The area enclosed by the curves xy + 4y = 16 and x + y = 6 is equal to : (1) 28 - 30 loge 2 (2) 30 -

Q: Why do we subtract the lower curve from the upper curve in the integral?

Area is a physical quantity, and by definition, it must be positive. When we calculate the area between two curves, we take the integral of the difference between the upper function and the lower function. If you accidentally swap them, you will get a negative result, which tells you that you have the order reversed. Always ensure your integrand is positive!

Q: How do I know which curve is on top without graphing?

You don't need a perfect graph! Just pick a test point within your interval of integration, say $x = 0$. For the line, $y = 6 - 0 = 6$. For the hyperbola, $y = 16 / (0 + 4) = 4$. Since $6 > 4$, the line is clearly the upper curve in this region.

Q: Why does the answer involve $\ln 2$?

Great observation! When we integrate $\frac{1}{x+4}$, we get $\ln(x+4)$. When we plug in our limits, we get terms like $\ln(8)$ and $\ln(2)$. Since $8 = 2^3$, we can use the logarithmic property $\ln(a^b) = b \ln a$ to convert $\ln(8)$ into $3 \ln 2$. This is a classic JEE technique to match your result with the given options.

Question

The area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ is equal to :

Options:

(A) $28 - 30 \log_e 2$

(B) $30 - 28 \log_e 2$

(C) $30 - 32 \log_e 2$

(D) $32 - 30 \log_e 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before two distinct mathematical entities: a sleek, sweeping rectangular hyperbola defined by $xy + 4y = 16$ and a sharp, decisive straight line $x + y = 6$. Our mission is to find the space they trap between them—the area enclosed by their intersection.

First, let us bring these equations into a form that speaks to us. The hyperbola, when we isolate $y$, becomes:

$$y = \frac{16}{x + 4}$$

The line is even friendlier: $y = 6 - x$. These are our two actors.

### The Meeting Point

Before we can measure the area, we must know where the stage begins and ends. To find where these curves meet, we set their $y$ values equal:

$$\frac{16}{x + 4} = 6 - x$$

Multiplying both sides by $(x + 4)$, we get $16 = (6 - x)(x + 4)$. Expanding this, we find $16 = 6x + 24 - x^2 - 4x$, which simplifies to the quadratic equation:

$$x^2 - 2x - 8 = 0$$

Factoring this, we find $(x - 4)(x + 2) = 0$. Our boundaries are set at $x = -2$ and $x = 4$. These are the gates of our integration.

### The Integral of Discovery

Now, we must sum up the infinitesimal strips of area between these curves. We know from our test point that the line sits above the hyperbola in this interval.

Thus, our integral is:

$$\int_{-2}^{4} \left( (6 - x) - \frac{16}{x + 4} \right) dx$$

We integrate term by term: the integral of $6$ is $6x$, the integral of $-x$ is $-\frac{x^2}{2}$, and the integral of $-\frac{16}{x+4}$ is $-16 \ln|x+4|$. We evaluate this expression from $-2$ to $4$.

### The Final Calculation

Substituting the upper limit $x = 4$:

$$6(4) - \frac{4^2}{2} - 16 \ln(4 + 4) = 24 - 8 - 16 \ln 8 = 16 - 16 \ln 8$$

Substituting the lower limit $x = -2$:

$$6(-2) - \frac{(-2)^2}{2} - 16 \ln(-2 + 4) = -12 - 2 - 16 \ln 2 = -14 - 16 \ln 2$$

Subtracting the lower limit from the upper limit, we get:

$$(16 - 16 \ln 8) - (-14 - 16 \ln 2) = 30 - 16 \ln 8 + 16 \ln 2$$

Since $\ln 8 = 3 \ln 2$, this becomes:

$$30 - 16(3 \ln 2) + 16 \ln 2 = 30 - 48 \ln 2 + 16 \ln 2 = 30 - 32 \ln 2$$

The final area enclosed by the two curves is **$30 - 32 \ln 2$** square units.

Animated Solution for Mathematics - Definite Integration: The area enclosed by the curves $x y + 4 y = 16$ and $x + y = 6$ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Curves

Express $y$ for the Hyperbola

Express $y$ for the Straight Line

Find Intersection Points

Form the Quadratic Equation

Solve for $x$

Set up the Area Integral

Integrate the Expression

Substitute the Upper Limit

Substitute the Lower Limit

Calculate the Difference

Simplify Logarithmic Terms

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Meeting Point

The Integral of Discovery

The Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :