How to use bisection method to approximate values?

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Sajith Dharmasena
Sajith Dharmasena 2015-4-22
评论： Sajith Dharmasena 2015-4-25
Okay so I have a function that spits out y values for a range of x values. For example I'm numerically integrating the function y = integral( x^2 + 2x + 6) using trapezoidal method from 0 to 2. So for say, x = 0:0.05:2 I have the corresponding y values. But now using this result, I want to find x values for a range of y values. Does anybody know how to write the code for this using the bisection method? Your help will be greatly appreciated :)

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Nalini Vishnoi
Nalini Vishnoi 2015-4-24
Can you provide more information about the problem? As I understand it, if you are integrating over x and it ranges from 0 to 2, that would mean that there is a single unique value of y. The value would not change as long as the limits of the integration are fixed. Could you please clarify the following information:
1. Do the limits of the integral change?
2. How does y depend on x if the limits of integration are constant?
3. Is the relationship one to one meaning one x value correspond to one unique y value and vice versa?
You can check how to implement a general bisection problem in MATLAB here .
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Sajith Dharmasena
Sajith Dharmasena 2015-4-25
Yes the relationship is one to one. Okay so I'm numerically integrating a function using the midvalue method. So say I've defined my function as
function [y] = mid(b,N) where b is the upper limit of the integral and N is the number of sections. So as you can tell as I increase N, y will converge to its exact value (also the integral goes from 0 to b).
But what I really want is to find the b value for a given y value. So for example, say I got y = 2.3 with mid(1.5,400). I want to write a new function that will spit b = 1.5 if I input y = 2.3 and N = 400. But I'm not sure how to do it.

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