Question

just want to check my answers

Answer 1

\[If F(x)=\int\limits_{0}^{2x}\frac{ 1 }{ t^2+1 }dt--> then F'(x) =\frac{ 2 }{ 4x^2+1}\]

Answer 2

\(\large \color{red}{\checkmark}\)

Answer 3

\[If F(x)=\int\limits_{0}^{x^2}\frac{ 1 }{ t^2+1 }--> then F'(x)=\frac{ 2 }{ x^4+1 }\]

Answer 4

Hey, Honey, I'm curious whether or not you can explain how you get from t^2 to 4x^2. (Your answer is correct, but how'd you get it?)

Answer 5

second answer is inccorrect.. try again

Answer 6

I believe something's missing in your answer to the 2nd problem. What is it?

Answer 7

Remember when using the chain rule, you multiply by the derivative of the upper limit if it is not x by itself.

Answer 8

for the second one is it 2/4x^4+1 because that what i got the first time i did it in my calculator

Answer 9

Shami, thanks for sharing that. I do need to take exception, though. If the upper limit is 2x, we replace t in the integrand with 2x (preferably enclosed in parentheses, and only then do we multiply the result by the DERIVATIVE of 2x.

Answer 10

So, Honey, are you satisfied, or would you like further discussion/information?

I should point out that what I mentioned to Shami is a consequence of the Chain Rule.

Answer 11

i know that but i m really bad at integral that why i want to learn this and i m so confuse for the second question

Answer 12

Would you care to ask any more questions about either your first or second question?

Answer 13

yes what did i do wrong in the second question

Answer 14

\(\large \frac{d}{dx} \left( \int\limits_{0}^{\color{red}{x^2}}\frac{ 1 }{ t^2+1 }\right) = \frac{ 1 }{ (\color{red}{x^2})^2+1 } * \frac{d}{dx}(\color{red}{x^2})\)

Answer 15

u need to multiply it wid derivative of upper limit... as shamil mentioned

Answer 16

wait do that with top and bottom

Answer 17

top

Answer 18

x^2/x^4+1

Answer 19

\(\large \frac{d}{dx} \left( \int\limits_{0}^{\color{red}{x^2}}\frac{ 1 }{ t^2+1 } dt\right) = \frac{ 1 }{ (\color{red}{x^2})^2+1 } * \frac{d}{dx}(\color{red}{x^2})\)


\(\large = \frac{ 1 }{ (\color{red}{x^2})^2+1 } * \color{red}{2x}\)


\(\large = \frac{ \color{red}{2x}}{ (\color{red}{x^2})^2+1 } \)

Answer 20

no wait u said top so its x^2/x^2+1

Answer 21

u forgot to take derivative for top limit.. .

Answer 22

ok now i get why i got the answer worry i evaluate at the worry interval

Answer 23

hmm-

Answer 24

thank u

Answer 25

u wlc :)

Answer 26

i have few more to check can u help