Why does the video include the example with $e^{2x}$ and $e^x$?

The DCT justifies exchanging the definite integral over $[a,b]$ with the infinite summation, allowing term-by-term integration. Specifically, it establishes that $\int_a^b \sum f_n(x)\,dx = \sum \int_a^b f_n(x)\,dx$.

Conditions: The previously stated convergence, domination, and finite-integral conditions are assumed to hold.; The summation is over the sequence of functions $f_n$.; The integration is over the closed interval $[a,b]$.

The three hypotheses are: (1) a sequence of functions $f_n$ converges to a function $f$ on the interval $[a,b]$; (2) there exists a dominating function $g$ such that $|g(x)| \ge |f_n(x)|$ for all $n$ and all $x \in [a,b]$; and (3) the integral of the dominating function over the interval is finite, meaning $\int_a^b g(x)\,dx < \infty$.

Conditions: The theorem is presented in a calculus-oriented form rather than a full measure-theoretic proof.; The interval $[a,b]$ is a closed, bounded real interval.

The integral is computed by substituting the Maclaurin series for $\cos x$, using the DCT to interchange the sum and integral, integrating each monomial term using the power rule, simplifying the factorial denominator, recognizing the resulting series as the sine series, and evaluating it at the endpoints $0$ and $\pi/2$.

Conditions: The integration interval is $[0,\pi/2]$.; The cosine power series is $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.; The DCT is applicable because the series terms are dominated by $g(x)=1$ on $[0,\pi/2]$.

The sum and integral can be interchanged because the Dominated Convergence Theorem applies. The terms of the cosine power series are bounded in magnitude by $1$ on the interval $[0,\pi/2]$, making the constant function $g(x)=1$ a valid dominating function with a finite integral over the interval.

Conditions: The integration interval is $[0,\pi/2]$.; The series terms are $\frac{(-1)^n}{(2n)!}x^{2n}$.; The dominating function is $g(x)=1$.