Video: Finding the Expression of a Function Given Its Second Derivative Using Integration

If 𝑓″(π‘₯) = 3π‘₯⁡ + 3π‘₯Β³ + 5π‘₯ + 2, determine 𝑓(π‘₯).

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Video Transcript

If 𝑓 double prime of π‘₯ equals three π‘₯ to the power of five plus three π‘₯ cubed plus five π‘₯ plus two, determine 𝑓 of π‘₯. We know how to find the derivative of different functions. For example, if we take a polynomial 𝑔 of π‘₯ equals two π‘₯ cubed plus three π‘₯ squared plus five π‘₯ minus one, we can differentiate this term by term using the power rule which tells us to multiply the coefficient by the power and then reduce the power by one. We also know that constants differentiate to zero.

So if we apply this to each term, we find that 𝑔 prime of π‘₯ equals six π‘₯ squared plus six π‘₯ plus five. We can differentiate this again to find 𝑔 double prime of π‘₯. This gives us 12π‘₯ plus six. But here for this problem, we’re wanting to find the antiderivative, β€œanti-” meaning opposite. So we’ve been given 𝑓 double prime of π‘₯ and we’ve been asked to determine 𝑓 of π‘₯. So we need to work backwards. But how do we do this? Well, if we know how to differentiate, then we can definitely do the opposite, its integration. So we take our rules for differentiating and we reverse them.

The opposite of reducing the power by one is adding one to the power. The opposite of multiplying the coefficient by the power is dividing by the power. Note that here we multiplied by the old power. So for the reverse, we divide by the new power. And because constants differentiate to zero, we need to add a constant of integration. So let’s now find the antiderivative of 𝑓 double prime of π‘₯. Just like how we differentiate term by term, we can integrate term by term. We start with three π‘₯ to the power of five. And we increase the power by one to give us three π‘₯ to the power of six. And then we divide by the new power of six.

We can actually simplify this term as three over six is one over two. We follow the same rules for our next term. We add one to the power to get three π‘₯ to the power of four and divide by the new power of four. Similarly, we increase the power of five π‘₯ to get five π‘₯ squared and divide by the new power. And now we integrate the constant, two. Remember how, in the first example, we looked at five π‘₯ differentiated to give us five. So five integrates to give us five π‘₯. In the same way two integrates to give us two π‘₯. We’re not quite finished with this step because we’ve got to add a constant of integration, 𝐢. So that gives us 𝑓 prime of π‘₯ but we’re not finished yet because we’re trying to determine 𝑓 of π‘₯. So we must integrate one more time.

Following the same procedure as before, π‘₯ to the power of six over two integrates to π‘₯ to the power of seven over two divided by seven. We could write this as π‘₯ to the power of seven over two multiplied by one over seven, which is just π‘₯ to the power of seven over 14. And now we move on to our next term. Adding one to the power and dividing by the new power gives us three π‘₯ to the power of five over four all divided by five, which we can rewrite as three π‘₯ to the power of five over 20.

Now, we move on and integrate five π‘₯ squared over two and then we simplify to five π‘₯ cubed over six. And then two π‘₯ integrates to two π‘₯ squared over two. And the twos cancel to give us π‘₯ squared. And now we also have to integrate our constant 𝐢. In the same way that two integrated to two π‘₯, 𝐢 integrates to 𝐢π‘₯. And finally, we need to add a constant of integration, which this time we’ll call 𝐷. So that gives us our final answer. 𝑓 of π‘₯ equals π‘₯ to the power of seven over 14 plus three π‘₯ to the power of five over 20 plus 5π‘₯ cubed over six plus π‘₯ squared plus 𝐢π‘₯ plus 𝐷.

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