Samacheer Kalvi Class 12 Business Maths Solutions Chapter 3 Integral Calculus II Exercise 3.1

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.1

Question 1. Using Integration, find the area of the region bounded the line given is 2y + x = 8, the x axis and the lines x = 2, x = 4.
Answer: First, we rewrite the given line equation to express \(y\) in terms of \(x\).
The equation is \( 2y + x = 8 \).
Subtract \(x\) from both sides: \( 2y = 8 - x \).
Divide by 2:
\( \implies y = \frac{8-x}{2} \)
\( \implies y = 4 - \frac{x}{2} \)
The region is bounded by the x-axis and the lines \( x = 2 \) and \( x = 4 \). So, the limits of integration are from 2 to 4.
To find the area, we integrate \(y\) with respect to \(x\) from \(x = 2\) to \(x = 4\).
The required Area \( A = \int_{2}^{4} y \,dx \)
Substitute the expression for \(y\):
\( A = \int_{2}^{4} (4 - \frac{x}{2}) \,dx \)
Now, we integrate each term:
\( A = \left[ 4x - \frac{1}{2} \left( \frac{x^2}{2} \right) \right]_{2}^{4} \)
\( \implies A = \left[ 4x - \frac{x^2}{4} \right]_{2}^{4} \)
Next, we evaluate this expression at the upper limit (4) and subtract its value at the lower limit (2).
\( A = \left[ 4(4) - \frac{4^2}{4} \right] - \left[ 4(2) - \frac{2^2}{4} \right] \)
\( A = \left[ 16 - \frac{16}{4} \right] - \left[ 8 - \frac{4}{4} \right] \)
\( \implies A = [16 - 4] - [8 - 1] \)
\( \implies A = 12 - 7 \)
\( \implies A = 5 \) sq.units
The calculation helps us understand the fundamental theorem of calculus, applying it to find the area under a curve.
In simple words: First, rewrite the line equation to find \(y\). Then, use integral to calculate the space between the line and the x-axis, from \(x=2\) to \(x=4\). This calculation shows the area is 5 square units.

🎯 Exam Tip: Always make sure to correctly rearrange the equation for \(y\) in terms of \(x\) (or \(x\) in terms of \(y\)) before setting up the integral, and double-check your limits of integration.

Question 2. Find the area bounded by the lines y – 2x – 4 = 0, y = 0, y = 3 and the y-axis.
Answer: First, we rewrite the given line equation to express \(x\) in terms of \(y\).
The equation is \( y - 2x - 4 = 0 \).
Add \(2x\) and 4 to both sides: \( y - 4 = 2x \).
Divide by 2:
\( \implies x = \frac{y-4}{2} \)
\( \implies x = \frac{y}{2} - 2 \)
The region is bounded by the y-axis, and the lines \( y = 1 \) and \( y = 3 \) (as indicated by the calculation limits and "y varies from 1 to 3"). So, the limits of integration are from 1 to 3.
To find the area, we integrate \(x\) with respect to \(y\) from \(y = 1\) to \(y = 3\).
The required Area \( A = \int_{1}^{3} x \,dy \)
Substitute the expression for \(x\):
\( A = \int_{1}^{3} \left( \frac{y}{2} - 2 \right) \,dy \)
Now, we integrate each term:
\( A = \left[ \frac{1}{2} \left( \frac{y^2}{2} \right) - 2y \right]_{1}^{3} \)
\( \implies A = \left[ \frac{y^2}{4} - 2y \right]_{1}^{3} \)
Next, we evaluate this expression at the upper limit (3) and subtract its value at the lower limit (1).
\( A = \left( \frac{3^2}{4} - 2(3) \right) - \left( \frac{1^2}{4} - 2(1) \right) \)
\( A = \left( \frac{9}{4} - 6 \right) - \left( \frac{1}{4} - 2 \right) \)
\( \implies A = \left( \frac{9-24}{4} \right) - \left( \frac{1-8}{4} \right) \)
\( \implies A = \frac{-15}{4} - \frac{-7}{4} \)
\( \implies A = \frac{-15+7}{4} = \frac{-8}{4} = -2 \)
Since area cannot be negative, we take the absolute value of the result.
Therefore, Area \( A = |-2| = 2 \) sq.units. This situation shows that sometimes the integral might yield a negative value if the region is on the 'negative' side of the axis, and we take its absolute value for the actual area.
In simple words: First, rewrite the line equation to find \(x\). Then, calculate the integral from \(y=1\) to \(y=3\). The result is -2, but because area can't be negative, the final area is 2 square units.

🎯 Exam Tip: When calculating area, if the definite integral yields a negative value, it means the region is below the x-axis (or to the left of the y-axis). Always take the absolute value of the result for the final area.

