Video: Fundamental Theorem of Calculus

Find, if possible, an antiderivative 𝐹 of 𝑓(π‘₯) = 1/(2π‘₯ βˆ’ 1) that satisfies the conditions 𝐹(0) = 1 and 𝐹(1) = βˆ’1.

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Video Transcript

Find, if possible, an antiderivative capital 𝐹 of 𝑓 of π‘₯ equals one over two π‘₯ minus one that satisfies the conditions capital 𝐹 of zero is one and capital 𝐹 of one is negative one.

Another way of thinking about the antiderivative is evaluating the indefinite integral. So, what we’re actually going to do is integrate one over two π‘₯ minus one with respect to π‘₯ and consider the conditions on it in a moment. To integrate one over two π‘₯ minus one, we’re going to use a substitution. We’re going to let 𝑒 be equal to two π‘₯ minus one, which means that d𝑒 by dπ‘₯ equals two. Now, we do know that d𝑒 by dπ‘₯ is not a fraction, but we do treat it a little like one for the purposes of integration by substitution and say that a half d𝑒 equals dπ‘₯.

Let’s perform some substitutions. We replace two π‘₯ minus one with 𝑒 and dπ‘₯ with a half d𝑒. And we can, of course, take out the constant factor of one-half, and we have half times the integral of one over 𝑒. Well, remember, the integral of one over π‘₯ with respect to π‘₯ is the natural log of the absolute value of π‘₯. So, we can say that the integral of one over two π‘₯ minus one dπ‘₯ is a half times the natural log of 𝑒 plus 𝑐. We replace 𝑒 with two π‘₯ minus one, and we’ve obtained our antiderivative. Distributing the parentheses, we see that the antiderivative 𝑓 is equal to a half times the natural log of two π‘₯ minus one plus capital 𝐢.

We’re now going to consider conditions. But we also recall the fact that two π‘₯ minus one cannot be equal to zero. So, we’re actually going to have a piecewise function for our antiderivative. The first part of the function we’re interested in is when two π‘₯ minus one is greater than zero. Solving for π‘₯, and we obtain π‘₯ to be greater than one-half. The second is when one minus two π‘₯ is greater than zero. Solving for π‘₯ this time, and we obtain π‘₯ to be less than one-half. Let’s clear some space and formalize this.

We currently have that our antiderivative, capital 𝐹 of π‘₯, is equal to a half times the natural log of one minus two π‘₯ plus some constant when π‘₯ is less than one-half, and a half times the natural log of two π‘₯ minus one plus some constant when π‘₯ is greater than one half. We’re now going to use our conditions on the antiderivative to evaluate 𝑐. The first is capital 𝐹 of zero equals one. So, when π‘₯ is equal to zero, 𝐹 is equal to one.

Since π‘₯ is less than one-half, we’re going to use the first part of our function. We substitute π‘₯ equals zero and capital 𝐹 equals one in. And we get one equals a half times the natural log of one minus two times zero. Well, the natural log of one minus two times zero is the natural log of one, which is of course zero. So, we obtain 𝑐 here to be one.

The second condition we have is that when π‘₯ is equal to one, capital 𝐹 equals negative one. This time π‘₯ is greater than one-half, so we’re going to use the second part of the function. We see that negative one is equal to a half times the natural log of two times one minus one plus 𝑐. Once again, the natural log of two times one minus one is zero. So, we see that 𝑐 in the second part is equal to negative one. And so, the antiderivative capital 𝐹 does indeed exist. Such that capital 𝐹 of π‘₯ is equal to a half times the natural log of one minus two π‘₯ plus one for π‘₯ is less than one-half and a half times the natural log of two π‘₯ minus one minus one for π‘₯ is greater than one-half.

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