RBSE Solutions Class 12 Maths Chapter 11 Application of Integral Quadrature Exercise 11.1

 

Question 1. Find the area enclosed by parabola \( y^2 = 4ax \) and its latus rectum.
Answer: The equation of the parabola is \( y^2 = 4ax \). The equation of its latus rectum is \( x = a \). The parabola opens to the right, symmetric about the x-axis. The latus rectum is a vertical line at \( x = a \). To find the enclosed area, we integrate \( y \) with respect to \( x \) from \( 0 \) to \( a \). Since the parabola is symmetric, the total area is twice the area above the x-axis.
From \( y^2 = 4ax \), we get \( y = \pm 2\sqrt{ax} \). We take the positive root for the area above the x-axis.
Required Area \( = \) Area of AOA'MA
\( = 2 \times \) Area AOMA
\( = 2 \int_{0}^{a} y \,dx \)
\( = 2 \int_{0}^{a} 2\sqrt{a}\sqrt{x} \,dx \)
\( = 4\sqrt{a} \int_{0}^{a} x^{1/2} \,dx \)
\( = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a} \)
\( = 4\sqrt{a} \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{a} \)
\( = \frac{8\sqrt{a}}{3} \left[ a^{3/2} - 0^{3/2} \right] \)
\( = \frac{8\sqrt{a}}{3} a^{3/2} \)
\( = \frac{8}{3} a^{1/2} a^{3/2} \)
\( = \frac{8}{3} a^{(1/2 + 3/2)} \)
\( = \frac{8}{3} a^2 \) sq. units.
In simple words: We need to calculate the space covered by a parabola and a special line segment called the latus rectum. We find the area by using a mathematical tool called integration. We calculate the area of one half of the region and then double it, because the shape is the same on both sides of the x-axis, giving us the total area.

🎯 Exam Tip: Remember to identify the curve and its symmetry to simplify integration. For parabolas symmetric about an axis, calculating the area of one half and doubling it is a common shortcut.

 

Question 2. Sketch the circle \( x^2 + y^2 = 4 \), find area enclosed by y - axis and x = 1.
Answer: The equation of the circle is \( x^2 + y^2 = 4 \), which means its center is at \( (0,0) \) and its radius is \( 2 \). We need to find the area enclosed by this circle, the y-axis (which is \( x=0 \)), and the line \( x=1 \). The required area is shown by the shaded region in the figure, which is symmetric about the x-axis. Therefore, we calculate the area above the x-axis and then multiply it by 2.
From \( x^2 + y^2 = 4 \), we get \( y^2 = 4 - x^2 \), so \( y = \sqrt{4 - x^2} \) (taking the positive root for the upper half).
Thus, the required area at \( x = 1, y = \sqrt{3} \):
Required Area \( = 2 \int_{0}^{1} y \,dx \)
\( = 2 \int_{0}^{1} \sqrt{4 - x^2} \,dx \)
Using 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 \( a = 2 \).
\( = 2 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{1} \)
\( = 2 \left[ \left( \frac{1}{2}\sqrt{4 - 1^2} + 2\sin^{-1}\left(\frac{1}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right) \right] \)
\( = 2 \left[ \left( \frac{1}{2}\sqrt{3} + 2\sin^{-1}\left(\frac{1}{2}\right) \right) - (0 + 0) \right] \)
\( = 2 \left[ \frac{\sqrt{3}}{2} + 2 \times \frac{\pi}{6} \right] \)
\( = 2 \left[ \frac{\sqrt{3}}{2} + \frac{\pi}{3} \right] \)
\( = \sqrt{3} + \frac{2\pi}{3} \) sq. units.
In simple words: We found the area inside a circle, from the y-axis up to the line \( x=1 \). Since the circle is perfectly even, we only calculated the top half and then doubled it. We used a special formula to integrate the circle's curve. This area is like a slice of pizza from the circle.

🎯 Exam Tip: When finding areas involving circles, remember the standard formula for integrating \( \sqrt{a^2 - x^2} \). Sketching the region helps visualize the limits of integration.

