Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.8

 

Question 1. Find the area of the region bounded by \( 3x - 2y + 6 = 0 \), \( x = -3 \), \( x = 1 \) and x axis.
Answer: The given equation of the line is \( 3x - 2y + 6 = 0 \). We can rewrite this to express \( y \) in terms of \( x \):
\( 2y = 3x + 6 \)
\( y = \frac { 1 }{ 2 }(3x + 6) \)
The x-intercept is when \( y=0 \), so \( 3x + 6 = 0 \Rightarrow 3x = -6 \Rightarrow x = -2 \).
The area is bounded by the line, the x-axis, and the vertical lines \( x = -3 \) and \( x = 1 \). We need to consider two parts because the line crosses the x-axis at \( x = -2 \). The area from \( x = -3 \) to \( x = -2 \) is below the x-axis, so we take the absolute value of the integral. The area from \( x = -2 \) to \( x = 1 \) is above the x-axis.
The graph shows the shaded region. We can see how the line changes from negative to positive y-values.

The area required is given by: \[ \text{Area} = \left| \int_{-3}^{-2} \frac{1}{2}(3x+6)dx \right| + \int_{-2}^{1} \frac{1}{2}(3x+6)dx \] \[ = \frac{1}{2} \left| \left[ \frac{3x^2}{2} + 6x \right]_{-3}^{-2} \right| + \frac{1}{2} \left[ \frac{3x^2}{2} + 6x \right]_{-2}^{1} \] \[ = \frac{1}{2} \left| \left( \frac{3(-2)^2}{2} + 6(-2) \right) - \left( \frac{3(-3)^2}{2} + 6(-3) \right) \right| + \frac{1}{2} \left[ \left( \frac{3(1)^2}{2} + 6(1) \right) - \left( \frac{3(-2)^2}{2} + 6(-2) \right) \right] \] \[ = \frac{1}{2} \left| \left( \frac{12}{2} - 12 \right) - \left( \frac{27}{2} - 18 \right) \right| + \frac{1}{2} \left[ \left( \frac{3}{2} + 6 \right) - \left( \frac{12}{2} - 12 \right) \right] \] \[ = \frac{1}{2} \left| (6 - 12) - \left( \frac{27 - 36}{2} \right) \right| + \frac{1}{2} \left[ \left( \frac{3 + 12}{2} \right) - (6 - 12) \right] \] \[ = \frac{1}{2} \left| (-6) - \left( -\frac{9}{2} \right) \right| + \frac{1}{2} \left[ \frac{15}{2} - (-6) \right] \] \[ = \frac{1}{2} \left| -6 + \frac{9}{2} \right| + \frac{1}{2} \left[ \frac{15}{2} + 6 \right] \] \[ = \frac{1}{2} \left| \frac{-12 + 9}{2} \right| + \frac{1}{2} \left[ \frac{15 + 12}{2} \right] \] \[ = \frac{1}{2} \left| -\frac{3}{2} \right| + \frac{1}{2} \left[ \frac{27}{2} \right] \] \[ = \frac{1}{2} \left( \frac{3}{2} \right) + \frac{1}{2} \left( \frac{27}{2} \right) \] \[ = \frac{3}{4} + \frac{27}{4} \] \[ = \frac{30}{4} \] \[ = 7.5 \text{ sq. units} \] Thus, the total area enclosed by the given boundaries is 7.5 square units. This method ensures all areas, whether above or below the x-axis, are counted as positive contributions to the total area.
In simple words: First, we change the line equation to find y. Because the line crosses the x-axis, we calculate the area in two parts. We add up the size of both parts to get the total area, which is 7.5 square units.

🎯 Exam Tip: When finding the area between a curve and an axis, always check for axis crossings within the given interval. If the curve crosses the axis, split the integral at the crossing points and take the absolute value of any sections below the axis to ensure the area is positive.

 

Question 2. Find the area of the region bounded by \( 2x – y + 1 = 0 \), \( y = -1 \), \( y = 3 \) and y axis.
Answer: The given equation of the straight line is \( 2x – y + 1 = 0 \). We rewrite this to express \( x \) in terms of \( y \):
\( y = 2x + 1 \)
\( 2x = y - 1 \)
\( x = \frac { y-1 }{ 2 } \)
The y-intercept is when \( x=0 \), so \( y-1 = 0 \Rightarrow y = 1 \).
The area is bounded by the line, the y-axis, and the horizontal lines \( y = -1 \) and \( y = 3 \). We need to consider two parts because the line crosses the y-axis at \( y = 1 \). The area from \( y = -1 \) to \( y = 1 \) is to the left of the y-axis (negative x-values), so we take the absolute value of the integral. The area from \( y = 1 \) to \( y = 3 \) is to the right of the y-axis (positive x-values).
The diagram shows the shaded region. This approach helps in situations where integrating with respect to y is simpler than with respect to x.

