f(x)=∫₁²ˣ√(1 + t³)dt\nf(x)=□

Explanation:

Step1: Apply the chain - rule and fundamental theorem of calculus

Let $u = 2x$, then $F(x)=\int_{1}^{u}\sqrt{1 + t^{3}}dt$. By the fundamental theorem of calculus and the chain - rule, if $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)$.

Step2: Identify $f(t)$, $u(x)$ and their derivatives

Here, $f(t)=\sqrt{1 + t^{3}}$, $u(x)=2x$, and $u^{\prime}(x) = 2$.

Step3: Calculate $F^{\prime}(x)$

Substitute $u(x)$ into $f(t)$ and multiply by $u^{\prime}(x)$. So $F^{\prime}(x)=2\sqrt{1+(2x)^{3}}$.

Answer:

$2\sqrt{1 + 8x^{3}}$