Q.State the first fundamental theorem of integral 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: …

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 $h(0)=\sin^{-1}(0)=0$.

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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

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

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

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

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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

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

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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 integ …

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

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