Question

Calculus - I have a few of these problems answered, just checking my answers.

1. if f(x)+g(x)dx=8 [1,7] and f(x)dx=13 what does g(x) =?

Answer 1

I got -5 as my answer because of the property of integrals

Answer 2

You're correct!

Answer 3

okay next one.

2. if F(x)=f(t)dt where f(t) = \[\int\limits_{1}^{t ^{2}} 1/\sqrt{1+u{2}} du find F"(2).\]

Answer 4

i got cos(5x)^2*5+-cos(cosx)^2*-sinx

Answer 5

Is this what you mean?\[f(t)=\int_1^{t^2}\frac{du}{\sqrt{1+u^2}}\]

Answer 6

And are we finding \(F(t)=\int f(t)dt\) or \(F''(2)\)? Could you be a little clearer?

Answer 7

i got cos(5x)^2*5+-cos(cosx)^2*-sinx

Answer 8

can i send u a pdf?
for one on one?

Answer 9

i got cos(5x)^2*5+-cos(cosx)^2*-sinx

Answer 10

ill give u medals for all the help

Answer 11

Yep you can attach a file here.

Answer 12

i got cos(5x)^2*5+-cos(cosx)^2*-sinx

Answer 13

i got cos(5x)^2*5+-cos(cosx)^2*-sinx

Answer 14

#2

Answer 15

still there?

Answer 16

Sorry. That integral isn't a regular trig integral.\[\int \frac{du}{\sqrt{1+u^2}}=\sinh^{-1}(u)+C\]Have you seen those before in your class?

Answer 17

eh in the book perhaps, not lecture

Answer 18

hmm ooh actually you don't need to know that. Notice this:\[F'(x)=\frac{d}{dx}\int_1^x f(t)dt=f(x)\]So\[F''(t)=f'(t)\]Can you figure out what \(f'(t)\) is (it's similar to the above but slightly more involved...you need the chain rule since there's a \(t^2\) in the bounds)?

Answer 19

2
\[\sqrt{1+u^{4}}\]

Answer 20

Very close but not quite...one little thing you need to add. Also, it should be in terms of \(t\), but that's a subtlety and won't actually affect your answer.

Answer 21

2\[\sqrt{1+t^{4}}+c\]

Answer 22

Actually no constant of integration since it's a definite integral. What's \(\frac{d}{dt}t^2\)?

Answer 23

2..... that's why my 2 is above the radical

Answer 24

That's not true.\[\frac{d}{dt}t^2=2t\]

Answer 25

oops crap!

Answer 26

:P

Answer 27

So what's the final answer?

Answer 28

2t\[\sqrt{1+t^{4}}\]

Answer 29

The question asked \(F''(2)\) not \(F''(t)\).

Answer 30

lol that little stuff is going to tear me up! thanks for being so patient! usually im not this bad i promise .... 2*2= 4 ..... 2^4=16+1=17 so 4/17=final

Answer 31

close

Answer 32

LMFAO sqrt!!!!!! 4/\[\sqrt{17}\]

Answer 33

okay okay im done being stupid putting my thinking cap

Answer 34

yakeyglee want to help with #3? i got cos(5x)^2*5+-cos(cosx)^2*-sinx