(a) Iff′′(θ)=sinθ+cosθf″(θ)=sin?θ+cos?θthen we have thatf′(θ)=∫f′′(θ)dθ=∫(sinθ+cosθ)dθ=−cosθ+sinθ+C1f′(θ)=∫f″(θ)dθ=∫(sin?θ+cos?θ)dθ=−cos?θ+sin?θ+C1and using the fact that f′(0)=2f′(0)=2 we can solve for C1C1 as−cos0+sin0+C1=2C1=3−cos?0+sin?0+C1=2C1=3sof′(θ)=−cosθ+sinθ+3f′(θ)=−cos?θ+sin?θ+3which gives usf(θ)=∫f′(θ)dθ=∫(−cosθ+sinθ+3)dθ=−sinθ−cosθ+3θ+C2f(θ)=∫f′(θ)dθ=∫(−cos?θ+sin?θ+3)dθ=−sin?θ−cos?θ+3θ+C2where we can use the fact that f(0)=5f(0)=5 to solve for C2C2 as−sin0−cos0+3(0)+C2=5C2=6−sin?0−cos?0+3(0)+C2=5C2=6and therefore the desired function is given byf(θ)=−sinθ−cosθ+3θ+6f(θ)=−sin?θ−cos?θ+3θ+6

(b) Iff′′(x)=6+6x+36x2f″(x)=6+6x+36x2then we have thatf′(x)=∫f′′(x)dx=∫(6+6x+36x2)dx=6x+3x2+12x3+C1f′(x)=∫f″(x)dx=∫(6+6x+36x2)dx=6x+3x2+12x3+C1and using the fact that f′(1)=14f′(1)=14 we can solve for C1C1 as6(1)+3(1)2+12(1)3+C1=14C1=−76(1)+3(1)2+12(1)3+C1=14C1=−7sof′(x)=−7+6x+3x2+12x3f′(x)=−7+6x+3x2+12x3which gives usf(x)=∫f′(x)dx=∫(−7+6x+3x2+12x3)dx=−7x+3x2+x3+3x4+C2f(x)=∫f′(x)dx=∫(−7+6x+3x2+12x3)dx=−7x+3x2+x3+3x4+C2where we can use the fact that f(0)=2f(0)=2 to solve for C2C2 as−7(0)+3(0)2+(0)3+3(0)4+C2=2C2=2−7(0)+3(0)2+(0)3+3(0)4+C2=2C2=2and therefore the desired function is given byf(x)=2−7x+3x2+x3+3x4