1
$\begingroup$

I have the following practice problem that I'd like to know how to solve before taking my test, can someone explain what is necessary?

$$D_x \int_{0}^{2x} \left[15 \sqrt{2t^2 + 3t + 4} \right] dt$$

  • 1
    Is that supposed to be $D_x$, the derivative with respect to $x$?2012-12-08
  • 1
    Yes, I believe it should be.2012-12-08
  • 0
    Do you see what to do if the $2x$ were replaced by an $x$ (use the fundamental theorem of calculus)? Do that, but use the chain rule to account for the $2x$.2012-12-08

2 Answers 2

2

$$\frac{d}{dx}\,\int_{a(x)}^{b(x)}f(t)\,dt = f(b(x))\,b'(x) - f(a(x))\,a'(x)$$

In this case, $b(x)=2x$ and $a(x) = 0$.

  • 0
    Forgive me, I'm not sure I understand the notation you've used. does $f(x, t)$ just mean plugging x and t into function f?2012-12-08
  • 0
    @DoesTheLimExist $f$ is a function of $x$ and $t$, yes. e.g. $f(x, t)=x^2+t^2$ then $f(1, 2)=1^2+2^2$2012-12-08
  • 0
    I didn't downvote. I'm currently trying to understand all of what your equation shows. I'm not quite as advanced as your understanding of calculus :)2012-12-08
  • 0
    @DoesTheLimExist Does this make more sense?2012-12-08
  • 0
    Wouldn't that leave us with the answer: $15 \sqrt{2(2t)^2 + 3(2t) + 4} * 2 - 15 \sqrt{2(0)^2 + 3(0) + 4} * 0$?2012-12-09
1

You know from the fundamental theorem of calculus that $$\frac{d}{du}\int_0^uf(t)~dt=f(u)\;.\tag{1}$$ If $u=2x$ and $f(t)=15\sqrt{2t^2+3t+4}$, $(1)$ becomes

$$\frac{d}{du}\int_0^u15\sqrt{2t^2+3t+4}~dt=15\sqrt{2u^2+3u+4}=15\sqrt{8x^2+6x+4}\;.$$

If $$F(x)=\int_0^{2x}15\sqrt{2t^2+3t+4}~dt\;,$$

you now know $\dfrac{dF}{du}$; how do you get $\dfrac{dF}{dx}$ from this?

  • 0
    I see that you substituted u in, but how did you get $8x^2+6x+4$ out of that?2012-12-08
  • 0
    @DoesTheLimExist $u=2x$2012-12-08
  • 0
    @DoesTheLimExist: $2u^2+3u+4=2(2t)^2+3(2t)+4=8t^2+6t+4$2012-12-08
  • 0
    @DoesTheLimExist: No: $(2t)^2=(2t)(2t)=4t^2$, and twice that is $8t^2$.2012-12-08
  • 0
    Isn't $2(2t)^2 = 4t^2$ though? Aside from that, would $\frac{dF}{du} = \frac{dF}{dx}$, meaning would they be the same thing?2012-12-08
  • 0
    @DoesTheLimExist: (i) No; see previous comment. (ii) No, of course not: $u$ and $x$ **aren’t** the same thing. You need the chain rule here.2012-12-08
  • 0
    @DoesTheLimExist: Recall the chain rule in the form $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\;.$$2012-12-08
  • 0
    I was always under the impression the chain rule was: f(g(x) * g'. Perhaps they are the same thing, but the different format confuses me slightly. FYI when it takes me awhile to respond, I haven't left, I'm trying to read up on the topic so I'm able to reply.2012-12-09
  • 0
    @DoesTheLimExist: They’re different notations for the same thing (except that your version should be $f\,'\big(g(x)\big)\cdot t'$). The fractional notation is a little more convenient in this problem, however.2012-12-09
  • 0
    I'm not sure I understand. I'm used to doing integrals like so: f(b) * f'(b) - f(a) * f'(a). I got really confused when u substitution came into play and what not. Why can't this equation be solved this way? (Every time I read more to understand, the more I get confused)2012-12-09