1

Please I need help. How do I find the error estimate of the trapezoidal rule of the function $$ \int_{0}^{1}\frac{x^{2}}{1+y^{3}}dy $$ using $$ -\frac{h^{2}}{12}[f'(b)-f'(a)] $$ where $$x,y\in(0,1)$$ I know the formula, the confusing part is finding $f'(b)$ and $f'(a)$. I was thinking $f'(x_{b},y_{b}) = f'(1,1)$ and $f'(x_{a},y_{a}) = f'(0,0)$. Please if I am wrong, correct me. Thanks

Iphy Kelvin
  • 21
  • 1
    Welcome to Math SE. Hint: $x$ is a constant in the integral, so it's not changing (i.e., not $x_a$ and $x_b$), and also the derivative is wrt just $y$. – John Omielan May 05 '20 at 06:30
  • yeah the derivative is wrt to just y, so if x is not changing, what then will be the value of x in f'(b) and f'(a) since x is within (0,1). Thanks. – Iphy Kelvin May 05 '20 at 06:40
  • Note that $f$ is a function of only the variable $y$, i.e., $f(y) = \frac{x^2}{1 + y^3}$. As I stated, $x$ is treated as being a constant parameter, so its value in $f'(a)$ and $f'(b)$ will be just $x$, i.e., some undetermined value in $(0,1)$. In particular, note your error estimate will end up being a function of $x$, i.e., it won't be a constant. – John Omielan May 05 '20 at 06:49
  • Thanks so much. I get it now. So can I apply this same idea to an integral equation with kernel similiar to this equation? – Iphy Kelvin May 05 '20 at 06:50
  • You're welcome. I'm glad you understand this now. However, although I think I know, I'm not completely sure if I understand what you mean by "kernel". Note the basic concept, of course, applies to other integrals where there's one or more variables which aren't being integrated. If you have anything else to ask regarding this, or anything else, which is not part of this current question, I suggest you ask a new question about it. – John Omielan May 05 '20 at 06:53

0 Answers0