The area required is given by: \[ \text{Area} = \left| \int_{-1}^{1} \frac{y-1}{2}dy \right| + \int_{1}^{3} \frac{y-1}{2}dy \] \[ = \frac{1}{2} \left| \left[ \frac{y^2}{2} - y \right]_{-1}^{1} \right| + \frac{1}{2} \left[ \frac{y^2}{2} - y \right]_{1}^{3} \] \[ = \frac{1}{2} \left| \left( \frac{1^2}{2} - 1 \right) - \left( \frac{(-1)^2}{2} - (-1) \right) \right| + \frac{1}{2} \left[ \left( \frac{3^2}{2} - 3 \right) - \left( \frac{1^2}{2} - 1 \right) \right] \] \[ = \frac{1}{2} \left| \left( \frac{1}{2} - 1 \right) - \left( \frac{1}{2} + 1 \right) \right| + \frac{1}{2} \left[ \left( \frac{9}{2} - 3 \right) - \left( \frac{1}{2} - 1 \right) \right] \] \[ = \frac{1}{2} \left| \left( -\frac{1}{2} \right) - \left( \frac{3}{2} \right) \right| + \frac{1}{2} \left[ \left( \frac{9-6}{2} \right) - \left( -\frac{1}{2} \right) \right] \] \[ = \frac{1}{2} \left| -\frac{4}{2} \right| + \frac{1}{2} \left[ \frac{3}{2} + \frac{1}{2} \right] \] \[ = \frac{1}{2} |-2| + \frac{1}{2} \left[ \frac{4}{2} \right] \] \[ = \frac{1}{2}(2) + \frac{1}{2}(2) \] \[ = 1 + 1 \] \[ = 2 \text{ sq. units} \] The total area bounded by the given line and axes is 2 square units. Integrating with respect to y is a useful technique when the region is better defined by horizontal boundaries.
In simple words: First, we change the line equation to find x. Since the line crosses the y-axis, we split the area into two parts. We add up the size of both parts to get the total area, which is 2 square units.

🎯 Exam Tip: When integrating with respect to y, remember to express the curve's equation as \( x = f(y) \) and use the y-coordinates as the limits of integration. Also, check if the curve crosses the y-axis within the interval.

 

Question 3. Find the area of the region bounded by the curve \( 2 + x − x² + y = 0 \), x axis, \( x = -3 \) and \( x = 3 \)
Answer: The given curve is \( 2 + x - x^2 + y = 0 \). We can rewrite this as:
\( y = x^2 - x - 2 \)
To find the x-intercepts, we set \( y = 0 \):
\( x^2 - x - 2 = 0 \)
\( (x - 2)(x + 1) = 0 \)
So, the curve crosses the x-axis at \( x = -1 \) and \( x = 2 \).
The area is bounded by the curve, the x-axis, and the vertical lines \( x = -3 \) and \( x = 3 \). We need to split the integral into three parts:
1. From \( x = -3 \) to \( x = -1 \): The curve is above the x-axis.
2. From \( x = -1 \) to \( x = 2 \): The curve is below the x-axis, so we take the absolute value of the integral.
3. From \( x = 2 \) to \( x = 3 \): The curve is above the x-axis.
The graph illustrates the parabola and the areas to be calculated. The vertex of the parabola is at \( x = 0.5 \), \( y = -2.25 \).

