Question

sketch the region enclosed by the given curves and find its area.

Answer 1

\[y= \cos \pi x\]

\[y= 4x^2-1\]

Answer 2

@Nnesha

Answer 3

is it \[y = \cos(\pi)x\]
or \[\large\rm y=\cos(\pi x)\]
jst making sure...

Answer 4

if first one is correct then i will use vertical formula \[\rm \int\limits_{ }^{ } (upper ~function ) - (lower ~function )dx\]
for 2nd one \[\rm \int\limits_{ }^{ } (right~function)-(left~function )dx \]

Answer 5

https://www.wolframalpha.com/input/?i=y%3Dcos+pix++and+y%3D4x%5E2-1
https://www.wolframalpha.com/input/?i=y%3Dcos+(pi)x++and+y%3D4x%5E2-1

Answer 6

i believe first one is correct
set y=cospix and y=4x^2-1 equal to each other and then solve for x
that will you intersection points and then use first formula
upper - lower

Answer 7

woops suppose to be the 2nd one

Answer 8

im confuse .. is there a way to solve this ?

Answer 9

Are you sure its the 2nd one?

Answer 10

yes .. let me upload the pic .. one sec

Answer 11

@agent0smith

Answer 12

Hi

Answer 13

hii can you please help me D:

Answer 14

Have you found the intersection points yet ?

Answer 15

`set y=cospix and y=4x^2-1 equal to each other and then solve for x that will you intersection points `

Answer 16

no cuz i have no idea .. i set it equal to each other lol

Answer 17

Yeah solving an equation with trig functions is a pain in general.
But trust that your teacher gives you only the easy ones - the ones for which the intersection points can be found by eyeballing / guessing.

Answer 18

yeah lol i think my algebra is throwing me off

so thiis what i got

\[\cos \pi x +1= 4x^2\]

Answer 19

cos(pix) = 4x^2 - 1

Answer 20

^you can't solve that algebraically. Gotta use a calculator.

Answer 21

first stare at the equation for some time

Answer 22

plugin x = 0 and see if both sides balance out

Answer 23

I guess you can kinda eyeball it, but.... a graph saves a lot of time: http://www.wolframalpha.com/input/?i=cos(pi+x)+%3D+4x%5E2+-1

Answer 24

It's not easy to tell from looking at it (until you already know) that the solutions are \( x= \pm 1/2 \)

Answer 25

oh yeh its much easier loool

Answer 26

So now just do integral of top function minus bottom function, from -1/2 to 1/2, or 0 to 1/2 since it's symmetric.

And sorry @ganeshie8 but i feel like they aren't expecting students to eyeball this equation and mathemagically realize it has to be x=1/2.

Answer 27

loool

Answer 28

\[\large \int\limits_{-1/2}^{1/2}[\cos (\pi x) - (4x^2 - 1)]dx\]or\[\large 2\int\limits\limits_{0}^{1/2}[\cos (\pi x) - (4x^2 - 1)]dx\]

Answer 29

iight so how would i integrate the cos pi x ?

Answer 30

What's the deriv. of cos (pi x)?

Answer 31

wait idk D:

but i got the integral of the cos pi x
which is