Find the area bounded by the x-axis, part of the curve y = (1 + 8/x²) and the ordinates at x = 2 and

Q: Wait, why did we get a negative value when evaluating the lower limit? Does that mean the area is negative?

Great observation! In the Fundamental Theorem of Calculus, we evaluate $F(b) - F(a)$. Sometimes, the antiderivative $F(x)$ itself might be negative at a specific point, but the difference $F(b) - F(a)$ will always yield the positive area we expect. Trust the process!

Q: Why do we reject the negative root for $a$?

Excellent question. In this problem, our boundaries are $x=2$ and $x=4$. The vertical line $x=a$ must physically exist *between* these two boundaries to divide the area. Since $a$ must be positive and within the interval $[2, 4]$, we discard the negative root.

Q: Is there a shortcut to integrating $\frac{8}{x^2}$?

Think of it as $8x^{-2}$. Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get $\frac{8x^{-1}}{-1} = -\frac{8}{x}$. It's a standard integral, but always double-check your signs—that's where most students lose marks!

Question

Find the area bounded by the x-axis, part of the curve $y = \left( 1 + \frac{8}{x^2} \right)$ and the ordinates at $x = 2$ and $x = 4$. If the ordinate at $x = a$ divides the area into two equal parts, find $a$.

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Curve

Imagine you are standing on a coordinate plane, looking at the function $y = 1 + \frac{8}{x^2}$. As $x$ increases, the term $\frac{8}{x^2}$ rapidly diminishes, meaning the curve gracefully descends, asymptotically approaching the horizontal line $y=1$.

We are tasked with finding the area trapped between this curve, the x-axis, and the vertical boundaries $x=2$ and $x=4$. This is not just a calculation; it is a measurement of a specific, enclosed geometric space.

### The Power of the Integral

To capture this area, we invoke the Fundamental Theorem of Calculus. The total area $A$ is defined by the definite integral:

$$A = \int_{2}^{4} \left( 1 + \frac{8}{x^2} \right) \, dx$$

When we integrate $1$, we get $x$. When we integrate $\frac{8}{x^2}$, we treat it as $8x^{-2}$, which integrates to:

$$\frac{8x^{-1}}{-1} = -\frac{8}{x}$$

Thus, our antiderivative is $x - \frac{8}{x}$. Evaluating this from $2$ to $4$ requires us to find the difference between the values at the upper and lower limits.

At $x=4$, we have $4 - \frac{8}{4} = 4 - 2 = 2$. At $x=2$, we have $2 - \frac{8}{2} = 2 - 4 = -2$. The total area is $2 - (-2) = 4$. We have successfully quantified the space!

### The Dividing Knife

Now, the problem introduces a twist. We need to find a vertical line $x=a$ that slices this region into two equal parts. Since the total area is $4$, each half must have an area of $2$.

We are looking for $a$ such that:

$$\int_{2}^{a} \left( 1 + \frac{8}{x^2} \right) \, dx = 2$$

Using our previously found antiderivative, we evaluate:

$$\left[ x - \frac{8}{x} \right]_{2}^{a} = 2$$

Substituting the limits, we get $(a - \frac{8}{a}) - (2 - \frac{8}{2}) = 2$. This simplifies to $(a - \frac{8}{a}) - (-2) = 2$, which becomes $a - \frac{8}{a} + 2 = 2$.

### The Elegant Cancellation

Look closely at the equation $a - \frac{8}{a} + 2 = 2$. The constant $2$ appears on both sides, allowing them to cancel out perfectly.

We are left with $a - \frac{8}{a} = 0$, or $a = \frac{8}{a}$. Multiplying by $a$, we arrive at:

$$a^2 = 8$$

Taking the square root, we find $a = \sqrt{8} = 2\sqrt{2}$. Since $a$ must lie between $2$ and $4$, we accept the positive root, **$a = 2\sqrt{2} \approx 2.828$**.

You have just solved a classic problem of area bisection using the elegance of calculus. Keep this logic in your toolkit—it is the foundation of much more complex physics and engineering problems to come.

Animated Solution for Mathematics - Definite Integration: Find the area bounded by the x-axis, part of the curve $y = (1 + \frac{8}{x ^{2}})$ and the ordinates at $x = 2$ and $x = 4$ . If the ordinate at $x = a$ divides the area into two equal parts, find $a$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Region

Setting up the Integral

Integrating the Function

Evaluating the Upper Limit

Calculating Total Area

The Dividing Ordinate $x = a$

Setting up the Half-Area Equation

Substituting Limits

Simplifying the Equation

Solving for $a$

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry of the Curve

The Power of the Integral

The Dividing Knife

The Elegant Cancellation

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