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Consider the function f(x) whose second derivative is f''(x) = 8x + 4\sin(x). If f(0) = 3 and...

Question:

Consider the function {eq}f(x) {/eq} whose second derivative is {eq}\; f''(x) = 8x + 4\sin(x) {/eq}.

If {eq}f(0) = 3 {/eq} and {eq}f'(0) = 2 {/eq}, what is {eq}f(x) {/eq}?

Integration:


In calculus, Integration is the reverse of differentiation. Here, in our problem, we integrate the given function two times where also we will find the value of the constant two times as the values of the function are given in the problem.

Answer and Explanation:


Given

{eq}\; f''(x) = 8x + 4\sin(x) {/eq}.

and {eq}f(0) = 3 {/eq} and {eq}f'(0) = 2 {/eq}.


We have to find {eq}f(x) {/eq}.



Now

{eq}\; f''(x) = 8x + 4\sin(x) {/eq}.


Integrating both sides:

{eq}\begin{align} f'(x) &=\int \left ( 8x + 4\sin(x) \right )dx\\ &=\int 8xdx+\int 4\sin (x)dx\\ &=8\int xdx+4\int \sin (x)dx\\ &=8\left ( \dfrac{x^2}{2} \right )+4\left ( -\cos (x) \right )+C\\ &=4x^2-4\cos (x)+C\\ \end{align} {/eq}


Applying the condition {eq}x=0, f'(0)=2 {/eq}

{eq}\begin{align} f'(x) &=4x^2-4\cos (x)+C\\ f'(0) &=4(0)^2-4\cos (0)+C\\ 2 &=0-4+C\\ 2+4 &=C\\ C &=6 \end{align} {/eq}


Applying this value, we have:

{eq}f'(x)=4x^2-4\cos (x)+6 {/eq}



Again integrating both sides:

{eq}\begin{align} f(x) &=\int \left (4x^2-4\cos (x)+6 \right )dx\\ &=4\left ( \dfrac{x^3}{3} \right )-4\sin (x)+3x+C_1\\ &=\dfrac{4x^3}{3}-4\sin (x)+6x+C_1 \end{align} {/eq}


Applying the condition {eq}x=0, f(0)=3 {/eq}

{eq}\begin{align} f(x) &=\dfrac{4x^3}{3}-4\sin (x)+6x+C_1\\ f(0) &=\dfrac{4(0)}{3}-4\sin (0)+6(0)+C_1\\ 3 &=0-0+0+C_1\\ C_1 &=3 \end{align} {/eq}


Applying this value, we have:

{eq}\color{blue}{f(x)=\dfrac{4x^3}{3}-4\sin (x)+6x+3} {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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