dlthompson81 said:Here is the problem I got.
Find the area between 0 and pi.
y = 3sin(x) + sin(3x)
I'm really not even sure where to start with this. Any help?
dlthompson81 said:So the next step would be finding the anti derivative right?
So 3cos(x)+cos(3x) right? No - try again - pay attention to sign and chain rule. Differentiate your proposed expression and see if it gives back your original expression.
Then plug pi in, then plug 0 in and subtract it from pi?
[3cos(pi) + cos(3pi)] - [3cos(0)-cos(3*0)] = -8 <<< Why do you have different signs here - pay attention to your work
That is what I got when I worked it. Where did I go wrong? This problem comes from the chapter called integration by substitution. Not sure if that makes a difference or not, but we worked some like these in a previous chapter. Thanks for the help.
Subhotosh Khan said:dlthompson81 said:Here is the problem I got.
Find the area between 0 and pi.
y = 3sin(x) + sin(3x)
I'm really not even sure where to start with this. Any help?
\(\displaystyle Area\ \ = \ \ \int_0^{\pi}[3*sin(x) \ + \ sin(3*x)]dx\)
\(\displaystyle = -3\cdot \left [\cos(x)\right]_0^{\pi} \ \ - \ \ \frac{1}{3}\cdot \left [\cos(3x)\right]_0^{\pi}\)
\(\displaystyle =-3\cdot \left [\cos(\pi) \ - \ \cos(0)\right] \ \ - \ \ \frac{1}{3}\cdot \left [\cos(3\pi) \ - \ \cos(0)\right]\)
\(\displaystyle =-3\cdot \left [-1 \ - \ \ 1\right] \ \ - \ \ \frac{1}{3}\cdot \left [-1 \ - \ \ 1\right]\)
\(\displaystyle = \ 6 \ \ + \ \ \frac{2}{3} \ \ = \ \ \frac{20}{3} \ \ = \ \ 6.67\)
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