Respuesta :
Assuming the conditions are as given and are to be considered boundary values, you have upon integrating twice
[tex]f''(x)=9+\cos x[/tex]
[tex]\implies f'(x)=9x+\sin x+C_1[/tex]
[tex]\implies f(x)=\dfrac92x^2-\cos x+C_1x+C_2[/tex]
When [tex]f(0)=-1[/tex], you have
[tex]-1=\dfrac92(0)^2-\cos0+C_1(0)+C_2[/tex]
[tex]-1=-1+C_2[/tex]
[tex]C_2=0[/tex]
and when [tex]f\left(\dfrac{7\pi}2\right)=0[/tex], you get
[tex]0=\dfrac92\left(\dfrac{7\pi}2\right)^2-\cos\dfrac{7\pi}2+C_1\left(\dfrac{7\pi}2\right)+C_2[/tex]
[tex]0=\dfrac{441\pi^2}8+\dfrac{7\pi}2C_1[/tex]
[tex]C_2=-\dfrac{63\pi}4[/tex]
If you meant to include a prime on either of the initial values, then I'll let another user tackle that. The only difference would be the values you find for [tex]C_1[/tex] and [tex]C_2[/tex].
[tex]f''(x)=9+\cos x[/tex]
[tex]\implies f'(x)=9x+\sin x+C_1[/tex]
[tex]\implies f(x)=\dfrac92x^2-\cos x+C_1x+C_2[/tex]
When [tex]f(0)=-1[/tex], you have
[tex]-1=\dfrac92(0)^2-\cos0+C_1(0)+C_2[/tex]
[tex]-1=-1+C_2[/tex]
[tex]C_2=0[/tex]
and when [tex]f\left(\dfrac{7\pi}2\right)=0[/tex], you get
[tex]0=\dfrac92\left(\dfrac{7\pi}2\right)^2-\cos\dfrac{7\pi}2+C_1\left(\dfrac{7\pi}2\right)+C_2[/tex]
[tex]0=\dfrac{441\pi^2}8+\dfrac{7\pi}2C_1[/tex]
[tex]C_2=-\dfrac{63\pi}4[/tex]
If you meant to include a prime on either of the initial values, then I'll let another user tackle that. The only difference would be the values you find for [tex]C_1[/tex] and [tex]C_2[/tex].
For the given function the integrated function is,
[tex]f(x)=\dfrac{9x^2}{2} -cosx+\dfrac{63\pi}{4} x[/tex]
Given-
the given equation is,
[tex]f"(x)=9+cosx[/tex]
Integrating both side, we get,
[tex]f'(x)=9x+sinx+C_1[/tex]
Integrating again both side, we get
[tex]f(x)=\dfrac{9x^2}{2} -cosx+C_1x+C_2[/tex]
From the given question we have,
[tex]f(0)=-1[/tex]
use this value for the above function, we get,
[tex]f(0)=\dfrac{9(0)^2}{2} -cos0+C_1\times0+C_2[/tex]
[tex]-1=-1+C_2[/tex]
[tex]C_2=0[/tex]
from the given question we have,
[tex]f(\dfrac{7\pi }{2} )=0[/tex]
With this equation and value of [tex]C_2[/tex] find the value of the [tex]C_1[/tex].
[tex]f(x)=\dfrac{9x^2}{2} -cosx+C_1x+C_2[/tex]
[tex]0=\dfrac{9}{2}\times({\dfrac{7\pi }{2}})^2 -cos\dfrac{7\pi }{2}+C_1\dfrac{7\pi }{2}+0[/tex]
[tex]0={\dfrac{441\pi^2 }{8}} -0+C_1\dfrac{7\pi }{2}[/tex]
[tex]C_1\times\dfrac{7\pi }{2}={\dfrac{441\pi^2 }{8}}[/tex]
[tex]C_1=\dfrac{63\pi}{4}[/tex]
Put the value of the [tex]C_1[/tex] and [tex]C_2[/tex] in the integrated function we get,
[tex]f(x)=\dfrac{9x^2}{2} -cosx+C_1x+C_2[/tex]
[tex]f(x)=\dfrac{9x^2}{2} -cosx+\dfrac{63\pi}{4} x+0[/tex]
[tex]f(x)=\dfrac{9x^2}{2} -cosx+\dfrac{63\pi}{4} x[/tex]
Hence, for the given function the integrated function is,
[tex]f(x)=\dfrac{9x^2}{2} -cosx+\dfrac{63\pi}{4} x[/tex]
For more about the integration follow the link below-
https://brainly.com/question/18651211
