Question

can someone tell me how to graph
y=sin^3(x)+cos^2(x)+2 on [-pi,pi] on geogebra or mathematica?

Answer 1

i need to find the area under the curve and the x axis

Answer 2

geogebra doesn't like the notation sin^3(x)+cos^2(x)+2

but it will accept (sin(x))^3+(cos(x))^2+2

Answer 3

to find the area under the curve, you use the "Integral" function

if f(x) = (sin(x))^3+(cos(x))^2+2

then you can say

Integral[f, -pi, pi]

Answer 4

hhmmm thanks what about this function

Answer 5

http://www.wolframalpha.com/input/?i=sin^3%28x%29%2Bcos^2%28x%29%2B2+domain+-pi..pi

Answer 6

sec^2(x),y=5cosx on -pi/4 <=x<=pi/4

Answer 7

to say sec^2(x), you would type in (sec(x))^2

Answer 8

I tried what you said on geogebra it says invalid input

Answer 9

nevermind i got it now :D

Answer 10

Plot[Sin[x]^3 + Cos[x]^2 + 2, {x, -Pi, Pi}]

Answer 11

can you check my answer please?

Answer 12

for this problem

Answer 13

y=sec^2(x),y=5cos(x), -pi/4<=x<=pi/4 I have to check the area btwn the curves

Answer 14

I got -5.07

Answer 15

but I thought i area is always positive

Answer 16

if the curve lies below the x-axis, then the "area" is negative

the magnitude of the area, the true actual area, is positive, but it's negative to reflect that the region is below the x axis

Answer 17

hmm so the true actual area is 5.07

Answer 18

thanks alot! good night

Answer 19

well I mean if you take the absolute value of the area, then yeah you'll get 5.07 (assuming you got -5.07 beforehand)