Question

integrate ln(2+t^5)dt t=4, t=x Find dy/dx

Answer 1

do you mean \[\huge \int_4^x \ln (2 + t^5)dt\]

Answer 2

then derivative of that?

Answer 3

the upper is 4 and lower is x

Answer 4

yup derivative of that

Answer 5

so \[\huge \frac{dy}{dx} \left(\int_x^4 \ln |2+t^5| dt \right)\]

Answer 6

im thinking integration by parts...

Answer 7

\[ \Large \int_a^b \ln (t^5+2)dt = t\ln (t^5+2)-\int_a^b \frac{5t^5}{t^5+2}dt \]
Seems to go in a pretty messy direction doesn't it?

Answer 8

yes

Answer 9

@lgbasallote use Fundamental theorem of Calculus.

Answer 10

heh? im not the asker why you telling me?

Answer 11

ok let me do this.

Answer 12

using fundamental theorem
\[\Large \frac{d}{dx}\int\limits_{a}^{x}f(t)=f(x)\]

Answer 13

now
\[\Large \frac{d}{dx}\int\limits_{4}^{x}\ln|2+t^5|dt=\ln|2+x^5|\]

Answer 14

@Sunoikeo did you get it?

Answer 15

What I don't understand @sami-21

\[ \Large \int_a^xf(t)dt=F(x)-F(a) \]
If we differentiate that, shouldn't the derivative of the F(a) term be still around?

Answer 16

Or is that short because the entire function will be a function of x and a is a constant?

Answer 17

no that's not the answer for the problem

Answer 18

@Spacelimbus you are right . this is part two. i used FTC part 1 .

Answer 19

i tried ln|2+x^5|

Answer 20

nevermind I just answered my own question, it's evaluated at a constant value.

Answer 21

ah okay, I will keep following this, hopefully I didn't distract

Answer 22

*

Answer 23

your upper limit is x ? correct??

Answer 24

@Sunoikeo what is upper limit?

Answer 25

ok if the upper limit is 4 then using property of definite integrals change limits
.
\[\Large \frac{d}{dx}\int\limits_{x}^{4}\ln|2+t^5|dt= \frac{d}{dx}-\int\limits_{4}^{x}\ln|2+t^5|dt=-\ln|2+x^5|\]

Answer 26

I agree with this @sami-21 , just never applied the fundamental theorem of Calculus like this, there we would be done?

Answer 27

the answer was -ln(2+x^5)