Question 3. Calculate the area bounded by the parabola y² = 4ax and its latus rectum.
Answer: The given parabola is \( y^2 = 4ax \). This parabola is symmetrical about the x-axis and opens towards the positive x-axis.
Its vertex is at the origin \( (0, 0) \).
The focus of the parabola is \( (a, 0) \).
The equation of the latus rectum is \( x = a \). This is a vertical line at \(x = a\).
We need to find the area bounded by \( y^2 = 4ax \) and \( x = a \). Since the parabola is symmetric about the x-axis, we can find the area of the upper half (from \(y=0\) to \(y=\sqrt{4ax}\)) and then multiply it by 2.
From \( y^2 = 4ax \), we get \( y = \pm \sqrt{4ax} = \pm 2\sqrt{a}\sqrt{x} \). For the upper half, we take \( y = 2\sqrt{a}\sqrt{x} \).
The limits of integration for \(x\) are from \(0\) (vertex) to \(a\) (latus rectum).
The required Area \( A = 2 \int_{0}^{a} y \,dx \)
Substitute the expression for \(y\):
\( A = 2 \int_{0}^{a} 2\sqrt{a}\sqrt{x} \,dx \)
Take the constant out of the integral:
\( A = 4\sqrt{a} \int_{0}^{a} x^{1/2} \,dx \)
Now, we integrate \(x^{1/2}\):
\( A = 4\sqrt{a} \left[ \frac{x^{1/2+1}}{1/2+1} \right]_{0}^{a} \)
\( A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a} \)
\( \implies A = 4\sqrt{a} \left[ \frac{2x^{3/2}}{3} \right]_{0}^{a} \)
\( \implies A = \frac{8\sqrt{a}}{3} [x^{3/2}]_{0}^{a} \)
Next, we evaluate this expression at the upper limit (\(a\)) and subtract its value at the lower limit (0).
\( A = \frac{8\sqrt{a}}{3} (a^{3/2} - 0^{3/2}) \)
\( A = \frac{8\sqrt{a}}{3} (a^{3/2}) \)
\( \implies A = \frac{8}{3} a^{1/2} \cdot a^{3/2} \)
\( \implies A = \frac{8}{3} a^{(1/2)+(3/2)} \)
\( \implies A = \frac{8}{3} a^{4/2} \)
\( \implies A = \frac{8}{3} a^2 \) sq.units
This shows how the latus rectum, a key feature of the parabola, defines a specific area when combined with the parabolic curve.

🎯 Exam Tip: For symmetric curves like parabolas, you can calculate the area of one symmetric half and then multiply by two, which often simplifies the calculations. Always identify the correct limits based on the given boundaries.

Question 4. Find the area bounded by the line y = x and x-axis and the ordinates x = 1, x = 2
Answer: The given line is \( y = x \). The region is bounded by the x-axis and the vertical lines \( x = 1 \) and \( x = 2 \).
Since the line \( y = x \) is above the x-axis for \(x\) between 1 and 2, we can directly integrate \(y\) with respect to \(x\).
The limits of integration are from 1 to 2.
The required Area \( A = \int_{1}^{2} y \,dx \)
Substitute the expression for \(y\):
\( A = \int_{1}^{2} x \,dx \)
Now, we integrate \(x\):
\( A = \left[ \frac{x^2}{2} \right]_{1}^{2} \)
Next, we evaluate this expression at the upper limit (2) and subtract its value at the lower limit (1).
\( A = \frac{2^2}{2} - \frac{1^2}{2} \)
\( A = \frac{4}{2} - \frac{1}{2} \)
\( \implies A = \frac{3}{2} \) sq.units
This area calculation represents a trapezoid, showing the connection between geometry and calculus in finding bounded regions.

🎯 Exam Tip: For simple linear functions above the x-axis, the integral calculates the direct area. Always visualize the region to determine if the function is consistently above or below the axis.

Question 5. Using integration, find the area of the region bounded by the line y − 1 = x, the x-axis and the ordinates x = -2, x = 3.
Answer: First, we rewrite the given line equation: \( y - 1 = x \)
\( \implies y = x + 1 \)
The region is bounded by the x-axis and the vertical lines \( x = -2 \) and \( x = 3 \).
The line \( y = x + 1 \) crosses the x-axis when \( y = 0 \), so \( x + 1 = 0 \implies x = -1 \).
This means the line is below the x-axis from \( x = -2 \) to \( x = -1 \) and above the x-axis from \( x = -1 \) to \( x = 3 \).
Therefore, we need to split the integral into two parts. For the part below the x-axis, we use \( -y \,dx \).
The required Area \( A = \int_{-2}^{-1} (-y) \,dx + \int_{-1}^{3} y \,dx \)
Substitute \( y = x + 1 \):
\( A = \int_{-2}^{-1} -(x+1) \,dx + \int_{-1}^{3} (x+1) \,dx \)
Now, we integrate each part separately:
For the first integral:
\( \int -(x+1) \,dx = -\left( \frac{x^2}{2} + x \right) \)
Evaluate from -2 to -1:
\( -\left[ \left( \frac{(-1)^2}{2} + (-1) \right) - \left( \frac{(-2)^2}{2} + (-2) \right) \right] \)
\( = -\left[ \left( \frac{1}{2} - 1 \right) - \left( \frac{4}{2} - 2 \right) \right] \)
\( = -\left[ \left( -\frac{1}{2} \right) - (2 - 2) \right] \)
\( = -\left[ -\frac{1}{2} - 0 \right] = \frac{1}{2} \)
For the second integral:
\( \int (x+1) \,dx = \frac{x^2}{2} + x \)
Evaluate from -1 to 3:
\( \left[ \left( \frac{3^2}{2} + 3 \right) - \left( \frac{(-1)^2}{2} + (-1) \right) \right] \)
\( = \left[ \left( \frac{9}{2} + 3 \right) - \left( \frac{1}{2} - 1 \right) \right] \)
\( = \left[ \left( \frac{9+6}{2} \right) - \left( \frac{1-2}{2} \right) \right] \)
\( = \left[ \frac{15}{2} - \left( -\frac{1}{2} \right) \right] \)
\( = \frac{15}{2} + \frac{1}{2} = \frac{16}{2} = 8 \)
Finally, add the areas from both parts:
\( A = \frac{1}{2} + 8 = \frac{1+16}{2} = \frac{17}{2} \) sq.units
This method is crucial for areas that cross the axis of integration, ensuring all parts contribute positively to the total area.

