I have two functions: $F=6t$ and $x=t^3/3$
I want to find the integral over $t$ from $0$ to $4$.
I am a beginner at integration, so I converted $F$ in terms of $x$ i.e. $6\, (3x)^{1/3}$ and integrated it from $x=64/3$ to $x=0$ (I did this by putting t=4 in the eq. of $x$). The answer was $384$.
What's bothering me is that, when i converted $x$ in terms of $F$ i.e. $(F/6)^3\times 1/3$ and integrated it over from $F=24$ to $F=0$ (I did this by putting $t=4$ in the eq. of $F$). I got $128$.
What I am not getting is why are the answers different?
You have $F(t)=6t$ and $x(t)=t^3/3 \implies F(x)=6(3x)^{1/3}$ and $x(F)=\frac{1}{3}\left(\frac{F}{6}\right)^3$
Your question says that you need to calculate (correct me if I'm wrong), which you did and got as $$\int_{x(0)=0}^{x(4)=\frac{64}{3}}F \cdot\mathrm{d}x=\int_{0}^{\frac{64}{3}}6(3x)^{1/3}\mathrm{d}x=384 \tag{1}$$
Next you did, because you wanted to integrate with respect to $F$ this time, $$\int_{F(0)=0}^{F(4)=24}x \cdot\mathrm{d}F=\int_{0}^{24}\frac{1}{3}\left(\frac{F}{6}\right)^3\mathrm{d}F=128 \tag{2}$$
Okay, now look at $(1)$, if you want to integrate with respect to $F$, you will have to use change of variables from $x$ to $F$. Since you know $$x = \frac{1}{3}\left(\frac{F}{6}\right)^3 \implies \frac{\mathrm{d}x}{\mathrm{d}F}=\frac{F^2}{216} \implies \mathrm{d}x = \frac{F^2}{216} \mathrm{d}F$$ So from $(1)$, you have $$\int_{0}^{\frac{64}{3}}F \cdot\mathrm{d}x=\int_{0}^{24}F\cdot\frac{F^2}{216} \mathrm{d}F=\int_{0}^{24}\left(\frac{F}{6}\right)^3\mathrm{d}F \tag{3}$$
Well, as we know $(1)=(3)$, but clearly $(3) \neq (2)$. I hope you can now see where you went wrong