Find dy/dx for an integral?
Can someone walk me through how to find dy/dx (one of the problems I'm reviewing in my Calculus book):
$$\int_{1/x}^{2} t\sqrt{t-4} dt $$
I know I need my (x) value to be in the numerator so I can flip it and put a negative sign in front:
$$-\int_{2}^{1/x} t\sqrt{t-4} dt $$
Which is equivalent (as I understand it) to:
$$-\int_{2}^{x^{-1}} t\sqrt{t-4} dt $$
Do I then just start plugging in the x^-1 for my t, or do I let u=? something???
3 Answers
$\begingroup$Here is the most general approach.
Let $$y = \int\limits_{\ell(x)}^{u(x)}f(t)\text{ d}t\text{.}$$ I use $u$ and $\ell$ to mean "upper" and "lower." By the fundamental theorem of calculus, we know that $$\int\limits_{\ell(x)}^{u(x)}f(t)\text{ d}t = F(u(x)) - F(\ell(x))$$ where $F$ is an antiderivative of $f$, i.e., $F^{\prime} = f$. Now, $$\begin{align} \dfrac{\text{d}y}{\text{d}x} &= \dfrac{\text{d}}{\text{d}x}\left[F(u(x)) - F(\ell(x))\right] \\ &= F^{\prime}(u(x))u^{\prime}(x)- F^{\prime}(\ell(x))\ell^{\prime}(x) \text{ (chain rule)}\\ &= f(u(x))u^{\prime}(x) - f(\ell(x))\ell^{\prime}(x)\text{ since } F^{\prime} = f\text{.} \end{align}$$ Hence, if $y = \displaystyle \int_{1/x}^{2}t\sqrt{t-4}\text{d}t$, then setting $u = 2$, $\ell = 1/x$, and $f = t\sqrt{t-4}$, we have $$\dfrac{\text{d}y}{\text{d}x} = f(u)u^{\prime}-f(\ell)\ell^{\prime} = f(2)\cdot 0 - f(1/x)\cdot(x) = -f(1/x)(-1/x^2) = -\dfrac{1}{x}\sqrt{1/x-4}(-1/x^2) = \dfrac{1}{x^3}\sqrt{\dfrac{1}{x}-4}\text{.}$$
$\endgroup$ 2 $\begingroup$Let $$F(x)=\int t \sqrt{t-4} dt$$ $$f(x)=F'(x)=x\sqrt{x-4}$$ Then $$y=\int_{\frac1x}^2 t \sqrt{t-4} dt=F(2)-F\left(\frac1x\right)$$ $$\frac{\text d y}{\text d x}=\frac{\text d \left(-F\left(\cfrac1x\right)\right)}{\text d x}$$ $$=-F'\left(\frac1x\right)\frac{\text d \left(\cfrac1x\right)}{\text d x}$$ $$=-f\left(\frac1x\right)\left(-\frac1{x^2}\right)$$ $$=\frac{\cfrac1x\sqrt{\cfrac1x-4}}{x^2}$$ $$=\frac{\sqrt{\cfrac1x-4}}{x^3}$$
$\endgroup$ 3 $\begingroup$With $F$ an antiderivative of the integrand,
$$y=F(2)-F\left(\frac1x\right).$$
Then with the chain rule,
$$\frac{dy}{dx}=-\left(\frac1x\right)'F'\left(\frac1x\right)=\frac1{x^2}\frac1x\sqrt{\frac1x-4}.$$
$\endgroup$