The area required is: \[ \text{Area} = \int_{-3}^{-1} (x^2 - x - 2)dx + \left| \int_{-1}^{2} (x^2 - x - 2)dx \right| + \int_{2}^{3} (x^2 - x - 2)dx \] \[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-3}^{-1} + \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^{2} \right| + \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{2}^{3} \] For the first integral: \( \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) \right) - \left( \frac{(-3)^3}{3} - \frac{(-3)^2}{2} - 2(-3) \right) \) \( = \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) - \left( -9 - \frac{9}{2} + 6 \right) \) \( = \left( \frac{-2 - 3 + 12}{6} \right) - \left( -3 - \frac{9}{2} \right) \) \( = \frac{7}{6} - \left( \frac{-6 - 9}{2} \right) = \frac{7}{6} - \left( -\frac{15}{2} \right) = \frac{7}{6} + \frac{45}{6} = \frac{52}{6} \) For the second integral (take absolute value): \( \left( \frac{2^3}{3} - \frac{2^2}{2} - 2(2) \right) - \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) \right) \) \( = \left( \frac{8}{3} - \frac{4}{2} - 4 \right) - \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) \) \( = \left( \frac{8}{3} - 2 - 4 \right) - \left( \frac{-2 - 3 + 12}{6} \right) \) \( = \left( \frac{8}{3} - 6 \right) - \frac{7}{6} = \left( \frac{8 - 18}{3} \right) - \frac{7}{6} = -\frac{10}{3} - \frac{7}{6} = -\frac{20}{6} - \frac{7}{6} = -\frac{27}{6} \) So, \( \left| -\frac{27}{6} \right| = \frac{27}{6} \) For the third integral: \( \left( \frac{3^3}{3} - \frac{3^2}{2} - 2(3) \right) - \left( \frac{2^3}{3} - \frac{2^2}{2} - 2(2) \right) \) \( = \left( 9 - \frac{9}{2} - 6 \right) - \left( \frac{8}{3} - 2 - 4 \right) \) \( = \left( 3 - \frac{9}{2} \right) - \left( \frac{8}{3} - 6 \right) \) \( = \left( \frac{6 - 9}{2} \right) - \left( \frac{8 - 18}{3} \right) = -\frac{3}{2} - \left( -\frac{10}{3} \right) = -\frac{3}{2} + \frac{10}{3} = \frac{-9 + 20}{6} = \frac{11}{6} \) Total Area \( = \frac{52}{6} + \frac{27}{6} + \frac{11}{6} = \frac{52 + 27 + 11}{6} = \frac{90}{6} = 15 \text{ sq. units} \) The total area bounded by the curve, the x-axis, and the given lines is 15 square units. Breaking the problem into smaller, manageable integrals helps ensure accuracy.
In simple words: The curve crosses the x-axis at two points. We split the area into three parts based on where it is above or below the x-axis. We add these three areas together to get the total area of 15 square units.

🎯 Exam Tip: Always factorize quadratic equations to find x-intercepts, as these points define where the curve changes from being above to below the axis, which is crucial for correctly setting up area integrals.

 

Question 4. Find the area of the region bounded by the line \( y = 2x + 5 \) and the parabola \( y = x² - 2x \).
Answer: First, we need to find the points where the line and the parabola intersect. We set their y-values equal to each other:
\( x^2 - 2x = 2x + 5 \)
\( x^2 - 4x - 5 = 0 \)
\( (x - 5)(x + 1) = 0 \)
So, the x-coordinates of the intersection points are \( x = 5 \) and \( x = -1 \).
For \( x = 5 \), \( y = 2(5) + 5 = 15 \). Intersection point: \( (5, 15) \).
For \( x = -1 \), \( y = 2(-1) + 5 = 3 \). Intersection point: \( (-1, 3) \).
The area is bounded between these two curves from \( x = -1 \) to \( x = 5 \). We need to determine which curve is above the other in this interval. For example, at \( x=0 \), the line is \( y = 5 \) and the parabola is \( y = 0 \). So the line is above the parabola.
The diagram clearly shows the line above the parabola, and the shaded region between them. This visual representation aids in understanding the integration limits.

