Question

Definite integral help. Wait for it :)

Answer 1

\[\int\limits_{\frac{ -\Pi }{ 4 }}^{\frac{ \Pi }{ 6 }}(-\sin x) dx\]
I got
\[\cos(\frac{ \Pi }{ 6 }) - \cos(\frac{ -\Pi }{ 4 })\]
Right so far?

Answer 2

first of all, you know you can "factor" out the -1 right away right?

Answer 3

Hes got it right either way so far, lol.

Answer 4

isn't antiderivative of (-sinx) = cosx?

Answer 5

\[- \int\limits_{-\pi/4}^{\pi/2} \sin(x) dx\]

The derivative of what function makes sin(x)?

Answer 6

I'm used to working with degrees so radians trip me a bit

Answer 7

that's right. my apologies. i never doubted you :)

Answer 8

lol thanks

Answer 9

1pi rad = 180deg

Answer 10

I know, but I automatically convert it in my brain, but I must get used to workign with rads

Answer 11

you'll get used to it I assure you

Answer 12

You should watch this video http://www.youtube.com/watch?v=ao4EJzNWmK8

Answer 13

\[\cos 30 = \frac{ \sqrt{3} }{ 2 } and \cos(-45) = \frac{ 1 }{ \sqrt{2} }?\]

Answer 14

Options are
1) \[\frac{ 2 }{ \sqrt{3} }-\sqrt{2}\]
2) \[\sqrt{2}-\frac{ 1 }{ \sqrt{3} }\]
3) \[\frac{ 1 }{ \sqrt{2} }-\frac{ 2 }{ \sqrt{3} }\]
4) \[2-(\frac{ \sqrt{2}+\sqrt{3} }{ 2 })\]

Answer 15

I get
\[\frac{ \sqrt{3} }{ 2 }-\frac{ 1 }{ \sqrt{2} }\]
Am I wrong?

Answer 16

did i goof?

Answer 17

Its Pi/ 6 the top one. You goofed.

Answer 18

haha. also i did the subtracting in the wrong order XD

Answer 19

I see so. You confused me there for a bit.

Answer 20

Is my answer right?

Answer 21

I think I can help

Answer 22

\[\int\limits_{-\pi/4}^{\pi/6} -\sin(x)\]
\[= \cos(\pi/6) - \cos(-\pi/4)\]
= sqr(3)/2 - 1/sqrt(2)

Answer 23

So I'm right. Problem is I just cannot get my answer into a form that's one of the options.

Answer 24

let's take a look

Answer 25

Option 3 is like you. Wrong way around lol. I presume in a definite integral the upper bound is always more than the lower?

Answer 26

it doesn't need to be

Answer 27

I'm pretty sure I saw in Khan academy video that upper must be higher than lower.

Answer 28

\[\int\limits_{a}^{b}f(x) dx = -\int\limits_{b}^{a} f(x) dx\]
There's a little identity you'll probably run into soon :)

Answer 29

Makes sense but then we have a missing (-1) right?

Answer 30

That's the same as saying (a-b) = -(b-a) so nothing too complicated there :)

Answer 31

hmm
i think we have the correct solution but just need to move things around a little.

Answer 32

lol nothing immediately comes to mind...

Answer 33

I put all of them in excel and none of the four is equal to my answer. Is my answer definitely right?

Answer 34

yes: http://www.wolframalpha.com/input/?i=integral+from+-pi%2F4+to+pi%2F6+of+-sin%28x%29+dx

Answer 35

Can't believe I forgot about wolfram alpha.

Or maybe I misread the question:
It says f(x)= -sin x and x is element of [ -pi/4, pi/6]
Calculate area under graph of f.

Answer 36

in that solution you multiply the top and bottom of 1/sqrt(2) by sqrt(2)/sqrt(2) and that makes the answer the same.

It's ok to do this because sqrt(2)/sqrt(2) = 1.

Answer 37

nope sounds good to me

Answer 38

I put my answer into excel and get same as wolfram alpha = 0.158918623
1) gives me -0.259513024
2) gives me 0.836863293
3) gives me -0.447593757
4) gives me 0.426867815

So it looks like all the answers are wrong.

Answer 39

that would make sense :P

Answer 40

So do I complain to my lecturer? lol

Answer 41

i think so :)

Answer 42

Ach i have so much trouble with my university, but at the moment it's the only place I can study :(. Thanks for your help though.

Answer 43

hey, if its just the odd mistake, no big deal. But either way, i hope your studies are meaningful and fulfilling :) All the best

Answer 44

I've learnt a lot so far especially calculus, so I'm thankful for that. Thanks again :).

Answer 45

goodnight!

Answer 46

good morning lol

Answer 47

haha maybe for you ... see you around