Integral $\int_{0} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$

Question

Evaluate the Integral $$\int_{0} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$$

I got the answer to be $3(x-1)^{1/3}+C$ by integral techniques, but my friend says that it's false because the integral is improper. I couldn't understand what he meant so can you explain?

Answer 1

Formally, you must address the point $\;x=1\;$ , where the given function isn't defined (this is not so bad, but...) and in fact it isn't bounded there. You should evaluate the double limit:

$$\int_0^3\frac{dx}{(x-1)^{2/3}}=\lim_{\epsilon,\delta\to0}\left[\int_0^{1-\epsilon}\frac{dx}{(x-1)^{2/3}}+\int_{1+\delta}^3\frac{dx}{(x-1)^{2/3}}\right]$$

Check you still get the way you began to show...

Answer 2

You need to be careful around $x=1$ because that's where the integrated function is not continuous. That means that to be on the safe side you'd need to calculate the integral in interval $(0,1)$ and $(1,3)$ separately.

However, in this specific case, it is possible to construct an antiderivative that is defined on both sides of $x=1$ and continuous at $x=1$, and it is $3(x-1)^{1/3} + C$, so your result is fine. Just remember that you need to calculate a definite integral, so the final result is $$\big[3(x-1)^{1/3}\big]\big|_{x=0}^{x=3} = 3\big(2^{1/3}-(-1)^{1/3}\big) = 3\big(2^{1/3} +1\big)$$

This doesn't always work. For example, the antiderivative of $\frac{1}{x}$ is $\ln |x| + C$, but it's not continuous at $x=0$ no matter how you choose constant $C$ on both sides. That means that claiming for example that: $$ \int_{-1}^2 \frac{1}{x} dx = \big[\ln|x|\big]\big|_{x=-1}^{x=2}$$ is incorrect.

Answer 3

Since the function you are integrating is not defined at $x=1$, it is an improper integral. You should separate the integral into two parts and integrate separately.

Here is a partial solution :

$$\int_{0} ^{3} \dfrac {1} {(x-1)^{2/3}}dx=\int_{0} ^{1} \dfrac {1} {(x-1)^{2/3}}dx+\int_{1} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$$

Then you can take limits as

$$=\lim_{a\rightarrow 1^-}\int_{0} ^{a} \dfrac {1} {(x-1)^{2/3}}dx+ \lim_{b\rightarrow 1^+}\int_{b} ^{3} \dfrac {1} {(x-1)^{2/3}}dx$$ After this you can integrate, apply the Fundamental Theorem of Calculus and then evaluate the limits.

Answer 4

One should be wary that the integrand has a singularity at $x=1$ and thus it is wise to split the integral (call it $I$) between the subdomains $(0,1)$ and $(1,3)$, so we have $$I = \int_0^1 \frac{1}{(x-1)^{2/3}}dx + \int_1^3 \frac{1}{(x-1)^{2/3}}dx.$$

Since we have just removed a single point, nothing really changes. However, the singularity's nature may make these integrals diverge, so it should be helpful to treat them as improper integrals and follow a limit process with the integration limits. To this end, replace $1$ by $\alpha$ on the left-hand side integral and by $\beta$ on the right-hand side term. Then, take limits as $\alpha\to 1_-$ and $\beta\to 1_ +$.