implies b=F^{-1}(F(a)+\lambda)$$
In this case we can evaluate $b$, but if this is not the case, what can be done?
Thanks.
PS: More approaches are welcome, approaches not relying on $f$ primitive.
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$\begingroup$I'm not sure how satisfying it will be, but the work you've shown in your Question proves that talk about which $b$ attains which $\lambda$ is equivalent to talk about a specific antiderivative $F$ attaining values $\lambda$, namely the antiderivative of $f$ such that $F(a) = 0$. Since all antiderivatives of $f$ are obtained by adding the famous "+C" to any one of them, the latter constraint can always be arranged (provided the function $f$ is integrable on a suitable interval containing $a$).
The issue is whether solutions exist, and how to find them:
$$ F(b) = \lambda $$
Conceivably $f$ will be given in a form that we can easily evaluate, but expressions for $F$ are not available or are not easily evaluated. In such cases we are apt to fall back on numerical integration schemes to try and show solutions exists and narrow down the approximation of them.
The best "quadrature rule" depends on the smoothness of $f$ and on what precision is needed in locating $b$. Knowing that $F$ is increasing where $f$ is positive and decreasing where $f$ is negative, we can use knowledge of the sign of $f$ to help bracket intervals that might contain solutions $b$. Indeed, since the derivative of $F$ is simply $f$, using root finding methods that exploit derivatives is apt to make refining precision of approximate solutions go that much faster.
$\endgroup$ 1 $\begingroup$You are saying:
A simple case is when $f$ has primitive $F$ and $F$ has inverse $F^{-1}$
The first part of this is the same thing as saying that $f$ is integrable. Then we have the primitive defined by $F\left(x\right)=\int_{a}^{x}f\left(y\right)dy$. Here we also have $F(a)=0$ and in fact you are looking for solutions $b$ of the equation:
$F(b)=\lambda$.
I don't exclude that only the second part of the quote was meant to be mentioned as relevant.
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