Question

Find f. f ''(θ) = sin(θ) + cos(θ), f(0) = 2, f '(0) = 1 f(θ) =

Answer 1

Answer:

$f(theta)=sin(theta) - cos(theta)$ + C

This is my first time doing a double integral, so im only 90% sure in my answer

Step-by-step explanation:

You pretty much want to take the double integral of sinx + cosx

The anti-derivative of sinx = -cosx

The anti-derivative of cosx = sinx

So f' = -cosx + sinx

Now lets take the integral of f':

The anti-derivative of -cosx = sinx

The anti-derivative of sinx = -cosx

So, f(x) = sinx - cosx

Answer 2

Answer:    f(θ) = -sin(θ) - cos(θ) + 2θ + 3

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Work Shown:

I'll use x in place of theta since its easier to type on a keyboard.

f '' (x) = sin(x) + cos(x)

f ' (x) = -cos(x) + sin(x) + C ..... integrate both sides; dont forget the plus C

f ' (0) = 1

f ' (0) = -cos(0) + sin(0) + C

-cos(0) + sin(0) + C = 1

-1 + 0 + C = 1

C = 1+1

C = 2

So,

f ' (x) = -cos(x) + sin(x) + C

turns into

f ' (x) = -cos(x) + sin(x) + 2

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Now integrate both sides of the first derivative to get the original f(x) function

f ' (x) = -cos(x) + sin(x) + 2

f(x) = -sin(x) - cos(x) + 2x + D .... apply integral; D is some constant

f(0) = -sin(0) - cos(0) + 2(0) + D

f(0) = 0 - 1 + 0 + D

f(0) = D - 1

f(0) = 2

D-1 = 2

D = 2+1

D = 3

We have f(x) = -sin(x) - cos(x) + 2x + D update to f(x) = -sin(x) - cos(x) + 2x + 3

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So f '' (x) = sin(x) + cos(x) becomes f(x) = -sin(x) - cos(x) + 2x + 3 when f(0) = 2 and f ' (0) = 1

The last step is to replace every x with theta so that we get back to the original variable.

f(x) = -sin(x) - cos(x) + 2x + 3   turns into   f(θ) = -sin(θ) - cos(θ) + 2θ + 3