# Thread: area bounded by curves

1. ## area bounded by curves

hello , ive been stuck on this problem all night now. my heads starting to hurt lol. anyway im supposed to determine the area of the region by the graphs of the equations y=sin^2(pix) , y=0 , x=0 , x=1.

heres a link to what the equation looks like , just in case http://www.wolframalpha.com/input/?i=integrate+sin^2%28pi*x%29+dx+from+x%3D0+to+1

now, the answer i get is 3/4, which i think is wrong. i replaced sin^2(pix) with 1-cos^(pix).

2. You need to use the double angle formula: $\displaystyle cos(2x) = 1-2sin^2(x)$

Hence $\displaystyle sin^2(\pi x) = \frac{1}{2} - \frac{1}{2}cos(2 \pi x)$

3. Originally Posted by zerocool18
hello , ive been stuck on this problem all night now. my heads starting to hurt lol. anyway im supposed to determine the area of the region by the graphs of the equations y=sin^2(pix) , y=0 , x=0 , x=1.

heres a link to what the equation looks like , just in case http://www.wolframalpha.com/input/?i=integrate+sin^2%28pi*x%29+dx+from+x%3D0+to+1

now, the answer i get is 3/4, which i think is wrong. i replaced sin^2(pix) with 1-cos^(pix).

$\displaystyle \mbox{You can check that}\,\int \sin^2x\,dx\,=\frac{x-\sin x\cos x}{2}+C \,$$\displaystyle \,\mbox{, with }\,C \,\mbox{a constant, so a simple substitution will give you}$

$\displaystyle \int_0^1 \sin^2\pi x\,dx=$ $\displaystyle \frac{1}{\pi}\int_0^\pi\sin^2x\,dx$

Tonio

4. i think i got it , just one more question. Im supposed to use A=f(x)-g(x) to find the area. since the graph doesnt intersect anything, my equation should be the integral from 0 to 1 of sin^2(pix)-0 right? so i just have to integrate the sin^2(pix)?

5. Originally Posted by zerocool18
i think i got it , just one more question. Im supposed to use A=f(x)-g(x) to find the area. since the graph doesnt intersect anything, my equation should be the integral from 0 to 1 of sin^2(pix)-0 right? so i just have to integrate the sin^2(pix)?

Yes

Tonio