Перша частина ОТА: існування первісних

(Video) Fundamental Theorem of Calculus - Part 1

by PatrickJMT

Here's a video by patrickJMT where he explains the Fundamental Theorem of Calculus Part 1 and how you can solve problems with it.

The Fundamental Theorem of Calculus Part 1

If $f$ is continuous on $[a,b]$, then $$g(x)=\int_a^xf(t)dt\quad a\le x\le b$$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $$g'(x)=f(x)$$

Essentially, we're just taking the derivative of an integral. In other words, the derivative of an integral of a function is just the function. Basically, the two cancel each other out like addition and subtraction. Furthermore, we're just taking the variable in the top limit of the integral, $x$, and substituting it into the function being integrated, $f(t)$.

Problem 1 - Standard

Find the derivative of $$g(x)=\int_1^x(t^2-1)^{20}dt$$

All we need to do in this problem is replace the variable with $x$. $$g'(x)=(\color{red}{x}^2-1)^{20}$$

Problem 2 - Variable in Lower Limit

Find the derivative of $$h(x)=\int_x^2[\cos(t^2)+t]dt$$

In this problem the variable is in the lower limit. To apply the Fundamental Theorem of Calculus, the variable needs to be in the upper limit. Luckily, we can flip the limits of integration as long as we put a negative sign out front. $$\int_x^2[\cos(t^2)+t]dt=-\int_2^x[\cos(t^2)+t]dt$$

Now that the variable is in the upper limit we can find the derivative. $$h'(x)=-[\cos(\color{red}{x}^2)+\color{red}{x}]$$

Problem 3 - Expression in Upper Limit

Find the derivative of $$g(x)=\int_1^{\sqrt x}\frac{s^2}{s^2+1}ds$$

In this problem, we have our variable in the top limit but it's an expression. To solve these problems we need to apply the chain rule. You can substitute in the expression but then you also need to multiply it by the derivative of the expression. $$\begin{align} g'(x)&=\frac{(\color{red}{\sqrt x})^2}{(\color{red}{\sqrt x})^2+1}\cdot\color{red}{[\sqrt x]'}\\ g'(x)&=\frac{x}{x+1}\cdot\frac{1}{2\sqrt x}\\ g'(x)&=\frac{\sqrt x}{2(x+1)} \end{align}$$

Problem 4 - Expressions in Both Limits

Find the derivative of $$g(x)=\int_{\tan x}^{x^2}\frac{1}{\sqrt{2+t^4}}dt$$

In this problem we have a variable in both the upper and lower limit. What we can do is split the integral into two integrals at some point between the limits. $$\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx$$

Since, it doesn't matter where we break it up at, let's just choose zero. $$=\int_{\tan x}^0\frac{1}{\sqrt{2+t^4}}dt+\int_0^{x^2}\frac{1}{\sqrt{2+t^4}}dt$$

We need to flip the first integral because the variable is on the bottom. $$=-\int_{0}^{\tan x}\frac{1}{\sqrt{2+t^4}}dt+\int_0^{x^2}\frac{1}{\sqrt{2+t^4}}dt$$

Now we substitute the upper limit in and make sure to multiply by the derivative. $$\begin{align} g'(x)&=-\bigg[\frac{1}{\sqrt{2+\color{red}{\tan}^4\color{red}{x}}}\bigg]\cdot\color{red}{[\tan x]'}+\bigg[\frac{1}{\sqrt{2+\color{red}{(x^2)}^4}}\bigg]\cdot\color{red}{[x^2]'}\\ g'(x)&=-\frac{1}{\sqrt{2+\tan^4x}}\cdot\sec^2x+\frac{1}{\sqrt{2+x^8}}\cdot2x\\ g'(x)&=\frac{2x}{\sqrt{2+x^8}}-\frac{\sec^2x}{\sqrt{2+\tan^4x}} \end{align}$$

As mentioned previously, the first Fundamental Theorem of Calculus relates the derivative of a function to its antiderivative. Let's begin with an example:

Of course, we can calculate the derivative directly: $$\frac{d}{dx}\int_0^x 4t^3\, dt = \frac{d}{dx} t^4\bigg|_0^x=\frac{d}{dx} (x^4-0)=\frac{d}{dx} (x^4)=4x^3.$$

However, we can reduce our work greatly using the first Fundamental Theorem of Calculus, which states the following:

Theorem (FTC 1): Suppose $f(x)$ is a continuous function on the closed interval $[a,b]$, and define $F(x)$ on $[a,b]$ as $$F(x)=\int_a^x f(t)\, dt.$$ Then $F(x)$ is differentiable on $(a,b)$ and satisfies $F'(x)=f(x)$ for all $x$ in $(a,b)$.