The area required is the integral of the difference between the upper curve (line) and the lower curve (parabola) from \( x = -1 \) to \( x = 5 \): \[ \text{Area} = \int_{-1}^{5} (y_{\text{line}} - y_{\text{parabola}})dx \] \[ = \int_{-1}^{5} ((2x + 5) - (x^2 - 2x))dx \] \[ = \int_{-1}^{5} (-x^2 + 4x + 5)dx \] \[ = \left[ -\frac{x^3}{3} + \frac{4x^2}{2} + 5x \right]_{-1}^{5} \] \[ = \left[ -\frac{x^3}{3} + 2x^2 + 5x \right]_{-1}^{5} \] Now, we evaluate this at the limits: \( = \left( -\frac{5^3}{3} + 2(5)^2 + 5(5) \right) - \left( -\frac{(-1)^3}{3} + 2(-1)^2 + 5(-1) \right) \) \( = \left( -\frac{125}{3} + 50 + 25 \right) - \left( \frac{1}{3} + 2 - 5 \right) \) \( = \left( -\frac{125}{3} + 75 \right) - \left( \frac{1}{3} - 3 \right) \) \( = \left( \frac{-125 + 225}{3} \right) - \left( \frac{1 - 9}{3} \right) \) \( = \frac{100}{3} - \left( -\frac{8}{3} \right) \) \( = \frac{100}{3} + \frac{8}{3} = \frac{108}{3} = 36 \text{ sq. units} \) Therefore, the area bounded by the line and the parabola is 36 square units. This method of finding the area between two curves is a fundamental application of integral calculus.
In simple words: We find where the line and parabola cross each other. Then, we subtract the parabola's equation from the line's equation and integrate the result between these crossing points. This gives us the area of 36 square units.

🎯 Exam Tip: Always identify the upper and lower curves correctly by testing a point within the integration interval. A wrong identification will result in a negative area, or incorrect magnitude.

 

Question 5. Find the area of the region bounded between the curves \( y = \sin x \) and \( y = \cos x \) and the lines \( x = 0 \) and \( x = \pi \)
Answer: First, we find the intersection point of the two curves, \( y = \sin x \) and \( y = \cos x \):
\( \sin x = \cos x \)
\( \tan x = 1 \)
In the interval \( [0, \pi] \), this occurs at \( x = \frac{\pi}{4} \).
The area is bounded by the curves and the lines \( x = 0 \) and \( x = \pi \). We need to split the integral into two parts because the upper curve changes at the intersection point:
1. From \( x = 0 \) to \( x = \frac{\pi}{4} \): \( \cos x \ge \sin x \).
2. From \( x = \frac{\pi}{4} \) to \( x = \pi \): \( \sin x \ge \cos x \).
The diagram clearly shows the sine and cosine curves and the two shaded regions. Visualizing the curves helps in setting up the correct integrals.

The area required is: \[ \text{Area} = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)dx + \int_{\frac{\pi}{4}}^{\pi} (\sin x - \cos x)dx \] \[ = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} + [-\cos x - \sin x]_{\frac{\pi}{4}}^{\pi} \] Evaluate the first integral: \( = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - (\sin 0 + \cos 0) \) \( = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (0 + 1) \) \( = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \) Evaluate the second integral: \( = (-\cos \pi - \sin \pi) - \left( -\cos \frac{\pi}{4} - \sin \frac{\pi}{4} \right) \) \( = (-(-1) - 0) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \) \( = 1 - \left( -\frac{2}{\sqrt{2}} \right) = 1 - (-\sqrt{2}) = 1 + \sqrt{2} \) Total Area \( = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2} \text{ sq. units} \) The total area bounded by the sine and cosine curves and the given lines is \( 2\sqrt{2} \) square units. This problem highlights the importance of correctly identifying the "upper" curve in each sub-interval.
In simple words: We find where the sine and cosine curves meet. Then, we split the area into two parts, using the curve that is higher in each part. Adding these two areas together gives us a total area of \( 2\sqrt{2} \) square units.

🎯 Exam Tip: Remember the basic shapes of sine and cosine curves and their intersection points. Sketching the graphs is very helpful to correctly identify which function is greater in each interval.

 

Question 6. Find the area of the region bounded by \( y = \tan x \), \( y = \cot x \) and the lines \( x = 0 \), \( x = \frac { \pi }{2} \), \( y = 0 \).
Answer: First, we find the intersection point of the curves \( y = \tan x \) and \( y = \cot x \):
\( \tan x = \cot x \)
\( \tan x = \frac{1}{\tan x} \)
\( \tan^2 x = 1 \)
\( \tan x = 1 \) (since \( x \) is in the first quadrant, \( \tan x \) must be positive)
So, \( x = \frac{\pi}{4} \). At this point, \( y = \tan(\frac{\pi}{4}) = 1 \).
The region is bounded by the curves, the x-axis (\( y=0 \)), and the lines \( x = 0 \) and \( x = \frac{\pi}{2} \). We split the area into two parts:
1. From \( x = 0 \) to \( x = \frac{\pi}{4} \): The area is under \( y = \tan x \).
2. From \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \): The area is under \( y = \cot x \).
The diagram illustrates how the tan and cot functions behave, and how the shaded area is formed by integrating each function separately. This provides a clear visual for the solution.

