guys I can see that you are using MathJax for formatting equations. But in chat.stackexchange the MathJax is getting formatted into equations no matter how many times I refresh the page. Are you guys facing the same issue?
@HrishabhNayal evaluate the integrand (t-1)(t-2)t-5) in the intervals of roots....desmos.com/calculator/pimssr8mhy Then say integration==area under curve.. you're done
@JackRod Agar physics mei tumhara BTech karte hue interest bana hai to I guess you can check out this blog (physicsafterengineering.blogspot.com) and be a part of this community
@HrishabhNayal $$\frac{dF(x)}{dx} = \frac{d}{dt}\int_0^x (t-1)(t-2)(t-5)dt $$ $$\frac{dF(x)}{dx}= (x-1)(x-2)(x-5) $$ $$dF(x)=(x^3-8x^2+17x-10)dx$$ $$F(x)=\frac{x^4}{4}-\frac{8x^3}{3}+\frac{17x^2}{2}-10x$$ From here all you have to prove is $F(x)\neq 0$ when $x\in (0,5)$.
@rash I can't see how to easily show $F(x) \neq 0$ from here. All I can say is that I would have to factorise a cubic polynomial (after taking x acommon from $F(x)$) which is again a tedious task.
Also I think you accidently wrote $\frac{d}{dt}$ in first step. AFAIK it should be $$\frac{d}{dx}\int_0^x (t-1)(t-2)(t-3) \ dt$$
In the equation $3x^3-32x^2+102x-120$, neglect -120 first because I am gonna graph it. So, $3x^3-32x^2+102x$ will be $x(3x^2-32x+102)$. Roughly, find the roots of quad eqn cuz in JEE you don't wanna waste time.
But in between solving the quadratic, you will come to realise the discriminant is <0
so zero is the only root to $3x^3-32x^2+102x$
Sketching the graph you will get something like this
though by hand you can't exactly draw it like that, you will get that zero is the only root
and then minus 120 the whole graph will shift downwards by 120.
from drawing graphs, you can easily find that $F(x)\neq 0$ when $x\in (0,5)$.
@AnindyaPrithvi looking at the exact graph of the function in desmos, I don't think you can approximate area under the function. The area is almost going to infinity.
@rash I dont know what you graphed, but am pretty sure you did not look at my method.....secondly, your method is so lengthy that it will take atleast 15 minutes (without "desmos"), so do not quote "coz you dont wanna waste time in JEE"
@rash To compute $F(x)=\int_{0}^x g(x)$ for x>0, it's obvious that we are finding the area under the curve for g(x)...using sign changes, you can predict that from 0 to 1, it is negative and the only positive region is (1,2) ....within the range of x=5....then either integrate g(x) in the interval 1,2 and say that the magnitude is less than the integration in the interval 0,1....or use g(avg(roots)) to get a near maximum to predict the areas using triangle...
