What would you like to understand?

The final value of the integral is $1$. This is obtained by evaluating the recognized sine series at the endpoints of the interval: $\sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Conditions: The integration interval is $[0,\pi/2]$.; The integrated series has been correctly identified as the Maclaurin series for $\sin x$.; The Fundamental Theorem of Calculus is applied to evaluate the definite integral.

The final value of the integral is $1$. This is obtained by evaluating the recognized sine series at the endpoints of the interval: $\sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Conditions: The integration interval is $[0,\pi/2]$.; The integrated series has been correctly identified as the Maclaurin series for $\sin x$.; The Fundamental Theorem of Calculus is applied to evaluate the definite integral.

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 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 $[a,b]$.