Question

Find the area of the region that lies beneath the given curve.

http://www.webassign.net/cgi-bin/symimage.cgi?expr=y%3Dsin%5C%28%20x%5C%29%2C%200%3C%3Dx%3C%3Dpi

Answer 1

what? really? do you know a function whose derivative is sine? no hints this time.

Answer 2

-cos(x)

Answer 3

right?

Answer 4

yes. now plug in
\[\pi\] then plug in
\[0\] and subtract. you get
\[-(\cos(\pi)-\cos(0))=\cos(0)-\cos(\pi)=1-(-1)=2\]=

Answer 5

Brilliant steps/solution as always, sir!

Answer 6

DO NOT DO THE INTEGRAL YOUR WASTING YOUR TIME, remeber int of sin or cos from 0 to pi or what ever draw the pic and everyhalf circle or period is 2 up 2 down this should help do the problems in your head, like this one
\[\int\limits_{0}^{\pi}3\sin(x)-\cos(x)dx\]


since sinx from 0 to pi has a h"half circle" is 2 times 3=6
and cos from 0 to pi has even area on top and bottom so it cancels and the answer is 6

now if you do the antider you get -sin x which is 0 at bot those points and -3cosx from 0 to pi =-3(-1-1)=6 there you go i hope this helps