I have some wrong answers .. help?
Calculate the area between y-sin(x) and y=cos(x) , -pi<or equal to x <or equal to 2pi.
I have -1.9984.
Calculate the area between y=x^2 and y=3x+4.
I have -18.7.
Find the area of the region enclosed between y=2sin(x) and y=2cos(x) from x=0 and x=0.8pi
I have .6334465.
Can someone tell me what i did wrong?
So x^2 = y = 3x+4
this gives us x^2 = 3x+4
x^2-3x-4 = 0
So at x=4, and x=-1, these two graphs will cross. Then these are our boundaries, so we will integrate from -1 to 4.
Now we need to know which function is above the other, substituting in x=0 (because it is a point inside of our boundaries) we get that y=3x+4 is above.
So we know that we will take the integral of 3x+4 and then subtract the integral of x^2 (if you don't understand why to do this, look at the attached image, it should be rather clear.