Question

evaluate the integral

Answer 1

\[\int\limits_{1}^{2}(t+\frac{ 1 }{ t })^{2} dt\]

Answer 2

What's the problem?

Answer 3

i have to evaluate this integral

Answer 4

Yeah just expand the terms out then integrate

Answer 5

I would just foil it/expand it out, it becomes a simpler integral.

Answer 6

show me how?

Answer 7

You're currently in calculus level and you don't know how to expand?

Answer 8

\[\Large (t+\frac{ 1 }{ t })^{2} = (t+\frac{ 1 }{ t }) (t+\frac{ 1 }{ t })\]
multiply this out...

Answer 9

i dont know how to solve that. seriously

Answer 10

You remember how to expand something like this?

\[\Large (x+2)(x+2)\]

Answer 11

that is like x^2+4x+4

Answer 12

so what of the rest of the integral?

Answer 13

Do what you just did, with this instead

\[\Large (t+\frac{ 1 }{ t }) (t+\frac{ 1 }{ t })\]

Answer 14

t^2+2/t+1/t^2?

Answer 15

Not quite... \[\Large (t+\frac{ 1 }{ t }) (t+\frac{ 1 }{ t }) =\] \[\Large t^2 + \left( \frac{ 1 }{ t } \times t \right)+ \left( \frac{ 1 }{ t } \times t \right)+\frac{ 1 }{ t^2 }\]

Answer 16

so is that supposed to evaluate to something?

Answer 17

yes... simplify it.

Answer 18

ok

Answer 19

is it

Answer 20

t^2+1/t^2+t/2+1

Answer 21

No... maybe this will help\[\Large t^2 + \left( \frac{ 1 }{ t } \times \frac{ t }{ 1 } \right)+ \left( \frac{ 1 }{ t } \times \frac{ t }{ 1 } \right)+\frac{ 1 }{ t^2 }\]

Answer 22

\[t ^{2}+\frac{ 1 }{ t ^{2}}+2\]

Answer 23

Correct. \[\Large \int\limits\limits_{1}^{2}(t+\frac{ 1 }{ t })^{2} dt = \] \[\Large \int\limits\limits_{1}^{2} \left( t ^{2}+\frac{ 1 }{ t ^{2}}+2 \right) dt\]

Answer 24

but that isnt one of the answers

Answer 25

That's because you need to integrate it...

Answer 26

so how do i do that

Answer 27

Have you heard of the word antiderivative?

Answer 28

ive heard but dont know how to do it

Answer 29

so i need to find the anti derivative of what we just solved?

Answer 30

Yes

Answer 31

Write it this way to make it easier to integrate \[\Large \int\limits\limits\limits_{1}^{2} \left( t ^{2}+t^{-2}+2 \right) dt\]

Answer 32

t^3/3+2t-1/t+ c

Answer 33

Correct... sort of.

Answer 34

\[\frac{ t ^{3} }{ 3 }+2t-\frac{ 1 }{ t }+constant\]

Answer 35

What about your limits?

Answer 36

i dunno

Answer 37

They do not just go poof. Have you seem something that ends with \(|_a^b\) ?

Answer 38

no i guess not?

Answer 39

\[\left. \left(\frac{t^3}{3}+2t-\frac{1}{t}\right)\right |_1^2\]

Answer 40

A) 29/6
B) 37/6
C) 5/6
D) 15/2
doesnt seem like im getting close

Answer 41

Or maybe \[\Large \left[ \frac{t^3}{3}+2t-\frac{1}{t} \right]_{1}^{2}\]

Be patient. We can't just give you the answer if you have no idea how to get it.

Answer 42

so what do i do with that equation

Answer 43

The 1 is the lower limit, the 2 is te upper.

Answer 44

right so what does that mean or what do i do

Answer 45

When you evaluate a definiate integral, it gets you all the area back to the origin.