Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part B β€’ Question 77

If 𝑓′(π‘₯) = sin (π‘₯Β² βˆ’ 4π‘₯ + 5) and 𝑓(2) = βˆ’(1/2), what is the value of 𝑓(4)?

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Video Transcript

If 𝑓 prime of π‘₯ is equal to sin of π‘₯ squared minus four π‘₯ plus five and 𝑓 of two is equal to negative one-half, what is the value of 𝑓 of four?

In this question, we’ve been given 𝑓 prime of π‘₯, which is equal to 𝑓 of π‘₯ differentiated with respect to π‘₯. So we can find 𝑓 of π‘₯ by integrating 𝑓 prime of π‘₯. We could say that 𝑓 of π‘₯ is equal to the integral of 𝑓 prime of π‘₯ with respect to π‘₯. Since this is an indefinite integral, we mustn’t forget to add on our constant of integration or 𝑐. However, we’re not strictly asked to find 𝑓 of π‘₯ in this question. And this would, in fact, be a very difficult task, since the function of 𝑓 prime of π‘₯ is very tricky to integrate.

Instead, this question asks us to find the value of 𝑓 at a particular point. And that point is at π‘₯ equals four. If we combine this with the other piece of information given in the question, which is that 𝑓 of two is equal to negative one-half, along with the formula for 𝑓 of π‘₯ which we have found, then we will obtain that the integral from two to four of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of four minus 𝑓 of two. And you may be wondering where the constant of integration has gone.

However, if we rearrange our equation for 𝑓 of π‘₯ to make the integral the subject and then we make this a definite integral between two and four, then we will see that the two values of 𝑐 will cancel out with one another. And so we do not need to worry about them. Now, let’s get back to finding 𝑓 of four.

From the question, we’re given that 𝑓 of two is equal to negative one-half. So we find that our integral is equal to 𝑓 of four plus one-half. Next, we rearrange this equation to make 𝑓 of four the subject. We obtain that 𝑓 of four is equal to the integral from two to four of 𝑓 prime of π‘₯ with respect to π‘₯ minus one-half. Next, we can substitute in our equation for 𝑓 prime of π‘₯, leaving us with this.

Now, like I said earlier, this integral is very difficult to calculate. This is where we will use our graphing calculators to help us. We find that the integral between two and four of sin of π‘₯ squared minus four π‘₯ plus five, with respect to π‘₯, is equal to 0.8231290241. However, we can round this to three decimal places, which is just 0.823. For our final step, we subtract the half to give us the solution that 𝑓 of four is equal to 0.323.

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