The required area is: \[ \text{Area} = \int_{0}^{\frac{\pi}{4}} \tan x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx \] \[ = [\log|\sec x|]_{0}^{\frac{\pi}{4}} + [\log|\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \] Evaluate the first integral: \( = \log|\sec(\frac{\pi}{4})| - \log|\sec(0)| \) \( = \log|\sqrt{2}| - \log|1| \) \( = \log \sqrt{2} - 0 = \log \sqrt{2} \) Evaluate the second integral: \( = \log|\sin(\frac{\pi}{2})| - \log|\sin(\frac{\pi}{4})| \) \( = \log|1| - \log|\frac{1}{\sqrt{2}}| \) \( = 0 - \log(\frac{1}{\sqrt{2}}) = - \log(2^{-1/2}) = - (-\frac{1}{2})\log 2 = \frac{1}{2}\log 2 \) The source calculation shows `log √2 + log √2`. Let's re-check the `log|sin x|` part. \( = \log|1| - \log|\frac{1}{\sqrt{2}}| \) \( = 0 - (\log 1 - \log \sqrt{2}) = 0 - (0 - \log \sqrt{2}) = \log \sqrt{2} \) So, total Area \( = \log \sqrt{2} + \log \sqrt{2} = 2 \log \sqrt{2} \) Using logarithm properties, \( 2 \log \sqrt{2} = \log ((\sqrt{2})^2) = \log 2 \text{ sq. units} \) The total area bounded by the tangent and cotangent curves and the given lines is \( \log 2 \) square units. Recognizing fundamental integral forms is key for efficient problem-solving.
In simple words: We find where the tangent and cotangent curves cross. Then, we add the area under the tangent curve up to that crossing point to the area under the cotangent curve from that point onwards. This gives a total area of \( \log 2 \) square units.

🎯 Exam Tip: Remember the integral formulas for \( \tan x \) and \( \cot x \). Also, be careful with logarithm properties and absolute values, especially when evaluating at limits like 0 or \( \pi/2 \).

 

Question 7. Find the area of the region bounded by the parabola \( y² = x \) and the line \( y = x − 2 \).
Answer: First, we find the intersection points of the parabola \( y^2 = x \) and the line \( y = x - 2 \). Substitute \( x = y + 2 \) from the line equation into the parabola equation:
\( y^2 = y + 2 \)
\( y^2 - y - 2 = 0 \)
\( (y - 2)(y + 1) = 0 \)
So, the y-coordinates of the intersection points are \( y = 2 \) and \( y = -1 \).
For \( y = 2 \), \( x = 2 + 2 = 4 \). Intersection point: \( (4, 2) \).
For \( y = -1 \), \( x = -1 + 2 = 1 \). Intersection point: \( (1, -1) \).
The area is bounded between these two curves from \( y = -1 \) to \( y = 2 \). We need to determine which curve is to the right of the other in this interval. The line is \( x = y + 2 \) and the parabola is \( x = y^2 \). If we pick a point like \( y = 0 \) (between -1 and 2), then for the line \( x = 0+2 = 2 \), and for the parabola \( x = 0^2 = 0 \). Since \( 2 > 0 \), the line is to the right of the parabola.
The diagram shows the parabola opening to the right and the line intersecting it, with the area between them shaded. This clear visual helps in setting up the integral with respect to y.

The area required is the integral of the difference between the right curve (\( x_R \)) and the left curve (\( x_L \)) from \( y = -1 \) to \( y = 2 \): \[ \text{Area} = \int_{-1}^{2} (x_{\text{line}} - x_{\text{parabola}})dy \] \[ = \int_{-1}^{2} ((y+2) - y^2)dy \] \[ = \int_{-1}^{2} (-y^2 + y + 2)dy \] \[ = \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 2y \right]_{-1}^{2} \] Now, we evaluate this at the limits: \( = \left( -\frac{2^3}{3} + \frac{2^2}{2} + 2(2) \right) - \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) \right) \) \( = \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) \) \( = \left( -\frac{8}{3} + 2 + 4 \right) - \left( \frac{2 + 3 - 12}{6} \right) \) \( = \left( -\frac{8}{3} + 6 \right) - \left( -\frac{7}{6} \right) \) \( = \left( \frac{-8 + 18}{3} \right) + \frac{7}{6} \) \( = \frac{10}{3} + \frac{7}{6} \) \( = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \text{ sq. units} \) The total area bounded by the parabola and the line is \( \frac{9}{2} \) square units. This shows that choosing the correct integration variable (x or y) can simplify the problem significantly.
In simple words: We find where the parabola and the line cross each other. Then, we subtract the parabola's x-equation from the line's x-equation and integrate with respect to y between these crossing points. This gives an area of \( \frac{9}{2} \) square units.

