Question

need integral help please

Answer 1

post the question

Answer 2

^Patience

Answer 3

\[\int\limits_{}^{} \sqrt{x^3+6}\]

Answer 4

cal 1 or 2?

Answer 5

1 still

Answer 6

Are you sure you posted your question right? Im not seeing a way to do this with u-sub and you dont learn most other integration techniques until cal 2

Answer 7

we are allowed to use a calculator if that helps. I'll post the full thing..

Answer 8

Let f(x) be a continuous function such that f(1)=2 and f'(x)=sqrt(x^3+6). What is the value of f(5) ?

Answer 9

@sshayer

Answer 10

Sorry im not sure how to do this to be honest.

Answer 11

Okay thx for trying :D

Answer 12

@mathmale

Answer 13

@3mar Do you know how to do this?

Answer 14

if you can use a calculator, integrate from one to six

Answer 15

using wolfram i get about \(21.811\)

Answer 16

so \[f(6)-f(1)=21.811\] you know \(f(1)\) solve for \(f(6)\)

Answer 17

ok i messed up i thought it was \(f(6)\) but it is \(f(5)\) so that answer is incorrect

Answer 18

it is about \(15.798\)
http://www.wolframalpha.com/input/?i=integral+from+1+to+5+sqrt(x%5E2%2B6)

Answer 19

same idea though

Answer 20

I think we can find the linear approximation of f(5) from it

\(L(5) = f(1) +f'(1) (1-5)= 2+\sqrt{7}(-4)=2-4\sqrt{7}\)

Answer 21

\[L(x)=f(a)+f'(a)(x-a)\]
so wouldn't that be 5-1 in those brackets, @Loser66

Answer 22

@satellite73
The wolfram work posted was for sqrt(x^2+6), that explains the closed solution! lol
I also have 24.67 as Wolfram when using numerical integration with sqrt(x^+6) from 1 to 5.
@loser66
Linear approximation is an excellent idea for points close to f'(5), or when f'(x) changes very slowly. For points further away from it, the error would increase. Consider f'(1)=2.65 and f'(5)=11.44.

Answer 23

Yes. @amorfide thanks for pointing out.