Integration of -3t^2 wrt t, over interval 7 to x

[scrum]

I did this one

and got -3x^2. Which is correct

and for this next one

i thought it would be

arctan((-5/x))

but it's not and i don't know why.

 

[pka]

Re: Integration

\(\displaystyle h(x) = \int\limits_2^{\frac{{ - 5}}{x}} {\arctan (t)dt} \quad \Rightarrow \quad h'(x) = \arctan \left( {\frac{{ - 5}}{x}} \right)\left( {\frac{2}{{x^2 }}} \right)\)

\(\displaystyle \frac{{d\left( {\frac{{ - 5}}{x}} \right)}}{{dx}} = \left( {\frac{5}{{x^2 }}} \right)\)

We must use the chain rule.

 

[scrum]

Thanks for showing me that.

I'm kind if confused about what you did on the second step.
\(\displaystyle h(x) = \int\limits_2^{\frac{{ - 5}}{x}} {\arctan (t)dt} \quad \Rightarrow \quad h'(x) = \arctan \left( {\frac{{ - 5}}{x}} \right)\left( {\frac{2}{{x^2 }}} \right)\)

how did you get the 2 / x^2 ?

also the second bit means the arctan of (-5/x) but not (2 / x^2) right?

Sorry i'm confused. I'm working ahead so we haven't even had lecture for there lol.

 

[tkhunny]

1) It should be a 5, not a 2.
2) Where are we told that we are looking for derivatives? We seem to be just assuming that. You would do wwll to state it explicitly.

 

[scrum]

1) It should be a 5, not a 2.
2) Where are we told that we are looking for derivatives? We seem to be just assuming that. You would do wwll to state it explicitly.

oh sorry yea they are derivatives of those.