Question

Can someone please explain this to me? Thanks!

Answer 1

Answer:  Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$. We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}$

Answer is choice D

Answer 2

$F(x)=\displaystyle\int_\pi^{x^2}\sqrt{1+t^3}dt~\hfill u = x^2\implies \cfrac{du}{dx}=2x\\\\\\\cfrac{dF}{du}\cdot \cfrac{du}{dx}\implies \cfrac{dF}{dx}~\hfill \cfrac{d}{dx}\left[ \displaystyle\int_\pi^{x^2}\sqrt{1+t^3}dt\right]\implies \cfrac{d}{du}\left[ \displaystyle\int_\pi^{u}\sqrt{1+t^3}dt\right]\cfrac{du}{dx}\\\\\\\sqrt{1+u^3}\cdot \cfrac{du}{dx}\implies \sqrt{1+u^3}2x\implies 2x\sqrt{1+(x^2)^3}\implies \boxed{2x\sqrt{1+x^6}}$