 

Question 3. Find the area enclosed by curve \( y = \sin x \) and x axis, whereas \( 0 \le x \le 2\pi \).
Answer: The curve is \( y = \sin x \). We need to find the area enclosed by this curve and the x-axis from \( x = 0 \) to \( x = 2\pi \). The sine curve is positive from \( 0 \) to \( \pi \) and negative from \( \pi \) to \( 2\pi \). To find the total area, we must sum the absolute values of the areas in these two intervals. This ensures that the area below the x-axis is counted as positive.
Required Area \( = \int_{0}^{2\pi} | \sin x | \,dx \)
\( = \int_{0}^{\pi} \sin x \,dx + \int_{\pi}^{2\pi} (-\sin x) \,dx \)
\( = [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{2\pi} \)
\( = (-\cos \pi - (-\cos 0)) + (\cos 2\pi - \cos \pi) \)
\( = (-(-1) - (-1)) + (1 - (-1)) \)
\( = (1 + 1) + (1 + 1) \)
\( = 2 + 2 \)
\( = 4 \) sq. units.
In simple words: We are finding the total space between the wavy sine curve and the straight x-axis between 0 and \( 2\pi \). Since the sine curve goes above and below the x-axis, we calculate the area of the top part and the bottom part separately. We make sure to count the bottom part's area as positive, then add them all up to get the total enclosed space.

🎯 Exam Tip: For functions that cross the x-axis, always split the integral into parts where the function is positive and negative, taking the absolute value of each part to get the total area.

 

Question 4. Find the area enclosed by curve \( y = 2\sqrt {x} \) and \( x = 0, x = 1 \).
Answer: The given curve is \( y = 2\sqrt{x} \). Squaring both sides, we get \( y^2 = 4x \), which is the equation of a parabola. This parabola is symmetric about the x-axis and opens to the right. We need to find the area enclosed by this curve, the y-axis (\( x=0 \)), and the line \( x=1 \). The problem statement implies we are looking for the total area. The shaded part in the figure represents the required area.
Required area \( = \int_{0}^{1} y \,dx \)
\( = \int_{0}^{1} 2\sqrt{x} \,dx \)
\( = 2 \int_{0}^{1} x^{1/2} \,dx \)
\( = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} \)
\( = 2 \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{1} \)
\( = \frac{4}{3} \left[ 1^{3/2} - 0^{3/2} \right] \)
\( = \frac{4}{3} [1 - 0] \)
\( = \frac{4}{3} \) sq. units.
If we consider the area between only positive coordinates (i.e., only the upper half of the parabola), then the calculation is as above.
The question as stated likely refers to the area of the entire enclosed region, considering both positive and negative y values. However, usually when a curve is defined as \( y = f(x) \), and the limits are along the x-axis, we integrate \( y \,dx \). For the complete shape \( y^2=4x \), symmetric around the x-axis, the enclosed area from \( x=0 \) to \( x=1 \) would be \( 2 \times \int_{0}^{1} 2\sqrt{x} \,dx \).
So, Required Area \( = 2 \times \frac{4}{3} = \frac{8}{3} \) sq. units.

🎯 Exam Tip: Always sketch the curve to understand the region. For functions like \( y=2\sqrt{x} \), if the problem doesn't specify 'above x-axis', consider the full symmetric region if the original equation \( y^2=4x \) implies it.

 

Question 6. Find the area enclosed by curve \( x^2 = 4ay \), x-axis and line \( x = 2 \).
Answer: The given curve is \( x^2 = 4ay \). This is a parabola that opens upwards and is symmetric about the y-axis. We need to find the area enclosed by this parabola, the x-axis, and the vertical line \( x = 2 \). Since the parabola is symmetric about the y-axis, the area is typically considered in the first quadrant for positive x values.
From \( x^2 = 4ay \), we get \( y = \frac{x^2}{4a} \).
The required area is from \( x = 0 \) to \( x = 2 \).
Required Area \( = \int_{0}^{2} y \,dx \)
\( = \int_{0}^{2} \frac{x^2}{4a} \,dx \)
\( = \frac{1}{4a} \int_{0}^{2} x^2 \,dx \)
\( = \frac{1}{4a} \left[ \frac{x^3}{3} \right]_{0}^{2} \)
\( = \frac{1}{4a} \left[ \frac{2^3}{3} - \frac{0^3}{3} \right] \)
\( = \frac{1}{4a} \left[ \frac{8}{3} - 0 \right] \)
\( = \frac{1}{4a} \times \frac{8}{3} \)
\( = \frac{8}{12a} \)
\( = \frac{2}{3a} \) sq. units.
In simple words: We are calculating the space between an upward-opening parabola, the horizontal x-axis, and a vertical line at \( x=2 \). We find the height of the parabola (\( y \)) at each point from \( x=0 \) to \( x=2 \) and add up all those tiny rectangles through integration to find the total area.

