# Thread: area under a curve

1. ## area under a curve

could some one please help me and explain how to find the area under the curve Y=sin(x) between x=0 and x=1.7pi

Thanks

2. ## Re: area under a curve

Since you know that the sine curve has an intercept at \displaystyle \begin{align*} x = \pi \end{align*}, you will need to evaluate

\displaystyle \begin{align*} A = \left| \int_0^{\pi} \sin{x}\,dx \right| + \left| \int_{\pi}^{1.7\pi} \sin{x}\,dx \right| \end{align*}

3. ## Re: area under a curve

$\int_a^b \text{Sin}[x] \, dx = \text{Cos}[a]-\text{Cos}[b]$

4. ## Re: area under a curve

Originally Posted by MaxJasper
$\int_a^b \text{Sin}[x] \, dx = \text{Cos}[a]-\text{Cos}[b]$
MaxJasper, that's incorrect. The OP was asked to find the AREA. Since there is an x intercept, you need to take into account all the "signed areas", and add their absolute values.

5. ## Re: area under a curve

Originally Posted by Prove It
MaxJasper, that's incorrect. The OP was asked to find the AREA. Since there is an x intercept, you need to take into account all the "signed areas", and add their absolute values.
It is correct. It's up to him to find proper ranges and signs.

6. ## Re: area under a curve

Originally Posted by MaxJasper
It is correct. It's up to him to find proper ranges and signs.
And it's up to us to not be ambiguous, so that our guidance is of some use

7. ## Re: area under a curve

so am i correct in thinking this?

(-cospi)-(-cos0)=2
(-cos1.7pi)-(-cospi)=-1.59

therefore total area is 3.59 units^2

8. ## Re: area under a curve

Looks good to me