Understanding fundamental theory of calculus. Textbook unclear

Question

Here's the text I do not understand:

Example 27 is this: Prove that $$\int_a^b x \,dx= \frac{b^2 - a^2}{2}$$

I don't understand what this means:

Notice that $g'(x) = x$, that is, $g' = f$. In other words, if $g$ is defined as the integral of $f$ by Equation 1, then $g$ turns out to be an antiderivative of $f$, at least in this case. And if we sketch the derivative of the function $g$ shown in Figure 4 by estimating slopes of tangents, we get a graph like that of $f$ in Figure 2. So we suspect that $g' = f$ in Example 1 too.

I can see why $g'(x) = x$ because $\frac{d}{dx} \frac{x^2}{2} = x$. But, why does $g' = f$? That means the derivative of $g$ is equal to $f$ right? It does not mean the antiderivative of g is equal to $f$ right?

Answer 1

The book assumes (temporarily, for the sake of an example), that $f(x)=x.$ (It literally says $f(t)=t,$ but that’s the same thing.) After doing the integral, we see that $g’(x)=x.$ So now we have $$g’(x)=x=f(x),$$ From which we conclude $g’$ and $f$ are the exact same function, that is, $g’=f.$

This isn’t stated as a general claim yet; at that point, the book only claims it is true “at least in this case.” But here’s what it actually says about this case:

• $g$ is an antiderivative of $f.$
• $f$ is the derivative of $g.$

There are no other claims about anything being a derivative or antiderivative of anything else.

Be careful how you read the symbols and don’t read “$f$” when the text says “$g$.”

Answer 2

Yes, it means the derivative of $g$ is $f$. This is the Fundamental Theorem of Calculus, since $g$ is the antiderivative of $f$, and differentiating $g$ essentially cancels this, leading to $f$.

Thinking about this intuitively, you know that the integral of $f$ is the area under some function $f$. If you take the derivative of the area under $f$, you're measuring how fast the rate of change of area is.

Taking the limit, you'll realize that the change in area is essentially a rectangle with infinitely small width and a height equivalent to the current height of the function. Thus, the derivative of the integral of $f$ is just the function $f$ itself.

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Concisely, the first part of the Fundamental Theorem of Calculus is:

$$F(x)=\int_a^{x} f(t)\,dt \rightarrow {d\over dx} F(x)=f(x)$$

Given that $f$ is continuous and real-valued over $[a, b]$ and $x \in [a, b]$. This also leads to a corollary, called the second part of the Fundamental Theorem of Calculus:

$$F(x)=\int_{a}^{b} f(t)\,dt = F(b) - F(a)$$

This is where ${b^2-a^2\over 2}$ comes from. $x^2\over 2$ evaluated at $x=b$ and $x=a$ leads to ${b^2\over 2}-{a^2\over 2}$ or ${b^2-a^2\over 2}$.

Answer 3

As shown at the first line in the screenshot, "we take $f(t) = t$". By the Fundamental Theorem of Calculus, the derivative of $g$ is equal to $f$.