Question Video: Finding the Area of a Region Bounded by Two Reciprocal Functions | Nagwa Question Video: Finding the Area of a Region Bounded by Two Reciprocal Functions | Nagwa

# Question Video: Finding the Area of a Region Bounded by Two Reciprocal Functions Mathematics • Third Year of Secondary School

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Find the area of the region bounded above by π¦ = 1/π₯, bounded below by π¦ = 1/2π₯Β², and bounded on the side by π₯ = 1.

07:16

### Video Transcript

Find the area of the region bounded above by the curve π¦ is equal to one over π₯, bounded below by the curve π¦ is equal to one over two π₯ squared, and bounded on the side by the line π₯ is equal to one.

The question wants us to find the area of a region which is bounded above by the curve π¦ is equal to one over π₯, and bounded below by the curve π¦ is equal to one over two π₯ squared, and bounded on the side by the vertical line π₯ is equal to one. And we know how to calculate the area between two functions.

We recall if π and π are two continuous functions on the closed interval from π to π. And π of π₯ is greater than or equal to π of π₯ on this closed interval. Then, we can calculate the area bounded above by π of π₯, below by π of π₯, and on the sides by π₯ is equal to π and π₯ is equal to π as the integral from π to π of π of π₯ minus π of π₯ with respect to π₯.

In our case, our region is bounded above by the curve π¦ is equal to one divided by π₯. So, weβll set π is equal to one divided by π₯. And we notice this is a rational function and itβs continuous whenever its denominator is not equal to zero. Similarly, we see that our region is bounded below by the curve π¦ is equal to one divided by two π₯ squared. So, weβll set π of π₯ equal to this. And we notice this is also a rational function and itβs continuous when its denominator is not equal to zero.

So, we found regions on which our functions π and π are continuous. Now, letβs sketch a graph of our region so we can find the closed interval over which we need to integrate. Weβll start by sketching a graph of our reciprocal function π¦ is equal to one divided by π₯. We now want to sketch on the same graph the curve π¦ is equal to one divided by two π₯ squared. We know the general shape of this curve since it will be the same as one divided by π₯ squared. Except it will be stretched by a factor of a half in the vertical direction.

We can solve for any intersects between our curves π¦ is equal to one over π₯ and π¦ is equal to one over two π₯ squared by solving the equation one over π₯ is equal to one over two π₯ squared. Multiplying both sides of our equation by two π₯ squared gives us that two π₯ is equal to one, which we can solve to give us that π₯ is equal to one-half. So, our curves have their only intersect when π₯ is equal to one-half. Weβll also add π₯ is equal to one since this is where our region is bounded on the side.

Now, the question tells us our region is bounded above by one over π₯ and below by one over two π₯ squared. So, we could just sketch the curve π¦ is equal to one over two π₯ squared below on this region and then continue. However, for due diligence, letβs check that this is true.

First, we know when π₯ is less than or equal to zero, our curve one divided by two π₯ squared is positive. So, we donβt need to worry about this case. And for positive values of π₯, we know that one divided by π₯ will be greater than one divided by two π₯ squared when its denominator is smaller, so when π₯ is less than two π₯ squared. Similarly, one divided by π₯ will be smaller than one divided by two π₯ squared when its denominator is bigger.

Thereβs a few different ways of checking which is larger. Weβll consider a curve which is equal to the difference between the two values. When this curve is above the π₯-axis, two π₯ squared will be larger. And when itβs below the π₯-axis, two π₯ squared will be smaller. We factor out π₯ to get π₯ multiplied by two π₯ minus one. Solving this is equal to zero gives us that one of our factors must be zero. So, either π₯ is equal to zero or π₯ is equal to one-half. This tells that our π₯-intercepts are at zero and one-half.

Finally, since the leading sum of our polynomial is two π₯ squared, which is positive, we can sketch a graph of our curve. It will have a similar shape to π¦ is equal to π₯ squared. And for values of π₯ greater than zero, we can see our curve is below the π₯-axis when π₯ is between zero and one-half. And itβs above the π₯-axis when π₯ is greater than one-half. So, the difference is positive when π₯ is greater than one-half. This is the same as saying two π₯ squared is greater than π₯ when π₯ is greater than one-half. And the difference is negative when π₯ is between zero and one-half. So, two π₯ squared is less than π₯ when π₯ is between zero and one-half.

Finally, this tells us that our sketch should start above the curve one over π₯ and then go below the curve one over π₯. Now, from our sketch, we can see our region is bounded above by one over π₯, below by one over two π₯ squared, to the left by π₯ is equal to a half, and to the right by π₯ is equal to one. So, letβs clear some space and discuss if we can use our integral rule to calculate this area.

We know that π and π are continuous on any interval which does not contain π₯ is equal to zero. So, in particular, theyβre continuous on the closed interval from one-half to one. And we already discussed that π of π₯ is greater than or equal to π of π₯ when π₯ is greater than or equal to one-half. So, this is also true on the closed interval from one-half to one. Therefore, the conclusion tells us that we can calculate the area of our region by using the integral from one-half to one of π of π₯ minus π of π₯ with respect to π₯. This tells us our area is given by the integral from one-half to one of one over π₯ minus one over two π₯ squared with respect to π₯.

Weβre now ready to evaluate this integral. We recall 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 πΆ. And for constants π and π, where π is not equal to negative one, the power rule for integration tells us the integral of ππ₯ to the πth power with respect to π₯ is equal to ππ₯ to the power of π plus one all divided by π plus one plus the constant of integration πΆ. We add one to the exponent and divide by this new exponent.

Integrating one over π₯ gives us the natural logarithm of the absolute value of π₯. And then, we rewrite negative one divided by two π₯ squared as negative a half multiplied by π₯ to the power of negative two. So, integrating this, we add one to the exponent to get negative one and then divide through by this new exponent of negative one. This gives us one divided by two π₯. And since weβre calculating a definite integral, the constants of integration will cancel when we evaluate at the limits of our integral. So, we leave this out.

Evaluating at the limits of our integral, π₯ is equal to one-half and π₯ is equal to one, gives us the natural logarithm of the absolute value of one plus one divided by two times one minus the natural logarithm of the absolute value of half plus one divided by two times one-half. The absolute value of one is equal to one. And then, the natural logarithm of one is equal to zero.

We can simplify one divided by two times one to just be equal to one-half. Similarly, one divided by two times one-half gives us one. This gives us one-half minus the natural logarithm of the absolute value of a half minus one. We see that a half is positive, so the absolute value of one-half is just equal to one-half.

Finally, we could simplify this further by using the power rule for logarithms, which tells us π times the logarithm of π is equal to the logarithm of π to the power of π. Using this, we can simplify our answer to negative a half plus the natural logarithm of the reciprocal of one-half. And the reciprocal of one-half is just two, giving us negative a half plus the natural logarithm of two. Therefore, weβve shown the area of the region bounded above by the curve π¦ is equal to one over π₯, bounded below by the curve π¦ is equal to one over two π₯ squared, and bounded on the side by π₯ is equal to one is negative one-half plus the natural logarithm of two.

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