The last part of the theorem can be rewritten in Leibniz's notation as $$\frac{d}{dx}\int_a^x f(t)\, dt=f(x).$$

This theorem is important for various reasons: first, it tells us that any function continuous over a closed interval has an antiderivative. (In the theorem, $F(x)$ is the antiderivative of $f(x)$.)

This fundamental theorem also lets us calculate the derivative, at an endpoint, of any definite integral of a function over an interval with that endpoint. For example, the problem from above, of calculating $$\frac{d}{dx}\int_0^x 4t^3\, dt,$$ can be solved quickly using the Fundamental Theorem of Calculus: $$\frac{d}{dx}\int_0^x 4t^3\, dt=4x^3.$$ An even greater benefit from this theorem is that it allows us to find the derivative of any integral, regardless of how messy it is! For example, we know by FTC that $$\frac{d}{dx} \int_0^x \sqrt[5]{1+t\cos{t}+\pi^t}= \sqrt[5]{1+x\cos{x}+\pi^x}.$$ However, it would be impossible to compute that derivative directly. This is one of the most important benefits of the theorem.

Suppose $f(x)$ is a continuous function on $[a,b]$, and as above define $$F(x) = \int_a^x f(t)\, dt.$$ Observe that for any $h$, $$\frac{1}{h}(F(x+h)-F(x))=\frac{1}{h}\left(\int_a^{x+h} f(t)\, dt-\int_a^x f(t)\, dt\right)$$ $$=\frac{1}{h}\left[\left(\int_a^x f(t)\, dt+\int_x^{x+h} f(t)\, dt\right)-\int_a^x f(t)\, dt\right]$$ $$=\frac{1}{h}\int_x^{x+h} f(t)\, dt.$$ So, by the definition of derivatives, $$F'(x)=\lim_{h\to 0} \frac{1}{h}(F(x+h)-F(x))=\lim_{h\to 0} \frac{1}{h}\int_x^{x+h} f(t)\, dt.$$ Now, we assume that $h>0$. (The case $h<0$ can be shown very similarly.) In the interval $[x,x+h]$, we know that since $f(t)$ is continuous, it takes on both a minimum value $m_h$ and a maximum value $M_h$. In other words, for any $t$ in $[x,x+h]$, we have $$m_h\le f(t)\le M_h.$$ So, it follows that $$\int_x^{x+h} m_h\, dt\le \int_x^{x+h} f(t)\, dt\le \int_x^{x+h} M_h\, dt.$$ The left hand side is equal to $$\int_x^{x+h} m_h\, dt=m_ht\bigg|_x^{x+h}=m_h((x+h)-x)=m_hh,$$ and similarly the right side equals $M_hh$. Hence, $$m_hh\le \int_x^{x+h} f(t)\, dt\le M_hh,$$ and so dividing both sides by $h$ yields $$m_h\le \frac{1}{h}\int_x^{x+h} f(t)\, dt\le M_h.$$ Since $f(t)$ is continuous, as $h$ approaches $0$, all values of $f(t)$ in the interval $[x,x+h]$ approach $f(x)$, including $m_h$ and $M_h$. But if those two numbers approach $f(x)$, then any number in between the two also approach $f(x)$. Specifically, by taking the limit as $h\to 0$ on all sides of the inequality, we have $$f(x)=\lim_{h\to 0} m_h\le \lim_{h\to 0} \frac{1}{h}\int_x^{x+h} f(t)\, dt\le \lim_{h\to 0} M_h=f(x).$$ Therefore, we must have $$F'(x)=\lim_{h\to 0} \frac{1}{h}\int_x^{x+h} f(t)\, dt=f(x),$$ as desired.

Consider an arbitrary function, and denote $a$ and $b$ as two different values such that the function, $f$, is continuous on $[a,b]$. Finally, suppose that $t$ is some value in the interval $[a,b]$.

Let us define $F(x) = \displaystyle\int_{a}^{t}{f(x)} \ dx$. Basically, this will return the area under the curve $f(x)$.

Now, what the Fundamental Theorem of Calculus says is: $$F'(x) = f(x)$$ Translating this into English, we get that the rate of the area between $a$ and $t$ is essentially $f$ evaluated at $t$.

Intuitively, this makes sense; if the rate of the area change is decreasing, the function values will get smaller and smaller. However, this is not a formal proof. If you would like to know the formal proof, see the next section.

This gives us many insights into what we will study soon, and the idea that the Fundamental Theorem of Calculus gives is fantastic!

Happy Thanksgiving, reader! How about try out these problems? Lol, have fun!