Question: In a class of $60$ students, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. $6$ students major in both Mathematics and Physics, $15$ majors in both Physics and Biology but no one majors in both Mathematics and Biology.   
 In an English test, the average marks scored by the students majoring in Mathematics is $45$ and that of students majoring in Biology is $60$. However the combined average marks in English of students of these two majors is $50$. What is the maximum possible number of students who major only in Physics?

My solution approach :-   

Using the average marks given for Mathematics, English and both I figured out that $\Rightarrow$   
 Number of students in Mathematics $= 2$ (Number of students in Biology) $ = 2y$ ;where $y$ is the number of students in Biology.

Now number of students majoring ONLY in Physics came out to be $60-3y$.

From here onwards I am not able to proceed. Please help!

Thanks in advance !

Answer: 
Your question title says it: Integrate by parts.
$$\int\_{a}^{b}u dv = uv|\_{a}^{b}- \int\_{a}^{b} vdu$$
Let $u = t^n$, so $dv = e^{-st}dt$, and so $du = nt^{n-1}dt$ while $$v = \frac{e^{-st}}{-s}$$
The limit of the $uv$ term as $t \rightarrow 0$ and as $t \rightarrow \infty$ are both $0$. Now look at what you have left:
$$\int\_{0}^{\infty}\frac{1}{s}nt^{n-1}e^{-st} dt$$
Pull out constants and you have it.
$$\mathcal{L}(t^n)(s)=\frac{n}{s}\mathcal{L(t^{n-1})}$$
You can integrate by part or you can use the Laplace transform of a derivative:
$$s\mathcal{L}(t^n)(s)= \mathcal{L}{(nt^{n-1})}$$
$$s\mathcal{L}(t^n)(s)= \mathcal{L}\{(t^{n})'\}$$
$$s\mathcal{L}(t^n)(s)-0= \mathcal{L}\{(t^{n})'\}$$
This is just:
$$ \mathcal{L}\{y'\}=s\mathcal{L}(y)-y(0)$$
Where $y(t)=t^n$