evaluate the integral and interpret it as the area of a region. sketch the region.

Question

anonymous · Answer

$$\int\limits_{0}^{\Pi/2}$$ absolute value (sinx- cosx) dx

anonymous · Answer

|dw:1346869822932:dw|
it is twice the integral from zero to pi/2 of (cosx-sinx)

anonymous · Answer

? i still dont understand. i evaluated it and got 1 as an answer but its not the right answer

anonymous · Answer

sorry i meant twice the intergral from zero to pi/4. what is the answer?

anonymous · Answer

$$(3/2)\sqrt{3}-1$$
thats the answer

anonymous · Answer

well i don't know if that can be the correct answer. you have to consider the graph in two parts. from zero to pi/4 the function is negative since sin(x) is less than cos(x) so you have to intergrate the function as cos(x)-sin(x). from pi/4 to pi/2 the function is positive so you have to intergrate it as sin(x)-cos(x). however the two areas are the same so you could just multiply the area of the intergral from 0 to pi/4 of (cos(x)-sin(x))

amistre64 · Answer

$$\int |x|`dx=\frac12x|x|+C$$

amistre64 · Answer

say:
u = sinx -cosx
du = cosx - sinx dx$$-\int\frac{|u|}{u}du$$

amistre64 · Answer

which works out to be:$$-\frac{u^2}{|u|}$$now when x=0; u=sin(0)-cos(0)=-1
  when x=pi/2; u=sin(pi/2)-cos(pi/2)=1
$$-\left.\frac{u^2}{|u|}\right|_{-1}^{1}=-1+1=0$$
if i did it right

amistre64 · Answer

http://www.wolframalpha.com/input/?i=integrate+%7Csinx-cosx%7C+from+0+to+pi%2F2 the wolf says i did it wrong :)