what is difference between $\int_{0}^{1+x^3}\sqrt{t}$ and $\int_{0}^{x^3}\sqrt{1+t}$ and $\int_{0}^{x}\sqrt{1+t^3}$?

Question

Given $f(x)=\sqrt{1+x^3}$, which of the following is the correct integral?

• $\int_{0}^{1+x^3}\sqrt{t}dt$
• $\int_{0}^{x^3}\sqrt{1+t}dt$
• $\int_{0}^{x}\sqrt{1+t^3}dt$

How are they all not the same thing? Don't you just replace the $t$ inside of the integrals with the top values on the integral sign?

Answer 1

Let us define $$ f_1(x)=\int_0^{1+x^3}\sqrt t~dt, $$ $$ f_2(x)=\int_0^{x^3}\sqrt{1+t}~dt $$ and $$ f_3(x)=\int_0^x\sqrt{1+t^3}~dt. $$ With the substitution $s=1+t$ you get $$ f_2(x)=\int_1^{1+x^3}\sqrt s ~ds = \int_0^{1+x^3}\sqrt s~ds-\int_0^1\sqrt s~ds = f_1(x)-\frac23. $$ You see that $f_1$ and $f_2$ are not the same function.

I don't know if there exists a nice closed form of $f_3$, but we can easily see that it has to be different from $f_1$ and $f_2$. Using FTOC we get $$ f_3'(x)=\sqrt{1+x^3}. $$ If you compute $f_1(x)=\left[\frac23t^{3/2}\right]_0^{1+x^3}=\frac23(1+x^3)^{3/2}$, then using the chain rule implies $$ f_1'(x)=\sqrt{1+x^3}\cdot3x^2\neq\sqrt{1+x^3}=f_3'(x). $$ Because the derivative is unique, the functions $f_1$ and $f_3$ can't be the same. And further $f_1'=f_2'$, so $f_1$, $f_2$ and $f_3$ are all different functions!

Answer 2

We have that given $f(x)$ the integral over $[a,b]$ is given by

$$\int_a^b f(t) \,dt$$

then the correct solution for $f(x)=\sqrt{1+x^3}$ is $$\int_{0}^{3}\sqrt{1+t^3}\,dt$$

Answer 3

Note that $$\int_0^{x} f(t) dt = \int_{0}^{x} \sqrt{1+t^3}\, dt.$$ If you want to use $\sqrt{1+t}$ instead of using $\sqrt{1+t^3}$, you need to do the substitution properly. For example, if we substitute $y=t^3$ then $dt = \frac{1}{3 y^{2/3}} dy$. So $$\int_0^{x} f(t) dt = \int_{0}^{x} \sqrt{1+t^3}\, dt = \int_0^{x^3} \frac{\sqrt{1+y}}{3y^{2/3}} dy= \int_0^{x^3} \frac{\sqrt{1+t}}{3t^{2/3}} dt \neq \int_{0}^{x^3} \sqrt{1+t} dt.$$

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Perhaps, a simpler example would clarify how substitution works. Suppose $h(x) = x^2$ and you want to compute $\int_{0}^{2} h(x) dx$. It is easy to see that $$I = \int_{0}^{2} x^2 dx = \frac{8}{3}.$$ Now if we do what you wrote then $$I = \int_{0}^{2^2} x dx = 4 \neq \frac{8}{3}.$$ Which means that we are missing something in the integrand. Indeed, by setting $t = x^2 \implies dx = \frac{1}{2 \sqrt{t}} dt$, we get $$I = \int_{0}^{2} x^2 dx = \int_{0}^{4} t \cdot \frac{1}{2 \sqrt{t}} dt = \frac{8}{3}.$$ Hopefully, this would clarify the mistake you are doing.