Question

homework9: problem 9
(1 point)
let ( f^{prime prime}(x)=2 cos (x) ).
(a) find the most general formula for ( f^{prime}(x) ). if an arbitrary constant must be used here, use an upper - case \c\.
( f^{prime}(x)= )
(b) based on your answer to (a), find the most general formula for ( f(x) ). if another new arbitrary constant must be used here, use an upper - case \d\.
( f(x)= )
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Explanation:

Part (a)

Step1: Integrate \( f''(x) \) to get \( f'(x) \)

We know that the antiderivative of \( \cos(x) \) is \( \sin(x) \), and the antiderivative of a constant multiple \( a\cos(x) \) is \( a\sin(x) \) (by the constant multiple rule of integration). Also, when integrating, we add an arbitrary constant \( C \) (since the derivative of a constant is zero, so the most general antiderivative includes this constant).

Given \( f''(x) = 2\cos(x) \), we integrate with respect to \( x \):

\( f'(x)=\int f''(x)dx=\int 2\cos(x)dx \)

Using the constant multiple rule \( \int k\cdot g(x)dx = k\int g(x)dx \) (where \( k = 2 \) and \( g(x)=\cos(x) \)) and the fact that \( \int \cos(x)dx=\sin(x)+C \), we have:

\( \int 2\cos(x)dx = 2\int \cos(x)dx = 2\sin(x)+C \)

Step2: Write the formula for \( f'(x) \)

From the integration above, the most general formula for \( f'(x) \) is \( 2\sin(x)+C \)

Part (b)

Step1: Integrate \( f'(x) \) to get \( f(x) \)

We know that the antiderivative of \( \sin(x) \) is \( -\cos(x) \), the antiderivative of a constant \( C \) with respect to \( x \) is \( Cx \), and we add a new arbitrary constant \( D \) (since we are finding the most general antiderivative).

From part (a), \( f'(x)=2\sin(x)+C \). We integrate \( f'(x) \) with respect to \( x \):

\( f(x)=\int f'(x)dx=\int (2\sin(x)+C)dx \)

Using the sum rule of integration \( \int (g(x)+h(x))dx=\int g(x)dx+\int h(x)dx \), we split the integral:

\( \int (2\sin(x)+C)dx=\int 2\sin(x)dx+\int Cdx \)

For the first integral, using the constant multiple rule \( \int k\cdot g(x)dx = k\int g(x)dx \) (where \( k = 2 \) and \( g(x)=\sin(x) \)) and the fact that \( \int \sin(x)dx=-\cos(x)+D_1 \) (we will use a new constant later, so we can just keep track of the constants step - by - step), we have \( \int 2\sin(x)dx = 2\int \sin(x)dx=2(-\cos(x))=-2\cos(x) \)

For the second integral, \( \int Cdx = Cx \) (since the antiderivative of a constant \( C \) is \( Cx \))

Then, we add the new arbitrary constant \( D \) (because when we integrate the sum of functions, the total constant of integration is the sum of the constants from each integral, and we can combine them into a single constant \( D \) for simplicity in the most general form). So:

\( f(x)=-2\cos(x)+Cx + D \)

Step2: Write the formula for \( f(x) \)

From the integration above, the most general formula for \( f(x) \) is \( -2\cos(x)+Cx + D \)

Answer:

(a):
\( f'(x)=\boxed{2\sin(x)+C} \)