# Need help to understand the part of a calculus video related to Fundamental Theorem of Calculus

I cannot understand what professor means in this video at time 42:42. So we say the derivative of function $L$ is $1/x$. Then somehow we use definite integral to find antiderivative. Aren't antiderivative is found by using indefinite integrals? And definite derivatives are used to find the area under the curve? Also I do not understand why he uses $1$ and $x$ as lower and upper limits.

Thanks

## 2 Answers

The function $g(x)=1/x$ is continuous for $x>0.$ The Fundamental Theorem of Calculus says that therefore, for any constant $A>0,$ the function $G(x)=\int_A^xg(t)dt=\int_A^x(1/t)dt$ is an anti-derivative of $g(x).$ In order to have $G(1)=0$ we must have $0=\int_A^1(1/t)dt ,$ which is true iff $A=1.$ So G satisfies $G'(x)=g(x)=1/x$ and $G(1)=0.$

The Fundamental Theorem of Calculus also says that if any $H'(x)=g(x)$ then $H(x)$ must be $\int_B^xg(t)dt$ for some constant $B.$ So if also $H(1)=0$ then $H(x)=\int_1^xg(t)dt=G(x).$

That is:... (I). The function $G(x)=\int_1^xg(t)dt$ satisfies $G'(x)=g(x)$ and $G(1)=0.$... (II). If $H'(x)=g(x)$ and $H(1)=0$ then $H=G.$ So there is a unique solution.

The only reason that there is such a thing as an indefinite integral is the Fundamental Theorem: Indefinite integrals of continuous functions are just anti-derivatives.

It is a bit confusing on how it's worded in the video. From FTC, we know that:

$$\int_a^bf'(t)\ dt = f(b) -f(a)$$

Here, he's saying that he's going to integrate $L'(x)$ because going from the derivative to the function itself is part of the fundamental theorem of calculus. Notice from above how when we take the integral of the derivative (I'm saying this very roughly), you get the original function (minus the starting point $f(a)$)

So $$\int_a^bL'(t)\ dt$$

What should $a$ and $b$ be to get just $L(x)$? Well, from the fundamental theorem of calculus I wrote at the beginning, we're going to want to use $a = 1$ and $b = x$, and we'll see if that works:

$$\int_0^xL'(t)\ dt = L(x) - L(1)$$

And we know that $L(1) = 0$ so that's why you're just left with $L(x)$ on the right side!

But I want to address your question about the area and stuff, because that's confusing at first. Think about a function $s(t)$ that describes position as a function of time. $s'(t)$ would then be your velocity. When you do, for example, $$\int_0^5s'(t)\ dt = s(b) -s(a)$$

What you're saying is if I add up an infinite amount of infinitely tiny velocity times time, you end up with the total change in position!