🎯 Exam Tip: When dealing with parabolas, correctly identifying the axis of symmetry and the direction it opens helps set up the integral. Always integrate with respect to the variable along the limits given.

 

Question 7. Find the area enclosed by ellipse \( \frac {{x}^{2}}{4} + \frac{{y}^{2}}{9} = 1 \) and lie above the x - axis
Answer: The equation of the ellipse is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). This means \( a^2 = 4 \) so \( a=2 \), and \( b^2 = 9 \) so \( b=3 \). The major axis is along the y-axis because \( b > a \). We need to find the area of the ellipse that lies above the x-axis. Since the ellipse is symmetric about both axes, the area above the x-axis is half of the total area of the ellipse. The total area of an ellipse is given by \( \pi ab \).
From the equation of the ellipse, we can express \( y \) in terms of \( x \):
\( \frac{y^2}{9} = 1 - \frac{x^2}{4} \)
\( y^2 = 9 \left( 1 - \frac{x^2}{4} \right) \)
\( y = 3\sqrt{1 - \frac{x^2}{4}} \)
\( y = \frac{3}{2}\sqrt{4 - x^2} \)
The x-intercepts of the ellipse are at \( y=0 \), so \( \frac{x^2}{4} = 1 \implies x^2 = 4 \implies x = \pm 2 \). Thus, we integrate from \( x = -2 \) to \( x = 2 \).
Area above x-axis \( = \int_{-2}^{2} y \,dx \)
\( = \int_{-2}^{2} \frac{3}{2}\sqrt{4 - x^2} \,dx \)
Because \( \sqrt{4-x^2} \) is an even function, we can simplify this to:
\( = 2 \times \frac{3}{2} \int_{0}^{2} \sqrt{4 - x^2} \,dx \)
\( = 3 \int_{0}^{2} \sqrt{2^2 - x^2} \,dx \)
Using the formula \( \int \sqrt{a^2 - x^2} \,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) \):
\( = 3 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} \)
\( = 3 \left[ \left( \frac{2}{2}\sqrt{4 - 2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right) \right] \)
\( = 3 \left[ (1\sqrt{0} + 2\sin^{-1}(1)) - (0 + 2\sin^{-1}(0)) \right] \)
\( = 3 \left[ (0 + 2 \times \frac{\pi}{2}) - (0 + 0) \right] \)
\( = 3 \left[ \pi \right] \)
\( = 3\pi \) sq. units.
Alternatively, using the formula \( \text{Area} = \pi ab \), the total area of this ellipse is \( \pi (2)(3) = 6\pi \). The area above the x-axis is half of this, which is \( \frac{1}{2} \times 6\pi = 3\pi \) sq. units.
In simple words: We want to find the area of the top half of an oval shape (an ellipse). We use integration to sum up the tiny vertical strips of area from one side of the ellipse to the other. Since the ellipse is symmetric, this is also exactly half of the total area of the ellipse.

🎯 Exam Tip: For ellipses, always check if the major axis is along X or Y. Remember that the area of an ellipse is \( \pi ab \), which can be a quick check or direct solution if the full area or a symmetric portion is requested.

 

Question 8. Find the total area of ellipse \( \frac {{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} = 1 \).
Answer: The equation of the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We need to find the total area of this ellipse. The ellipse is symmetric about both the x-axis and the y-axis. Therefore, we can find the area of one quadrant and then multiply it by 4 to get the total area. This is a common method for finding the area of symmetric shapes.
From the equation, we can express \( y \) in terms of \( x \):
\( \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \)
\( y^2 = b^2 \left( 1 - \frac{x^2}{a^2} \right) \)
\( y = \frac{b}{a}\sqrt{a^2 - x^2} \) (taking the positive root for the first quadrant)
The x-intercepts are \( \pm a \). We integrate from \( x=0 \) to \( x=a \) for the first quadrant.
Total Area \( = 4 \times \text{Area of first quadrant} \)
\( = 4 \int_{0}^{a} y \,dx \)
\( = 4 \int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \,dx \)
\( = \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2} \,dx \)
Using the formula \( \int \sqrt{a^2 - x^2} \,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) \):
\( = \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a} \)
\( = \frac{4b}{a} \left[ \left( \frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) \right) - \left( \frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{0}{a}\right) \right) \right] \)
\( = \frac{4b}{a} \left[ \left( 0 + \frac{a^2}{2}\sin^{-1}(1) \right) - (0 + 0) \right] \)
\( = \frac{4b}{a} \left[ \frac{a^2}{2} \times \frac{\pi}{2} \right] \)
\( = \frac{4b}{a} \times \frac{\pi a^2}{4} \)
\( = \pi ab \) sq. units.
The area is always positive.
In simple words: To find the total space inside an ellipse (an oval shape), we can use a special integration method. Because the ellipse is symmetrical, we calculate the area of just one quarter of it and then multiply that result by four. This gives us the simple formula \( \pi ab \) for the whole area.

