# Video: Finding the Expression of a Function given Its Second Derivative and the Value of the Function and Its Derivative at a Point Using Integration

Determine the function π if πβ³(π₯) = βπ₯β΄ + 1, π(1) = 5, and πβ²(1) = 3.

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### Video Transcript

Determine the function π if π double prime of π₯ is equal to negative π₯ to the fourth power plus one, π evaluated at one is equal to five, and π prime evaluated at one is equal to three.

The question is asking us to find the function π, given that the second derivative of π is equal to negative π₯ to the fourth power plus one, π evaluated at one is five, and the first derivative of π evaluated at one is equal to three. We recall that π double prime of π₯ is the same as saying the second derivative of π with respect to π₯. And thatβs the same as saying the derivative of dπ dπ₯ with respect to π₯.

And the question tells us that this is equal to negative π₯ to the fourth power plus one. And we recall if weβre given a derivative function is equal to some function of π₯, we can solve this by using the opposite process to differentiation. Weβre going to integrate both sides of this equation with respect to π₯. Remember, integrating is the opposite process to differentiation. So, integrating the derivative function of dπ dπ₯ with respect to π₯ will just give us dπ dπ₯. So, we have that dπ dπ₯ is equal to the integral of negative π₯ to the fourth power plus one with respect to π₯.

We can rewrite dπ dπ₯ as π prime of π₯. We can then integrate this by using the power rule for integration. We add one to the exponent and then divide by this new exponent. To integrate negative π₯ to the fourth power, we add one to four to get five and then divide by this new value of five. And to integrate one, we rewrite it as π₯ to the zeroth power and then apply the power rule for integration. So, we now have that π prime of π₯ is equal to negative π₯ to the fifth power over five plus π₯ plus the constant of integration π one.

We recall the question tells us that π prime evaluated at one is equal to three. So, we can use this to find the value of π one. We substitute π₯ is equal to one into our equation for π prime of π₯. This gives us three is equal to negative one to the fifth power over five plus one plus π one. Rearranging and simplifying this, we see that π one is equal to eleven-fifths. So, weβve shown that π prime of π₯ is equal to negative π₯ to the fifth power over five plus π₯ plus eleven-fifths.

But the question wants us to find the function π, not the function π prime of π₯. So, to get from our derivative function π prime of π₯ to π, weβre going to need to use the opposite process to differentiation. Weβre going to need to integrate with respect to π₯. Using this and knowing that integration is the opposite process to differentiation, we have that π of π₯ is equal to the integral of π prime of π₯ with respect to π₯ which is also equal to the integral of negative π₯ to the fifth power over five plus π₯ plus 11 over five with respect to π₯. And again, we can evaluate this integral by using the power rule for integration.

Using the power rule for integration, we get negative π₯ to the sixth power over five times six plus π₯ squared over two plus eleven-fifths π₯ plus the constant of integration π two. We can simplify five times six to just give us 30. So, weβve now found an expression for our function π of π₯. However, it still has the unknown constant π two. And we recall the question told us that π evaluated at one was equal to five. Using this and substituting π₯ is equal to one gives us that five is equal to negative one times one to the sixth power over 30 plus one squared over two plus eleven-fifths times one plus π two. Rearranging, simplifying, and evaluating this expression, we can see that π two is equal to seven over three.

If we then substitute this value of π two back into our expression for π of π₯, we have shown that the function π where π double prime of π₯ is equal to negative π₯ to the fourth power plus one and π evaluated at one is equal to five and π prime of one is equal to three. Is given by π of π₯ is equal to negative π₯ to the sixth power over 30 plus π₯ squared over two plus 11π₯ over five plus seven-thirds.