Video: Finding an Area between a Curve and a Line

The curve shown is 𝑦 = 1/π‘₯. What is the area of the shaded region? Give an exact answer.

03:10

Video Transcript

The curve shown is 𝑦 equals one over π‘₯. What is the area of the shaded region? Give an exact answer.

Now we recall that the area of the region bounded by the curve 𝑦 equals 𝑓 of π‘₯, the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏, and the π‘₯-axis is given by the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. In this case, we’re told that the function 𝑓 of π‘₯ is one over π‘₯. And from the graph, we can see that the values of the limits for this integral are negative one for the lower limit and negative one-third for the upper limit. So we have the definite integral from negative one to negative one-third of one over π‘₯ with respect to π‘₯.

We then recall that the integral of one over π‘₯ with respect to π‘₯ is equal to the natural logarithm of the absolute value of π‘₯ plus the constant of integration. And that absolute value is really clear here because the two values for our limits are both negative. And we recall that the natural logarithm of a negative value is undefined. So we must make sure we include those absolute value signs.

So we’re taking the natural logarithm of a positive value. We don’t need the constant of integration 𝑐 here as we’re performing a definite integral. Substituting the limits gives the natural logarithm of the absolute value of negative one-third minus the natural logarithm of the absolute value of negative one. That’s the natural logarithm of one-third minus the natural logarithm of one. And at this point, we can recall that the natural logarithm of one is just zero. So our answer has simplified to the natural logarithm of one-third.

Now this may not be immediately obvious to you. But in fact, the natural logarithm of one-third is a negative value. We can see this if we recall one of our laws of logarithms, which is that the logarithm of π‘Ž over 𝑏 is equal to the logarithm of π‘Ž minus the logarithm of 𝑏. And so the natural logarithm of one-third is the natural logarithm of one minus the natural logarithm of three. And again, we recall that the natural logarithm of one is equal to zero. So our answer appears to be that this area is equal to negative the natural logarithm of three.

This doesn’t really make sense though as areas should be positive. What we see then is that when we use integration to evaluate an area below the π‘₯-axis, we will get a negative result. This doesn’t mean though that the area itself is negative. The negative sign is just signifying to us that the area is below the π‘₯-axis.

Really then, what we should’ve done is include absolute value signs around our integral sign at the beginning. And what this means is that although the value of the integral is negative, the natural logarithm of three, the value of the area is the absolute value of this. So that’s just the natural logarithm of three. The integral is negative to signify that the area is below the π‘₯-axis. But the area itself is positive. So our answer to the question, and it is an exact answer, is that this area is equal to the natural logarithm of three.

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