Q.Evaluate $\displaystyle\int_1^3 (2x+1)\,dx$ using the Fundamental Theorem of Calculus.

By the Fundamental Theorem of Calculus, $\dfrac{d}{dx}\int_0^x g(t)\,dt = g(x)$.

Here $g(t) = t\sin t$, so applying the first fundamental theorem of calculus directly:

$f'(x) = \frac{d}{dx}\int_0^x t\sin t\,dt = x\sin x$

By the Fundamental Theorem of Calculus, integrating a derivative just evaluates the original function at the limits.

Let $h(x) = \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)$. Since $\displaystyle\int_0^1 \frac{d}{dx}[h(x)]\,dx = h(1)-h(0)$ (Fundamental Theorem of Calculus), we only need the boundary values — no actual differentiation/integration is required.

At $x=1$: $\dfrac{2(1)}{1+1^2} = 1$, so $h(1)=\sin^{-1}(1)=\dfrac{\pi}{2}$.

At $x=0$: $\dfrac{2(0)}{1+0}=0$, so $h(0)=\sin^{-1}(0)=0$.

So the integral $= \dfrac{\pi}{2} - 0 = \dfrac{\pi}{2}$.

(Note: on $[0,1]$, $\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)=2\tan^{-1}x$ without any branch jump, so $h$ is smooth there and the FTC applies directly.)

Statement of the First Fundamental Theorem of Integral Calculus.

First fundamental theorem of integral calculus: Let $f$ be a continuous function on $[a,b]$, and let $F(x)$ be defined as

$F(x) = \int_a^x f(t)\,dt, \qquad x\in[a,b]$

Then $F$ is differentiable on $[a,b]$ and

$F'(x) = f(x) \quad \text{for all } x\in[a,b]$

i.e. $F$ is an antiderivative of $f$ — the derivative of the variable-upper-limit integral of a continuous function returns the function itself.

Differentiate both sides of the integral equation using the Fundamental Theorem of Calculus and solve for $f(x)$.

We are given $\displaystyle\int_0^x f(t)\,dt = x + \int_1^x t\,f(t)\,dt$.

Differentiate both sides with respect to $x$. By the Fundamental Theorem of Calculus, $\dfrac{d}{dx}\int_0^x f(t)dt = f(x)$ and $\dfrac{d}{dx}\int_1^x t f(t)\,dt = x f(x)$. The right side also has $\dfrac{d}{dx}(x)=1$.

So: $f(x) = 1 + x f(x)$.

Rearrange: $f(x) - x f(x) = 1 \Rightarrow f(x)(1-x) = 1 \Rightarrow f(x) = \dfrac{1}{1-x}$.

By the Second Fundamental Theorem of Calculus, $f'(x)=x\sin x$.

Given $f(x)=\displaystyle\int_0^x t\sin t\,dt$.

The Fundamental Theorem of Calculus (Part 2) states that if $f(x)=\displaystyle\int_a^x g(t)\,dt$ where $g$ is continuous, then

$f'(x)=g(x).$

Here $g(t)=t\sin t$, so replacing $t$ by $x$:

$f'(x)=x\sin x.$

apply the fundamental theorem of calculus directly

By the Fundamental Theorem of Calculus, if $f(x)=\displaystyle\int_0^x t\sin t\,dt$, then $f'(x)$ is just the integrand evaluated at $t=x$:

$f'(x)=x\sin x$

By the Fundamental Theorem of Calculus, differentiating $\int_0^x t\sin t\,dt$ simply substitutes $x$ for $t$.

By the second fundamental theorem of calculus, if $f(x) = \int_0^x g(t)\,dt$, then $f'(x) = g(x)$.

Here $g(t) = t\sin t$, so $f'(x) = x\sin x$.