🎯 Exam Tip: When given curves like \( y^2=x \) or \( x^2=y \), consider integrating with respect to y or x, respectively, to simplify the process. Always sketch the graph to correctly identify the "right" and "left" curves (for dy integration) or "upper" and "lower" curves (for dx integration).

 

Question 8. Father of a family wishes to divide his square field bounded by \( x = 0 \), \( x = 4 \), \( y = 4 \) and \( y = 0 \) along the curve \( y² = 4x \) and \( x² = 4y \) into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided
Answer: The field is a square bounded by \( x = 0 \), \( x = 4 \), \( y = 0 \), and \( y = 4 \). Its total area is \( 4 \times 4 = 16 \) square units. If it's divided into three equal parts, each part should have an area of \( \frac{16}{3} \) square units.
The two curves are \( y^2 = 4x \) (or \( y = 2\sqrt{x} \)) and \( x^2 = 4y \) (or \( y = \frac{x^2}{4} \)).
First, find the intersection points of these two curves:
Substitute \( y = \frac{x^2}{4} \) into \( y^2 = 4x \):
\( \left( \frac{x^2}{4} \right)^2 = 4x \)
\( \frac{x^4}{16} = 4x \)
\( x^4 = 64x \)
\( x^4 - 64x = 0 \)
\( x(x^3 - 64) = 0 \)
So, \( x = 0 \) or \( x^3 = 64 \implies x = 4 \).
If \( x = 0 \), then \( y^2 = 0 \implies y = 0 \). Point: \( (0, 0) \).
If \( x = 4 \), then \( y^2 = 4(4) = 16 \implies y = 4 \). Point: \( (4, 4) \).
The curves intersect at \( (0, 0) \) and \( (4, 4) \). These points are the corners of the square field.
The diagram shows the square field and how the two parabolas divide it into three distinct regions (A1, A2, A3). This helps visualize how the area is partitioned.

Let's calculate the areas of the three regions: Area A2 is the area between the two parabolas from \( x = 0 \) to \( x = 4 \). In this interval, \( y = 2\sqrt{x} \) is the upper curve and \( y = \frac{x^2}{4} \) is the lower curve.
\( A_2 = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) dx \) \[ = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{1}{4} \cdot \frac{x^3}{3} \right]_{0}^{4} \] \[ = \left[ \frac{4}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{4} \] \[ = \left( \frac{4}{3}(4)^{3/2} - \frac{(4)^3}{12} \right) - (0 - 0) \] \[ = \left( \frac{4}{3}(8) - \frac{64}{12} \right) \] \[ = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. units} \] Area A3 is the area under the curve \( y = \frac{x^2}{4} \) from \( x = 0 \) to \( x = 4 \):
\( A_3 = \int_{0}^{4} \frac{x^2}{4} dx \) \[ = \left[ \frac{1}{4} \cdot \frac{x^3}{3} \right]_{0}^{4} \] \[ = \left[ \frac{x^3}{12} \right]_{0}^{4} \] \[ = \frac{4^3}{12} - 0 = \frac{64}{12} = \frac{16}{3} \text{ sq. units} \] Area A1 is the area between the curve \( x = \frac{y^2}{4} \) and the y-axis from \( y = 0 \) to \( y = 4 \):
\( A_1 = \int_{0}^{4} \frac{y^2}{4} dy \) \[ = \left[ \frac{1}{4} \cdot \frac{y^3}{3} \right]_{0}^{4} \] \[ = \left[ \frac{y^3}{12} \right]_{0}^{4} \] \[ = \frac{4^3}{12} - 0 = \frac{64}{12} = \frac{16}{3} \text{ sq. units} \] Since \( A_1 = A_2 = A_3 = \frac{16}{3} \) square units, it is possible to divide the field into three equal parts. Each part will be \( \frac{16}{3} \) square units. This is a clever application of area calculation to a real-world division problem.
In simple words: The field is a square of 16 square units. Two curves divide this square. We found that the areas of the three parts created by these curves are all \( \frac{16}{3} \) square units. Since these are equal, it is possible to divide the field equally among the family members.

