Question

Find the area of the region bounded by the curves y = x2 and y = cos(x). Give your answer correct to 2 decimal places

Answer 1

y = x^2 and y = cos x

Answer 2

Is that what you meant?

Answer 3

A long long time ago x2 use to mean x^2. But nowadays I only see its interpretation as 2 times x.

Answer 4

Yes , otherwise they would write 2x (generally speaking).

Answer 5

not x2

Answer 6

@Krissy3039 Please show any work you have done.

Answer 7

to avoid confusion i still write xx :)

Answer 8

what about for x^(1/2)

Answer 9

X > maybe?

Answer 10

sqrt(x) should work tho

Answer 11

Learn some LaTeX!

Answer 12

agreed :p

Answer 13

$$
\Large y = x^2 , ~y = \cos x
$$

Answer 14

Okay, to the integral!!!

Answer 15

latex warps the brain ... you start writing by hand and the latex just doesnt look the same anymore.

Answer 16

...said the person who probably has legible handwriting. :-)
LaTeX has connected me with the world! People can read what I write!

Answer 17

It helps to graph the two curves to find the region that is bounded.
The region where the two areas overlaps is going to be the area that we want.
https://www.desmos.com/calculator/xfnjcxr8vq

Answer 18

it was x^2 sorry

Answer 19

looking at @perl 's graph you have two intersections to find . The solutions can be found by solving x^2=cos(x). There is no algebraic way to solve that equation though. You will have to approximate those intersections.

Answer 20

You can get in the neighborhood with a quadratic series approximation of the cosine.

\(1 - \dfrac{x^{2}}{2} = x^{2}\implies x = 0.816\) or so

With just a hair more difficulty, still only an algebra problem, you can use the quartic approximation.

\(1 - \dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} = x^{2} \implies x = 0.824\)

That's actually pretty close.

Answer 21

thats the x value for the intercept and the y value is .679

Answer 22

what would you do to find the area then

Answer 23

Of course, I meant, \(Quadratic \implies x = \dfrac{\sqrt{6}}{3}\;and\;Quartic \implies x = \sqrt{18-10\sqrt{3}}\). What would be the point of the approximation if the answer weren't prettier?

Where is your integral with the limits in place?

Answer 24

Those are only approximations. You will need to do better than that. That just gives you a place to start looking. The first is too far left and the second too far right. You can bracket the actual answer with that sort of information.

Let's see that integral.

Answer 25

wait what im confused

Answer 26

do you know how to setup the integral in that region where there is purple and green in @perl 's little page he posted above as a link

Answer 27

\[\int\limits_{\text{ \lower limit (the left intersection x value of course )}}^{\text{ upper limit (the right intersection x value of course )}}(\text{ top function } - \text{ bottom function } ) dx\]

Answer 28

okay so would it be x^3/3 - sin(x)?

Answer 29

@freckles

Answer 30

on the interval in which we look at cos(x) and x^2
isn't cos(x)>x^2?

Answer 31

so it should be the other way around
sin(x)-x^3/3

Answer 32

now you just have to find your intersection
and do the difference of the expression above evaluated at both

Answer 33

i got - 1.46787...

Answer 34

@freckles

Answer 35

what did you use as your interesection numbers?

Answer 36

-.8241 and .8241

Answer 37

hmm... so you did:
(sin(.8241)-(.8241)^3/3)-(sin(-.8241)-(-.8241)^3/3)

Answer 38

yes

Answer 39

then would the area be 1.09?

Answer 40

\[(\sin(.8241)-\frac{(.8241)^3}{3})-(\sin(-.8241)-\frac{(-.8241)^3}{3})\]
maybe you didn't exactly enter it into the calculator right

\[\sin(.8241)-\frac{(.8241)^3}{3}-(-\sin(.8241)-\frac{-(.8241)^3}{3}) \\ \text{ since } \sin(x) \text{ and } x^3 \text{ are odd functions } \\ 2 \sin(.8241)-2 \cdot \frac{(.8241)^3}{3} \\ =2(\sin(.8241)-\frac{2}{3}\sin(.8241))\]
yeah that sounds a lot closer

Answer 41

so should i go with 1.09 then?

Answer 42

it is a pretty good approximation

Answer 43

okay thank you!

Answer 44

\(\color{blue}{\text{Originally Posted by}}\) @perl
\[
\Large{
\int_{-.824}^{.824} \cos (x)-x^2 ~dx ~
\\~\\~\\ or \\~\\
2\cdot \int_{0}^{.824} \cos (x) -x^2 ~dx ~
=1.0947

}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 45

and i made a type-o above

Answer 46

\[\sin(.8241)-\frac{(.8241)^3}{3}-(-\sin(.8241)-\frac{-(.8241)^3}{3}) \\ \text{ since } \sin(x) \text{ and } x^3 \text{ are odd functions } \\ 2 \sin(.8241)-2 \cdot \frac{(.8241)^3}{3} \\ =2(\sin(.8241)-\frac{1}{3}(.8241)^3)\]
since I dragged that 2 out of there