Calculus help plz!
@mrm
alright, let me take a look at it
@AloneS
do you know that \[\int\limits g'(x) dx= g(x)+C \\ \] and also that \[\int\limits_a^b g'(x) dx=g(x)|_a^b =g(b)-g(a)\]
how do i apply this to my problem?
\[\text{ Let } F(x)=\int\limits_1^x g'(t) dt \\ \text{ Then we can write } F(x)=g(t)|_1^x \\ \text{ which means } F(x)=g(x)-g(1)\]
now to find the derivative of F just differentiate both sides
\[F(x)=\int\limits_1^x g'(t) dt \implies F'(x)=g'(x)\]
this is one of the fundamental theorems of calculus
im still a little confused. can you keep helping me?
may i ask what part what I said is confusing so i can explain that part more
what am i taking the derivative of?
you are taking the derivative of F aka the derivative of the integral
with the limits?
yes your integral has limits
the lower limit is x^3 while the upper limit is 1
can you tell me if you understand everything I said above?
yes, but i cant take the derivative of it with t
\[F(x)=\int\limits_1^x g'(t) dt= g(t)|_1^x =g(x)-g(1)\] so you understood this?
no
Sorry, I lost your question Help. You're in good hands with myininaya.
do you know to find a definite integral you must find an antiderivative of the integrand an antiderivative of g'(t) is g(t)
however what I'm trying to show is you don't actually need to know the antiderivative to find the derivative of a definite integral that is in terms of x
like here is another example: \[F(x)=\int\limits_{x^2}^{1} g'(t) dt= g(t)|_{x^2}^{1}=g(1)-g(x^2) \\ \text{ now \to find } F' \\ \text{ I differentiate both sides of } F(x)=g(1)-g(x^2)\]
\[F'(x)=0-(x^2)' \cdot g'(x^2) \\ F'(x)=-2x g'(x^2)\]
g'(x^2) can be evaluated easily since you know what g'(t) equals this was the integrand
can you try yours forgot about cos(t^4) for a sec and replace it with g'(t)
and follow what I did above except with your limits
i will try
Join our real-time social learning platform and learn together with your friends!