🎯 Exam Tip: For problems involving multiple curves and boundaries, always sketch the region carefully to identify the upper/lower or right/left curves and the correct limits of integration for each sub-region. This is critical for setting up the integrals correctly.

 

Question 9. The curve \( y = (x – 2)² + 1 \) has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Answer: The given curve is a parabola \( y = (x - 2)^2 + 1 \). This can be written as \( (y - 1) = (x - 2)^2 \), which is a parabola with its vertex at \( (2, 1) \). So, the minimum point P is \( (2, 1) \).
The slope of the chord PQ is given as 2. The equation of the line (chord) PQ passing through P\( (2, 1) \) with a slope of 2 is:
\( y - y_1 = m(x - x_1) \)
\( y - 1 = 2(x - 2) \)
\( y - 1 = 2x - 4 \)
\( y = 2x - 3 \)
Now, we find the intersection points of the parabola \( y = (x - 2)^2 + 1 \) and the line \( y = 2x - 3 \):
\( (x - 2)^2 + 1 = 2x - 3 \)
\( x^2 - 4x + 4 + 1 = 2x - 3 \)
\( x^2 - 4x + 5 = 2x - 3 \)
\( x^2 - 6x + 8 = 0 \)
\( (x - 2)(x - 4) = 0 \)
The x-coordinates of the intersection points are \( x = 2 \) and \( x = 4 \).
For \( x = 2 \), \( y = 2(2) - 3 = 1 \). This is point P\( (2, 1) \).
For \( x = 4 \), \( y = 2(4) - 3 = 5 \). This is point Q\( (4, 5) \).
The area bounded by the curve and the chord PQ is the integral of the difference between the upper curve (line) and the lower curve (parabola) from \( x = 2 \) to \( x = 4 \).
The diagram illustrates the parabola, the chord PQ, and the shaded area between them. It clearly shows the line is above the parabola in the interval [2, 4].

The area required is: \[ \text{Area} = \int_{2}^{4} ((2x - 3) - ((x - 2)^2 + 1))dx \] \[ = \int_{2}^{4} (2x - 3 - (x^2 - 4x + 4 + 1))dx \] \[ = \int_{2}^{4} (2x - 3 - x^2 + 4x - 5)dx \] \[ = \int_{2}^{4} (-x^2 + 6x - 8)dx \] \[ = \left[ -\frac{x^3}{3} + \frac{6x^2}{2} - 8x \right]_{2}^{4} \] \[ = \left[ -\frac{x^3}{3} + 3x^2 - 8x \right]_{2}^{4} \] Now, we evaluate this at the limits: \( = \left( -\frac{4^3}{3} + 3(4)^2 - 8(4) \right) - \left( -\frac{2^3}{3} + 3(2)^2 - 8(2) \right) \) \( = \left( -\frac{64}{3} + 48 - 32 \right) - \left( -\frac{8}{3} + 12 - 16 \right) \) \( = \left( -\frac{64}{3} + 16 \right) - \left( -\frac{8}{3} - 4 \right) \) \( = \left( \frac{-64 + 48}{3} \right) - \left( \frac{-8 - 12}{3} \right) \) \( = -\frac{16}{3} - \left( -\frac{20}{3} \right) \) \( = -\frac{16}{3} + \frac{20}{3} = \frac{4}{3} \text{ sq. units} \) The area bounded by the curve and the chord PQ is \( \frac{4}{3} \) square units. This example effectively combines coordinate geometry concepts with integral calculus.
In simple words: We first find the two points where the parabola and the line meet. Then, we subtract the parabola's equation from the line's equation and integrate the result between these two x-values. This gives us the area of \( \frac{4}{3} \) square units.

🎯 Exam Tip: When dealing with curves and lines, carefully find all intersection points as these will define the limits of integration. Always check which function represents the "upper" boundary in the region of interest.

