f ''(x) = 7 + cos x, f(0) = −1, f(9π/2) = 0.
To find f'(x) we integrate f''(x)
f'(x) = ∫ f''(x) = ∫ (7 + cos x) dx = 7x + sin x + A
To find f'x) we integrate f'(x)
f(x) = ∫ f'(x) = ∫ (7x + sin x + A) dx = 7*[tex] \frac{x^2}{2} [/tex] - cosx + Ax + B
Now we plug in the values and solve for A and B
f(x) = [tex] \frac{7x^2}{2} [/tex] - cosx + Ax + B
f(0) = −1
f(0) = [tex] \frac{7*0^2}{2} [/tex] - cos(0) + A(0) + B
-1 = - cos(0) + B
-1 = -1 + B
B = 0
f(9π/2) = 0
f([tex] \frac{\pi9}{2} [/tex]) = [tex] \frac{7}{2}(\frac{9\pi }{2}) ^2 [/tex] - cos([tex] \frac{\pi9}{2} [/tex])+ A([tex] \frac{\pi9}{2} [/tex]) + B
f([tex] \frac{\pi9}{2} [/tex]) = [tex] \frac{ 7*81\pi^2}{8} [/tex] - cos([tex] \frac{\pi9}{2} [/tex])+ A([tex] \frac{\pi9}{2} [/tex]) + B
0 = [tex] \frac{ 7*81\pi^2}{8} [/tex] - 0+ A([tex] \frac{\pi9}{2} [/tex]) + 0
A([tex] \frac{\pi9}{2} [/tex]) = [tex] \frac{ 7*81\pi^2}{8} [/tex]
multiply by [tex] \frac{2}{\pi9} [/tex] on both sides
A = [tex] \frac{63\pi}{4} [/tex]
So f(x) = [tex] \frac{7x^2}{2} [/tex] - cosx + [tex] \frac{63\pi}{4} [/tex] x + 0