🎯 Exam Tip: When a curve crosses the x-axis within the integration limits, calculate the area of each segment separately (taking the absolute value of any negative integral) and then sum them up. Do not integrate across the axis directly without splitting.

Question 6. Find the area of the region lying in the first quadrant bounded by the region y = 4x², x = 0, y = 0 and y = 4
Answer: The given parabola is \( y = 4x^2 \). We are looking for the area in the first quadrant, bounded by \( x = 0 \) (y-axis), \( y = 0 \) (x-axis), and \( y = 4 \).
Since the region is defined with respect to \(y\) limits, it's easier to integrate with respect to \(y\). We need to express \(x\) in terms of \(y\).
From \( y = 4x^2 \):
\( x^2 = \frac{y}{4} \)
Taking the square root (and positive root for the first quadrant):
\( \implies x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2} \)
The limits of integration for \(y\) are from \(0\) to \(4\).
The required Area \( A = \int_{0}^{4} x \,dy \)
Substitute the expression for \(x\):
\( A = \int_{0}^{4} \frac{\sqrt{y}}{2} \,dy \)
Take the constant out of the integral:
\( A = \frac{1}{2} \int_{0}^{4} y^{1/2} \,dy \)
Now, we integrate \(y^{1/2}\):
\( A = \frac{1}{2} \left[ \frac{y^{1/2+1}}{1/2+1} \right]_{0}^{4} \)
\( A = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4} \)
\( \implies A = \frac{1}{2} \left[ \frac{2y^{3/2}}{3} \right]_{0}^{4} \)
\( \implies A = \frac{1}{3} [y^{3/2}]_{0}^{4} \)
Next, we evaluate this expression at the upper limit (4) and subtract its value at the lower limit (0).
\( A = \frac{1}{3} (4^{3/2} - 0^{3/2}) \)
\( A = \frac{1}{3} ((\sqrt{4})^3 - 0) \)
\( A = \frac{1}{3} (2^3) \)
\( \implies A = \frac{1}{3} (8) = \frac{8}{3} \) sq.units
Integrating with respect to \(y\) is often simpler when the boundaries are horizontal lines or the y-axis itself, providing an alternative perspective to area calculation.

🎯 Exam Tip: When the region is defined by horizontal lines and the y-axis, it's often more efficient to integrate with respect to \(y\) by expressing \(x\) as a function of \(y\).

Question 7. Find the area bounded by the curve y = x² and the line y = 4
Answer: The given curve is a parabola \( y = x^2 \), which opens upwards with its vertex at the origin \( (0, 0) \).
The given line is \( y = 4 \), which is a horizontal line.
To find the area bounded by them, we first need to find their points of intersection.
Set the equations equal: \( x^2 = 4 \)
Take the square root of both sides: \( x = \pm 2 \)
So, the intersection points are \( (-2, 4) \) and \( (2, 4) \). These are the limits of integration for \(x\).
The line \( y = 4 \) is the upper boundary (\( y_{upper} \)) and the parabola \( y = x^2 \) is the lower boundary (\( y_{lower} \)) of the region.
The required Area \( A = \int_{-2}^{2} (y_{upper} - y_{lower}) \,dx \)
Substitute the expressions for \(y_{upper}\) and \(y_{lower}\):
\( A = \int_{-2}^{2} (4 - x^2) \,dx \)
Now, we integrate each term:
\( A = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} \)
Next, we evaluate this expression at the upper limit (2) and subtract its value at the lower limit (-2).
\( A = \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \)
\( A = \left( 8 - \frac{8}{3} \right) - \left( -8 - \frac{-8}{3} \right) \)
\( A = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) \)
\( A = 8 - \frac{8}{3} + 8 - \frac{8}{3} \)
\( A = 16 - \frac{16}{3} \)
Combine the terms by finding a common denominator:
\( A = \frac{16 \times 3}{3} - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} \)
\( \implies A = \frac{32}{3} \) sq.units
Finding the area between two curves is a common application of definite integrals, useful in various fields like engineering and physics.

🎯 Exam Tip: When finding the area between two curves, always determine which function is "above" the other in the specified interval. The integral will be of the upper function minus the lower function, integrated over the interval defined by their intersection points or given boundaries.