Q.If $f(x)=\displaystyle\int_{0}^{x}t\cos t\,dt$, then $\dfrac{df}{dx}=$

Evaluating $\int_a^b f(x)\,dx$ directly as a limit of Riemann sums is tedious even for simple $f$. Newton and Leibniz's Fundamental Theorems connect a definite integral to the anti-derivative of $f$, turning integration into a two-step procedure: find an anti-derivative, then substitute the limits.

Theorem 9.1 (First Fundamental Theorem). If $f(x)$ is continuous on $[a,b]$ and $F(x)=\displaystyle\int_a^x f(u)\,du$ for $a<x<b$, then $\dfrac{d}{dx}F(x)=f(x)$ — that is, $F$ is an anti-derivative of $f$. (This is the tool behind MCQs of the shape "if $f(x)=\int_0^x g(t)\,dt$, find $f'(x)$": the answer is simply $g(x)$.)

Theorem 9.2 (Second Fundamental Theorem). If $f(x)$ is continuous on $[a,b]$ and $F(x)$ is any anti-derivative of $f(x)$, then

$\int_a^b f(x)\,dx = F(b)-F(a),$ …