Question

d/dx from 1 to sqrt(x) of 12t^7 dt

Answer 1

what's the relation between t and x?

Answer 2

or is it 12x^7?

Answer 3

or sqrt(t)?

Answer 4

sqrt t

Answer 5

ok. in general use the following:\[\int\limits_{}^{}t^ndt=\frac{1}{t+1}t^{n+1}\]to solve these

Answer 6

thus,\[12\int\limits_{1}^{\sqrt{t}}t^7dt=\frac{12}{8}((\sqrt{t})^8-\sqrt{1})\]

Answer 7

forgot the exponent 8 on the sqrt(1) term but it=1 anyways

Answer 8

so just simplify it and don't forget your constant of integration

Answer 9

oh ic, thanks alot :)

Answer 10

np

Answer 11

\[\huge \frac{d}{dx}\int\limits_1^{\sqrt x} 12t^7 dt\]
This looks like one of those problems where they just want you to use the FTC, Part 1 (Fundamental Theorem of Calculus).

I don't think they want you to actually do the integration, since there is a d/dx in front. The limit is a sqrt x correct..? Or was it suppose to be t?

Answer 12

yes sqrt x

Answer 13

Here is a quick example.
\[\huge \frac{d}{dx}\int\limits_0^x \cos (e^t) dt \quad = \quad \cos(e^x)\]
You're integrating, but then you're differentiating, giving us back what we started with, but now in terms of a new variable, due to the limits of integration.

Answer 14

In your problem, the only thing you have to be very careful about - is the fact that your upper limit is not simply "x", it's sqrt(x), so we'll have to apply the chain rule when we differentiate.

Answer 15

\[\huge \frac{d}{dx}\int\limits_1^{\sqrt x}12t^7 dt\]
\[\huge =12(\sqrt x)^7\frac{d}{dx}(\sqrt x)\]
Understand what I did? I just plugged in the value of our upper limit for t, and then need to apply the chain rule, taking the derivative of the sqrt term.

Answer 16

It's probably not what you're used to doing with integrals, so it's a little tricky to get used to :) Applying a strange rule hehe

Answer 17

yes, I got it. thanks