Second Fundamental theorem of calculus

Question

Theorem 3.20 (Second Foundamental Theorem of Calculus)
Let $f$ be a continuous function on $[a, b]$ and $F$ any function on $[a, b]$, differentiable on $(a, b)$, continuous on $[a, b]$ such that $F'(x) = f(x)$ for all $x \in (a, b)$.
Then $$\int_a^b f(x)\mathrm dx = F(b) - F(a)$$

I need to use the second Fundamental theorem of calculus to work out:

$$\int_{0}^\frac{\pi}{8}\tan(2x)\mathrm dx$$

Firstly it is clear that $\tan(2x)$ is continuous on $\left[0,\frac{\pi}{8}\right]$

Now $F(x)=-\frac{1}{2}\ln|\cos(2x)|=-\frac{1}{2}\ln\cos(2x)$

where $F'(x)=f(x)$

To show that $F(x)$ is differentiable $\forall x \in(0,1)$ is it enough to say that as $f(x)$ is continuous on $(0,1)$ the derivative exists?

Answer 1

I hope I read your question correctly as I edit my original answer considerably. To quote your original question To show that F(x) is differentiable ∀x∈(0,1) is it enough to say that as f(x) is continuous on (0,1) the derivative exists? The question is whether or not the fact the derivative function is continuous is important. The reason it's important, which isn't really clear from the Second Fundamental Theorem of Calculus, is if f isn't continuous, it can't have an antiderivative. That's what the first fundamental theorem of calculus says. Without the assumption of continuity of f,while it can be the case that the integral of f(x) is F(x) and F is continuous on [a,b],it does not follow that F(x) is differentiable with F'(x) = f(x). In other words,without the continuity assumption for f(x), the relationship with F is one sided and that means you can't simply evaluate the antiderivative at the endpoints to obtain the integral value of f on the interval.

It's important to note something else,though-the fundamental theorem of calculus (both parts) only guaruntees the existence of an antiderivative. Even if all the conditions are met, it does not follow that that antiderivative can be expressed in closed form i.e. as a formula of basic functions such as polynomials, exponentials, etc. For example, the function e^(-x^2) satisfies all the given conditions-yet it's antiderivative cannot be expressed in closed form!Even the clever trick everyone and his brother learns in basic calculus using polar coordinates to evaluate the integral-notice the result is a number,not a function.

That answer your question?

Answer 2

The theorem tells you that $$\int_a^b f(x)\mathrm dx = F(b) - F(a)$$ where $F$ is any function that verifies $F'(x) = f(x)$ for every $x \in (a, b)$.

You have already found such an $F$, namely $$F(x) = -\frac12\ln\cos2x$$ If you really want to be sure that indeed $F'(x) = f(x)$ you can just differentiate using the theorems of differentiation. But that is usually not necessary, since you find $F(x)$ with the standard methods of antidifferentiation, which guarantee that $F'(x) = f(x)$ over a certain interval.

Nothing else is needed to evaluate the integral, you just compute the difference at the bounds of integrations: $$\int_0^{\pi/8}\tan2x\mathrm dx = F\left(\frac\pi8\right) - F(0) = \frac{\ln2}4$$