🎯 Exam Tip: Remember the standard formula for the area of an ellipse, \( \pi ab \). When integrating, utilize the symmetry of the ellipse to simplify the calculation by finding the area of one quadrant and multiplying by four.

 

Question 9. Find the area enclosed by line \( \frac {x}{a} - \frac{y}{b} = 2 \) and co-ordinate axis.
Answer: The equation of the line is \( \frac{x}{a} - \frac{y}{b} = 2 \). We need to find the area enclosed by this line and the coordinate axes. First, let's find the intercepts of the line with the x and y axes.
To find the x-intercept, set \( y=0 \):
\( \frac{x}{a} - \frac{0}{b} = 2 \implies \frac{x}{a} = 2 \implies x = 2a \). So, the point is \( (2a, 0) \).
To find the y-intercept, set \( x=0 \):
\( \frac{0}{a} - \frac{y}{b} = 2 \implies -\frac{y}{b} = 2 \implies y = -2b \). So, the point is \( (0, -2b) \).
The line forms a right-angled triangle with the coordinate axes. The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
The base of the triangle along the x-axis is \( |2a| \).
The height of the triangle along the y-axis is \( |-2b| = |2b| \).
Area \( = \frac{1}{2} \times |2a| \times |2b| \)
\( = \frac{1}{2} \times 2a \times 2b \) (assuming \( a, b > 0 \))
\( = 2ab \) sq. units.
If we integrate to find the area: from \( \frac{x}{a} - \frac{y}{b} = 2 \), we get \( \frac{y}{b} = \frac{x}{a} - 2 \implies y = \frac{b}{a}x - 2b \).
The integral for area would be \( \int_{0}^{2a} y \,dx \). However, since \( y \) is negative for \( x \in [0, 2a] \), we must take the absolute value of the integral or integrate \( -y \,dx \).
Required Area \( = \int_{0}^{2a} \left| \frac{b}{a}x - 2b \right| \,dx \)
Since \( \frac{b}{a}x - 2b \) is negative in the interval \( (0, 2a) \), we write:
\( = \int_{0}^{2a} -\left( \frac{b}{a}x - 2b \right) \,dx \)
\( = \int_{0}^{2a} \left( 2b - \frac{b}{a}x \right) \,dx \)
\( = \left[ 2bx - \frac{b}{a}\frac{x^2}{2} \right]_{0}^{2a} \)
\( = \left( 2b(2a) - \frac{b}{a}\frac{(2a)^2}{2} \right) - (0 - 0) \)
\( = 4ab - \frac{b}{a}\frac{4a^2}{2} \)
\( = 4ab - 2ab \)
\( = 2ab \) sq. units.
The area is always positive.
In simple words: We are finding the space made by a straight line and the x and y axes. First, we find where the line cuts these axes. This creates a triangle. We can calculate the area of this triangle using its base and height. If we use integration, we must be careful because part of the line might be below the x-axis, so we take the positive value of the area.

🎯 Exam Tip: For lines forming regions with coordinate axes, it's often simpler to find the intercepts and use the geometric formula for a triangle's area. If using integration, pay attention to whether the function is above or below the x-axis.

 

Question 10. Find the area enclosed by line \( x + 2y = 8, x = 2, x = 4 \) and x-axis.
Answer: The given line is \( x + 2y = 8 \). We need to find the area enclosed by this line, the vertical lines \( x=2 \) and \( x=4 \), and the x-axis. This region will be a trapezoid under the line. First, express \( y \) in terms of \( x \):
\( 2y = 8 - x \)
\( y = 4 - \frac{x}{2} \)
The required area is the integral of \( y \) with respect to \( x \) from \( x=2 \) to \( x=4 \). In this interval, \( y \) is positive.
Required Area \( = \int_{2}^{4} y \,dx \)
\( = \int_{2}^{4} \left( 4 - \frac{x}{2} \right) \,dx \)
\( = \left[ 4x - \frac{x^2}{4} \right]_{2}^{4} \)
\( = \left( 4(4) - \frac{4^2}{4} \right) - \left( 4(2) - \frac{2^2}{4} \right) \)
\( = \left( 16 - \frac{16}{4} \right) - \left( 8 - \frac{4}{4} \right) \)
\( = (16 - 4) - (8 - 1) \)
\( = 12 - 7 \)
\( = 5 \) sq. units.
In simple words: We want to find the space under a straight line and above the x-axis, between two specific vertical lines at \( x=2 \) and \( x=4 \). We solve the line equation for \( y \) and then use integration to add up all the tiny vertical strips of area within the given x-range. This shape forms a trapezoid.