 

Question 10. Find the area of the region common to the circle \( x² + y² = 16 \) and the parabola \( y² = 6x \)
Answer: First, we find the intersection points of the circle \( x^2 + y^2 = 16 \) and the parabola \( y^2 = 6x \).
Substitute \( y^2 = 6x \) into the circle equation:
\( x^2 + 6x = 16 \)
\( x^2 + 6x - 16 = 0 \)
\( (x + 8)(x - 2) = 0 \)
So, \( x = -8 \) or \( x = 2 \).
Since \( y^2 = 6x \), \( x \) must be non-negative (because \( y^2 \) is always non-negative). Therefore, \( x = -8 \) is impossible.
We use \( x = 2 \). Substitute this back into \( y^2 = 6x \):
\( y^2 = 6(2) = 12 \)
\( y = \pm\sqrt{12} = \pm 2\sqrt{3} \)
The intersection points are \( (2, 2\sqrt{3}) \) and \( (2, -2\sqrt{3}) \).
The radius of the circle \( x^2 + y^2 = 16 \) is \( r = \sqrt{16} = 4 \).
The common region is symmetric about the x-axis. So, we can find the area of the upper half and multiply by 2.
The upper half of the common area consists of two parts:
1. Area under the parabola \( y = \sqrt{6x} \) from \( x = 0 \) to \( x = 2 \).
2. Area under the circle \( y = \sqrt{16 - x^2} \) from \( x = 2 \) to \( x = 4 \).
The diagram clearly illustrates the circle and parabola, along with the two regions (A and B) that make up the common area. This visual aids in setting up the definite integrals.

Total Area \( = 2 \times \left( \int_{0}^{2} \sqrt{6x} \, dx + \int_{2}^{4} \sqrt{16 - x^2} \, dx \right) \)
For the first integral (parabola part): \( \int_{0}^{2} \sqrt{6x} \, dx = \sqrt{6} \int_{0}^{2} x^{1/2} \, dx \) \( = \sqrt{6} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} = \sqrt{6} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2} \) \( = \sqrt{6} \cdot \frac{2}{3} (2^{3/2} - 0) = \frac{2\sqrt{6}}{3} (2\sqrt{2}) = \frac{4\sqrt{12}}{3} = \frac{4 \cdot 2\sqrt{3}}{3} = \frac{8\sqrt{3}}{3} \) For the second integral (circle part), we use the formula \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \). Here \( a = 4 \): \( \int_{2}^{4} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{2}^{4} \) \( = \left( \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) - \left( \frac{2}{2}\sqrt{16 - 2^2} + 8\sin^{-1}\left(\frac{2}{4}\right) \right) \) \( = \left( 2\sqrt{0} + 8\sin^{-1}(1) \right) - \left( 1\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right) \right) \) \( = \left( 0 + 8 \cdot \frac{\pi}{2} \right) - \left( 2\sqrt{3} + 8 \cdot \frac{\pi}{6} \right) \) \( = 4\pi - \left( 2\sqrt{3} + \frac{4\pi}{3} \right) \) \( = 4\pi - 2\sqrt{3} - \frac{4\pi}{3} = \frac{12\pi - 4\pi}{3} - 2\sqrt{3} = \frac{8\pi}{3} - 2\sqrt{3} \) Now, sum the two parts and multiply by 2: Total Area \( = 2 \times \left( \frac{8\sqrt{3}}{3} + \frac{8\pi}{3} - 2\sqrt{3} \right) \) \( = 2 \times \left( \frac{8\sqrt{3} - 6\sqrt{3}}{3} + \frac{8\pi}{3} \right) \) \( = 2 \times \left( \frac{2\sqrt{3}}{3} + \frac{8\pi}{3} \right) \) \( = \frac{4\sqrt{3}}{3} + \frac{16\pi}{3} \) This can also be written as \( = \frac{4}{3} ( \sqrt{3} + 4\pi ) \) or \( \frac{4}{3} (4\pi + \sqrt{3}) \text{ sq. units} \) The total area common to the circle and the parabola is \( \frac{4}{3} (4\pi + \sqrt{3}) \) square units. This problem requires knowledge of both algebraic manipulation and standard integral formulas.
In simple words: First, we find where the circle and parabola cross. Then, we find the area in two parts: one under the parabola and one under the circle. We add these areas together and multiply by two because the shape is symmetric. This gives the total common area.

🎯 Exam Tip: For regions involving circles and parabolas, always solve for intersection points carefully. Remember to check for symmetry to simplify calculations, and recall the standard integration formula for \( \sqrt{a^2 - x^2} \).

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