Question about substitution in $\int \limits_{0}^{x} \frac{3t^7}{e^{t^4/2}}dt$

Question

I got a question about solving the integral $\int \limits_{0}^{x} \frac{3t^7}{e^{t^4/2}}dt$ with substitution. In my book it's substituted by $s=x^4/2$, so the result is

$\int \limits_{0}^{x} \frac{3t^7}{e^{t^4/2}}dt=3\int \limits_{0}^{x^4/2}se^{-s}ds$

Here I don't understand what happend to the $t^7$. Why is it $s$ now, although it was substituted by $s=x^4/2$. Is it a mistake or how was the substitution done here?

Answer 1

From $s=t^4/2$ (your question has a typo), we have $ds = 2t^3 dt$. Then, $$\int_0^x \frac{3t^7}{e^{t^4/2}}dt = \int_0^{x^4/2} 3se^{-s}ds.$$ Note that we substituted $t^7 dt = \frac{t^4}{2}\times 2t^3dt = sds$. This is what happend to $t^7$.

Answer 2

I think it should read $s=t^4/2.$ Then $t^4=2s$ and $2t^3= \frac{ds}{dt}$, hence

$$t^7 dt= s ds.$$

Answer 3

I think what they meant in the question was the substitution $s=t^2 /4$ instead of $s=x^4 /2$ by using the former substitution we get $ds=2t^3dt$ substitute this and the integral reduces to 3$\int_{0}^{x} ($s /$e^s) $ds