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

Given that 𝑓″(π‘₯) = βˆ’5𝑒^(4π‘₯) + 2π‘₯⁡, find 𝑓(π‘₯).

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Video Transcript

Given that 𝑓 double prime of π‘₯ is equal to negative five 𝑒 to the power of four π‘₯ plus two π‘₯ to the fifth power, find 𝑓 of π‘₯.

Recall, first of all, that 𝑓 double prime of π‘₯ means the second derivative of 𝑓 with respect to π‘₯. So the function we’ve been given is the result of differentiating 𝑓 twice. In order to find the function 𝑓 of π‘₯, we need to go back the other way. And as integration is the reverse of differentiation, we therefore need to integrate this function twice. Integrating once will give us the first derivative of our function 𝑓 of π‘₯. So we have that 𝑓 prime of π‘₯ is equal to the integral of negative five 𝑒 to the power of four π‘₯ plus two π‘₯ to the fifth power with respect to π‘₯.

Now, there are two types of functions that we need to recall how to integrate in order to answer this question. The more straightforward of these is just a general polynomial term. And so, we recall that the integral or antiderivative of π‘₯ to the power of 𝑛, where 𝑛 is some real number not equal to negative one, is π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus a constant of integration 𝐢. We increase the exponent by one and then divide by the new exponent.

The second type of function we need to recall how to integrate is an exponential function or, more specifically, a function of the form 𝑒 to the power of π‘Žπ‘₯ for some constant π‘Ž. We recall that this integral is equal to one over π‘Ž 𝑒 to the power of π‘Žπ‘₯ plus a constant of integration 𝐢. The factors of negative five and positive two just act as multiplicative constants. So let’s perform this integral. We have negative five, and then we integrate 𝑒 to the power of four π‘₯, which we’ve seen is equal to one-quarter 𝑒 to the power of four π‘₯. We then have plus two, and we integrate π‘₯ to the power of five, which we’ve just seen is π‘₯ to the power of six over six. We’ll only include one single constant of integration which we’ll call 𝐢.

Simplifying the constants, and we have that 𝑓 prime of π‘₯ is equal to negative five over four 𝑒 to the power of four π‘₯ plus one-third, that simplified from two-sixths, π‘₯ to the sixth power plus some constant 𝐢. Remember, though, that we’re looking to find the function 𝑓 of π‘₯. And, so far, we have 𝑓 prime of π‘₯. So we need to integrate a second time. We have then that 𝑓 of π‘₯ is equal to the integral of negative five over four 𝑒 to the power of four π‘₯ plus one-third π‘₯ to the power of six plus 𝐢 with respect to π‘₯. And we can now apply the two rules of integration that we’ve already written down for a second time. Integrating the first two terms, we have negative five over four multiplied by a quarter 𝑒 to the power of four π‘₯ plus one-third multiplied by one-seventh π‘₯ to the seventh power.

We then need to integrate that constant 𝐢, and this will give us 𝐢π‘₯. We can think of 𝐢 as 𝐢π‘₯ to the power of zero. So when we increase the exponent, we get π‘₯ to the power of one. And then we divide by one. We must also remember to include a new constant of integration which we’ll call 𝐷. Simplifying each of the constants, and we have our answer to the problem. 𝑓 of π‘₯ is equal to negative five over 16 𝑒 to the power of four π‘₯ plus one over 21 π‘₯ to the seventh power plus 𝐢π‘₯ plus 𝐷, where 𝐢 and 𝐷 are constants.

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