The area is given by the integral,
![\displaystyle\int_0^{\ln5}2\pi x(y)\sqrt{1+\left(\dfrac{\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cln5%7D2%5Cpi%20x%28y%29%5Csqrt%7B1%2B%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dy%7D%5Cright%29%5E2%7D%5C%2C%5Cmathrm%20dy)
We have
![x=\dfrac{e^y+e^{-y}}2\implies\dfrac{\mathrm dx}{\mathrm dy}=\dfrac{e^y-e^{-y}}2](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7Be%5Ey%2Be%5E%7B-y%7D%7D2%5Cimplies%5Cdfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dy%7D%3D%5Cdfrac%7Be%5Ey-e%5E%7B-y%7D%7D2)
So now compute the integral:
![\displaystyle\frac\pi2\int_0^{\ln5}(e^y+e^{-y})\sqrt{4+(e^y-e^{-y})^2}\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%5Cpi2%5Cint_0%5E%7B%5Cln5%7D%28e%5Ey%2Be%5E%7B-y%7D%29%5Csqrt%7B4%2B%28e%5Ey-e%5E%7B-y%7D%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Substitute
and
:
![\displaystyle\frac\pi2\int_0^{\frac{24}5}\sqrt{4+u^2}\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%5Cpi2%5Cint_0%5E%7B%5Cfrac%7B24%7D5%7D%5Csqrt%7B4%2Bu%5E2%7D%5C%2C%5Cmathrm%20du)
Another substitution,
and
:
![\displaystyle\frac\pi2\int_0^{\tan^{-1}\frac{12}5}\sqrt{4+(2\tan v)^2}\,2\sec^2v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%5Cpi2%5Cint_0%5E%7B%5Ctan%5E%7B-1%7D%5Cfrac%7B12%7D5%7D%5Csqrt%7B4%2B%282%5Ctan%20v%29%5E2%7D%5C%2C2%5Csec%5E2v%5C%2C%5Cmathrm%20dv)
![\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sqrt{1+\tan^2v}\,\sec^2v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Cpi%5Cint_0%5E%7B%5Ctan%5E%7B-1%7D%5Cfrac%7B12%7D5%7D%5Csqrt%7B1%2B%5Ctan%5E2v%7D%5C%2C%5Csec%5E2v%5C%2C%5Cmathrm%20dv)
![\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Cpi%5Cint_0%5E%7B%5Ctan%5E%7B-1%7D%5Cfrac%7B12%7D5%7D%5Csec%5E3v%5C%2C%5Cmathrm%20dv)
There's a well-known formula for the integral of secant cubed, but if you don't know it off the top of your head (like me), you can integrate by parts:
![\displaystyle I=\int\sec^3v\,\mathrm dv=\sec v\tan v-\int\sec v\tan^2v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%3D%5Cint%5Csec%5E3v%5C%2C%5Cmathrm%20dv%3D%5Csec%20v%5Ctan%20v-%5Cint%5Csec%20v%5Ctan%5E2v%5C%2C%5Cmathrm%20dv)
Expand the remaining the integral in terms of powers of secant:
![\displaystyle\int\sec v\tan^2v\,\mathrm dv=\int\sec v(\sec^2v-1)\,\mathrm dv=\int\sec^3v\,\mathrm dv-\int\sec v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Csec%20v%5Ctan%5E2v%5C%2C%5Cmathrm%20dv%3D%5Cint%5Csec%20v%28%5Csec%5E2v-1%29%5C%2C%5Cmathrm%20dv%3D%5Cint%5Csec%5E3v%5C%2C%5Cmathrm%20dv-%5Cint%5Csec%20v%5C%2C%5Cmathrm%20dv)
so that
![I=\sec v\tan v-\left(I-\displaystyle\int\sec v\,\mathrm dv\right)](https://tex.z-dn.net/?f=I%3D%5Csec%20v%5Ctan%20v-%5Cleft%28I-%5Cdisplaystyle%5Cint%5Csec%20v%5C%2C%5Cmathrm%20dv%5Cright%29)
![2I=\sec v\tan v+\displaystyle\int\sec v\,\mathrm dv](https://tex.z-dn.net/?f=2I%3D%5Csec%20v%5Ctan%20v%2B%5Cdisplaystyle%5Cint%5Csec%20v%5C%2C%5Cmathrm%20dv)
![\implies I=\displaystyle\int\sec^3v\,\mathrm dv=\frac{\sec v\tan v}2+\frac12\ln|\sec v+\tan v|+C](https://tex.z-dn.net/?f=%5Cimplies%20I%3D%5Cdisplaystyle%5Cint%5Csec%5E3v%5C%2C%5Cmathrm%20dv%3D%5Cfrac%7B%5Csec%20v%5Ctan%20v%7D2%2B%5Cfrac12%5Cln%7C%5Csec%20v%2B%5Ctan%20v%7C%2BC)
Coming back to the area integral, we use the formula above to get
![\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv=\pi\left(\sec v\tan v+\ln|\sec v+\tan v|\right)\bigg|_0^{\tan^{-1}\frac{12}5}](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Cpi%5Cint_0%5E%7B%5Ctan%5E%7B-1%7D%5Cfrac%7B12%7D5%7D%5Csec%5E3v%5C%2C%5Cmathrm%20dv%3D%5Cpi%5Cleft%28%5Csec%20v%5Ctan%20v%2B%5Cln%7C%5Csec%20v%2B%5Ctan%20v%7C%5Cright%29%5Cbigg%7C_0%5E%7B%5Ctan%5E%7B-1%7D%5Cfrac%7B12%7D5%7D)
Next,
![\tan\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{12}5](https://tex.z-dn.net/?f=%5Ctan%5Cleft%28%5Ctan%5E%7B-1%7D%5Cdfrac%7B12%7D5%5Cright%29%3D%5Cdfrac%7B12%7D5)
![\sec\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{13}5](https://tex.z-dn.net/?f=%5Csec%5Cleft%28%5Ctan%5E%7B-1%7D%5Cdfrac%7B12%7D5%5Cright%29%3D%5Cdfrac%7B13%7D5)
![\tan0=0](https://tex.z-dn.net/?f=%5Ctan0%3D0)
![\sec0=1](https://tex.z-dn.net/?f=%5Csec0%3D1)
so the area is
![\pi\left(\dfrac{13}5\cdot\dfrac{12}5+\ln\left(\dfrac{13}5+\dfrac{12}5\right)-1\cdot0-\ln(1+0)\right)=\boxed{\left(\dfrac{156}{25}+\ln5\right)\pi}](https://tex.z-dn.net/?f=%5Cpi%5Cleft%28%5Cdfrac%7B13%7D5%5Ccdot%5Cdfrac%7B12%7D5%2B%5Cln%5Cleft%28%5Cdfrac%7B13%7D5%2B%5Cdfrac%7B12%7D5%5Cright%29-1%5Ccdot0-%5Cln%281%2B0%29%5Cright%29%3D%5Cboxed%7B%5Cleft%28%5Cdfrac%7B156%7D%7B25%7D%2B%5Cln5%5Cright%29%5Cpi%7D)
Answer:7y=-4x+1
Step-by-step explanation:
y=mx+c
-5=-4/7(-9)+c
c=1/7
y=-4/7x+1/7
7y=-4x + 1
-2 is the slope if that's what's asked ?
Answer:
True because a right triangle has to be 90 degrees and this diagram shows a right angle
Step-by-step explanation:
please i need brainliest
Answer:
x = 17 (When rounded to the nearest decimal place )