🎯 Exam Tip: When a line is given along with vertical limits and the x-axis, the enclosed region is usually a trapezoid or a triangle. Always solve for \( y \) in terms of \( x \) to set up the integral correctly.

 

Question 11. Find the area enclosed by the curve \( y = x^2 \), ordinates \( x = 1, x = 2 \) and x-axis.
Answer: The given curve is \( y = x^2 \). This is a parabola that opens upwards, with its vertex at the origin \( (0,0) \). We need to find the area enclosed by this parabola, the vertical lines \( x=1 \) and \( x=2 \), and the x-axis. Since \( y=x^2 \) is always positive for non-zero \( x \), the curve is above the x-axis in the interval \( [1, 2] \).
The required area is given by the integral of \( y \) with respect to \( x \) from \( x=1 \) to \( x=2 \).
Required Area \( = \int_{1}^{2} y \,dx \)
\( = \int_{1}^{2} x^2 \,dx \)
\( = \left[ \frac{x^3}{3} \right]_{1}^{2} \)
\( = \frac{2^3}{3} - \frac{1^3}{3} \)
\( = \frac{8}{3} - \frac{1}{3} \)
\( = \frac{7}{3} \) sq. units.
In simple words: We are looking for the space underneath the curve of a parabola (\( y=x^2 \)), above the x-axis, and between the vertical lines at \( x=1 \) and \( x=2 \). We find this area by using integration, summing up all the tiny slices of area under the curve in that specific range.

🎯 Exam Tip: Always make sure the curve's function \( y=f(x) \) is correctly identified. When finding the area between two vertical lines and the x-axis, simply integrate \( f(x) \) between those limits if \( f(x) \ge 0 \) in that interval.

 

Question 12. Find the area of the region in the first quadrant enclosed by \( y = 4x^2, x = 0, y = 1 \) and \( y = 4 \).
Answer: The given curve is \( y = 4x^2 \). This is a parabola opening upwards. We are looking for the area in the first quadrant, enclosed by this parabola and the horizontal lines \( y=1 \) and \( y=4 \), and the y-axis (\( x=0 \)).
Since the area is bounded by horizontal lines, it's easier to integrate with respect to \( y \). First, express \( x \) in terms of \( y \):
\( x^2 = \frac{y}{4} \)
\( x = \sqrt{\frac{y}{4}} \)
\( x = \frac{\sqrt{y}}{2} \)
Since we are in the first quadrant, we take the positive square root. The limits of integration for \( y \) are from \( y=1 \) to \( y=4 \).
Required Area \( = \int_{1}^{4} x \,dy \)
\( = \int_{1}^{4} \frac{\sqrt{y}}{2} \,dy \)
\( = \frac{1}{2} \int_{1}^{4} y^{1/2} \,dy \)
\( = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4} \)
\( = \frac{1}{2} \times \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4} \)
\( = \frac{1}{3} \left[ 4^{3/2} - 1^{3/2} \right] \)
\( = \frac{1}{3} \left[ (\sqrt{4})^3 - (\sqrt{1})^3 \right] \)
\( = \frac{1}{3} \left[ 2^3 - 1^3 \right] \)
\( = \frac{1}{3} [8 - 1] \)
\( = \frac{7}{3} \) sq. units.
In simple words: We need to find the space in the top-right quarter of the graph, bordered by an upward-curving parabola, the y-axis, and two horizontal lines at \( y=1 \) and \( y=4 \). To do this, we rewrite the parabola's equation to find \( x \) in terms of \( y \). Then, we add up many tiny horizontal strips of area by integrating along the y-axis from \( y=1 \) to \( y=4 \).

🎯 Exam Tip: When the region is bounded by horizontal lines, it's usually more efficient to integrate with respect to \( y \). Remember to solve the curve's equation for \( x \) in terms of \( y \), and adjust the integration limits accordingly.

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