Mathematics Asked on December 1, 2021
Problem:
Evaluate the following integral.
$$ int_{0}^{4} frac{dx}{sqrt{4-x}} $$
Answer:
Let $u = 4 – x$. We have:
begin{align*}
du &= -dx \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= int_{4}^{0} frac{- du}{sqrt{u}} \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= int_{0}^{4} frac{- du}{sqrt{u}} \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= int_{0}^{4} u^{-frac{1}{2}} ,, du \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= 2u^{ frac{1}{2} } Big|_0^4 = 2(4)^{ frac{1}{2} } – 2(0)^{ frac{1}{2} } \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= 2(2) – 0 \
int_{0}^{4} frac{dx}{sqrt{4-x}} &= 4
end{align*}
My answer matches the back of the book. However, the original function was not defined
at $x = 4$ (due to a divide by $0$ issue ) and I did not do anything special for this case. Should I have? Is my solution correct?
Indeed you are correct - the function "blows up" at ${x=4}$. However - a function can blow up to infinity at one of the endpoints like this and actually have finite area. In order to rigorously calculate the integral, you should use limits:
$${int_{0}^{4}frac{1}{sqrt{4-x}}dx=lim_{trightarrow 4^-}int_{0}^{t}frac{1}{sqrt{4-x}}dx}$$
This is how we deal with improper integrals in general (that is, integrals with a discontinuity at a point). You simply take a limit. That being said, as someone else has pointed out - your solution works fine. You can read more about them here: https://en.wikipedia.org/wiki/Improper_integral
Answered by Riemann'sPointyNose on December 1, 2021
To be thorough for improper integrals we would need to calculate $displaystyle limlimits_{epsilonto 0}int_epsilon^4 frac{mathrm{d}u}{sqrt{u}}$ and find that it has effectively a limit since $2sqrt{4}-2sqrt{epsilon}to 4-0=4$.
But we generally admit a certain numbers of results as "well known", such as $int x^{-a},mathrm{d}x$ is convergent in $0$ for $a<1$ and go directly for the final result without bothering with the improper character of the integral.
Prove convergence of improper integral using change of variable.
Answered by zwim on December 1, 2021
Your solution is fine, if a little informal.
If you want to evaluate it more formally, you need to recognize the integral as an improper integral, so the first step would become $$int_0^4 frac{dx}{sqrt{4-x}} := lim_{tto4 {}-} int_0^t frac{dx}{sqrt{4-x}}$$ but then the rest of the solution is almost identical to yours (with an extra limit hanging around until the end), so it doesn't add much.
Answered by Brian Moehring on December 1, 2021
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