Question

Need help with Calc1 - Integrals

Answer 1

F(x)=\[\int\limits_{\cos x}^{5x} \cos(u^{2})du\]

Answer 2

Find derivative I got cos(5x)^2 *5+-cos(cosx)^2*-sinx

Answer 3

You need to use the fundamental theorem of calculus here:
\[F'(x)=5\cos(25x^2)-\cos(\cos^2x)(-\sin x)=5\cos(25x^2)+\sin(x)\cos(\cos^2x).\]

Answer 4

You can derive a general formula given by:
\[{d \over dx}\int\limits_{g(x)}^{h(x)}f(x)dx=f(h(x))h'(x)-f(g(x))g'(x).\]

Answer 5

our answers are the same

Answer 6

That's good.

Answer 7

:) thanks then I'll ask another maybe u'll get to use ur brain a little more

Answer 8

Lol! Ask away!!

Answer 9

Evaluate the Integral :

\[\int\limits_{-1}^{2}x-2|x|dx\]

Answer 10

No idea what to do with the |x|

Answer 11

I'm sure you know how to do the first term, so I'll focus on the integration of |x| from -1 to 2.
To integrate the absolute value of x you need to redefine the function. You know that \(|x|=-x \text{ for } x<0\), so we can write the integral as:
\[\int\limits_{-1}^{2}|x|dx=\int\limits_{-1}^{0}-xdx+\int\limits_0^2xdx\]

Answer 12

true

Answer 13

Evaluate it and then multiply it by the constant \(-2\) and add to it the value you get from integrating the first term.

Answer 14

Makes sense?

Answer 15

lemme try

Answer 16

sure.

Answer 17

\[(1/2)x^{2}-3\] then i plug in each and do F(b)-F(a) right?

Answer 18

I didn't solve it actually, but you can always check your answers using wolfarmalpha :D

Answer 19

I don't know where the -3 came from though.

Answer 20

integrate x ---> (1/2)x^2 then i pluged in the numbers and added them together

Answer 21

If you plugged in the numbers, why do I still see x in there?

Answer 22

thats the first term
the -3 is the -2|x| part

Answer 23

Oh I see. You're right then, you just need to plug -1 to 2 in 1/2 x^2.

Answer 24

Are you sure about the -3? It should be -5.

Answer 25

lemme check ur prob right

Answer 26

ehhh im getting (3/2) * -2 = -6/2 = -3

Answer 27

\[\int\limits_{-1}^{2}\left| x \right|dx=\int\limits_{-1}^0-xdx+\int\limits_0^2xdx={-x^2 \over 2}|_{-1}^0+{x^2 \over 2}|_0^2=(0-{-1 \over 2})+({4 \over 2}-0)={5 \over 2}\]

Answer 28

\(\large =\frac{5}{2}\)

Answer 29

my bad forgot to do the 0 minus on the negative one!

Answer 30

I thought so!

Answer 31

My appologies so now its easy \[x^{2}/2-5 \] [-1, 2] ---> 4/2=2-5=-3 ( 1/2)-5 = -9/2

so -3-(-9/2) = 3/2

Answer 32

Wait! Not like this. Do first \(\frac{x^2}{2}|_{-1}^2=\frac{4}{2}-\frac{1}{2}=\frac{3}{2}\). So the final value of our integral is \(\frac{3}{2}-5=-\frac{7}{2}.\)

Answer 33

I gotta go now!

Answer 34

sorry for being difficult thanks!

Answer 35

no problem! glad to help :D