It looks like you want to compute the double integral
over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is
Hey heart rate was 100 BPM immediately after running.
Answer:
Not necessarily. There are many ways to write a basic equation with a negative answer. For example, -3-4 = -7. Here 4 is a positive number but because you subtract 4 to 3 or take away 4 from 3 being a negative number and so you get a negative answer. Another example is 6+(-9). There are a couple of ways you can resolve this. My method is to subtract 9 from 6 which gives you 3 and simply add a negative sign.
Let me know if you'd like more examples. Hope this helps!! Sorry if it is confusing I can explain you in a more simpler way if you'd like.
Step-by-step explanation:
Answer:
Step-by-step explanation:
The Fundamental Theorem of Calculus states that:
![\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20%5Cint_a%5Ex%20f%28t%29%5C%2C%20dt%20%20%5Cright%5D%20%3D%20f%28x%29)
Where <em>a</em> is some constant.
We can let:
By substitution:
Taking the derivative of both sides results in:
![\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20g%27%28s%29%20%3D%20%5Cfrac%7Bd%7D%7Bds%7D%5Cleft%5B%20%5Cint_6%5Es%20g%28t%29%5C%2C%20dt%5Cright%5D)
Hence, by the Fundamental Theorem:
500/4= 125
she has written 125 words and needs to write 375 more
