J jeca86 Junior Member Joined Sep 9, 2005 Messages 62 Jan 30, 2006 #1 evaluate the integral i dont know how to draw the sign but the 3pi/2 is at the top and the 0 is at the bottom. the rest is absolute value of sinx multiplied by dx. i got: cos(3pi/2)-cos0=-1 is this right?
evaluate the integral i dont know how to draw the sign but the 3pi/2 is at the top and the 0 is at the bottom. the rest is absolute value of sinx multiplied by dx. i got: cos(3pi/2)-cos0=-1 is this right?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 30, 2006 #2 Hello, jeca86! Evaluate the integral: \(\displaystyle \L\,\int^{\;\;\;\frac{3\pi}{2}}_0|\sin x|\,dx\) Click to expand... I assume you know what the graph of \(\displaystyle y\,=\,sin x\) looks like. The graph of \(\displaystyle \,y\,=\,|\sin x|\,\) from \(\displaystyle x\,=\,0\,\) to \(\displaystyle \,x\,=\,\frac{3\pi}{2}\,\) looks like this: Code: | *** ** | *:::::* *:| | *:::::::::* *:::| |*:::::::::::* *::::| |::::::::::::: :::::| --*-------------*-----+-- And you want the area under the curve . . . got it?
Hello, jeca86! Evaluate the integral: \(\displaystyle \L\,\int^{\;\;\;\frac{3\pi}{2}}_0|\sin x|\,dx\) Click to expand... I assume you know what the graph of \(\displaystyle y\,=\,sin x\) looks like. The graph of \(\displaystyle \,y\,=\,|\sin x|\,\) from \(\displaystyle x\,=\,0\,\) to \(\displaystyle \,x\,=\,\frac{3\pi}{2}\,\) looks like this: Code: | *** ** | *:::::* *:| | *:::::::::* *:::| |*:::::::::::* *::::| |::::::::::::: :::::| --*-------------*-----+-- And you want the area under the curve . . . got it?
