GSEB Class 12 Maths Solutions Chapter 8 Application of Integrals Exercise 8.1

Question 1. Find the area of the region bounded by the curve \( y^2 = x \) and the lines \( x = 1, x = 4 \) and the x-axis.
Answer: The curve \( y^2 = x \) is a parabola that has its vertex at the origin. The x-axis serves as the line of symmetry, which also forms the parabola's axis. The region is bounded by the parabola and the lines \( x = 1 \) and \( x = 4 \). We calculate the area by integrating with respect to \( x \).
The area LMQP is given by:
\( \text{Area LMQP} = \int_{1}^{4} y\,dx \)
For the part above the x-axis, \( y = \sqrt{x} \).
\( = \int_{1}^{4} \sqrt{x}\,dx \)
\( = \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{1}^{4} \)
\( = \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{1}^{4} \)
\( = \frac{2}{3} \left[ 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] \)
\( = \frac{2}{3} [ (2^2)^{\frac{3}{2}} - 1 ] \)
\( = \frac{2}{3} [ 2^3 - 1 ] \)
\( = \frac{2}{3} [8-1] \)
\( = \frac{2}{3} \times 7 \)
\( = \frac{14}{3} \) sq. units.
In simple words: We need to find the area under the curve \( y^2=x \) between vertical lines \( x=1 \) and \( x=4 \). First, we write \( y \) as \( \sqrt{x} \). Then, we use integration to calculate the space, which turns out to be \( \frac{14}{3} \) square units.

Exam Tip: When finding the area bounded by a parabola and vertical lines, remember to sketch the graph to correctly identify the upper curve and the limits of integration. For \( y^2 = x \), consider \( y = \sqrt{x} \) for the area above the x-axis.

Question 2. Find the area of the region bounded by \( y^2 = 9x, x = 2, x = 4 \) and the x-axis in the first quadrant.
Answer: The provided curve, \( y^2 = 9x \), forms a parabola with its vertex at (0, 0) and its axis positioned along the x-axis. It shows symmetry about the x-axis because it includes only even powers of \( y \). The lines \( x = 2 \) and \( x = 4 \) are straight lines that run parallel to the y-axis, located at positive distances of 2 and 4 units from it, respectively. We need to find the area in the first quadrant.
For the first quadrant, \( y = \sqrt{9x} = 3\sqrt{x} \).
The area is given by:
\( \text{Area} = \int_{2}^{4} 3\sqrt{x}\,dx \)
\( = 3 \int_{2}^{4} x^{\frac{1}{2}}\,dx \)
\( = 3 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{2}^{4} \)
\( = 3 \cdot \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{2}^{4} \)
\( = 2 \left[ 4^{\frac{3}{2}} - 2^{\frac{3}{2}} \right] \)
\( = 2 [ (2^2)^{\frac{3}{2}} - 2\sqrt{2} ] \)
\( = 2 [ 2^3 - 2\sqrt{2} ] \)
\( = 2 [ 8 - 2\sqrt{2} ] \)
\( = 16 - 4\sqrt{2} \) sq. units.
In simple words: We want to find the area in the first quarter of the graph, bordered by the parabola \( y^2=9x \), the x-axis, and two vertical lines \( x=2 \) and \( x=4 \). We calculate this by integrating \( 3\sqrt{x} \) from \( x=2 \) to \( x=4 \). The final area is \( 16 - 4\sqrt{2} \) square units.

Exam Tip: When dealing with areas in a specific quadrant, ensure you use the correct positive or negative root for \( y \) (or \( x \)) based on the quadrant. For \( y^2 = 9x \) in the first quadrant, use \( y = 3\sqrt{x} \).

Question 3. Find the area of the region bounded by \( x^2 = 4y, y = 2, y = 4 \) and the y-axis in the first quadrant.
Answer: The given curve \( x^2 = 4y \) represents a parabola with its vertex located at (0, 0). Moreover, because it contains only even powers of \( x \), it is symmetrical around the y-axis. The lines \( y = 2 \) and \( y = 4 \) are straight lines that are parallel to the x-axis, situated at positive distances of 2 and 4 units from it, respectively. We need to find the area in the first quadrant.
For the first quadrant, \( x = \sqrt{4y} = 2\sqrt{y} \).
The area is given by:
\( \text{Area} = \int_{2}^{4} x\,dy \)
\( = \int_{2}^{4} 2\sqrt{y}\,dy \)
\( = 2 \int_{2}^{4} y^{\frac{1}{2}}\,dy \)
\( = 2 \left[ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right]_{2}^{4} \)
\( = 2 \cdot \frac{2}{3} \left[ y^{\frac{3}{2}} \right]_{2}^{4} \)
\( = \frac{4}{3} \left[ 4^{\frac{3}{2}} - 2^{\frac{3}{2}} \right] \)
\( = \frac{4}{3} [ (2^2)^{\frac{3}{2}} - 2\sqrt{2} ] \)
\( = \frac{4}{3} [ 2^3 - 2\sqrt{2} ] \)
\( = \frac{4}{3} [ 8 - 2\sqrt{2} ] \)
\( = \frac{32 - 8\sqrt{2}}{3} \) sq. units.
In simple words: We are finding the area in the first quarter of the graph, bounded by the parabola \( x^2=4y \), the y-axis, and horizontal lines \( y=2 \) and \( y=4 \). This means we integrate \( 2\sqrt{y} \) with respect to \( y \) from 2 to 4. The final calculated area is \( \frac{32 - 8\sqrt{2}}{3} \) square units.

Exam Tip: When the region is bounded by the y-axis and horizontal lines, it's generally easier to integrate with respect to \( y \) (i.e., use \( \int x\,dy \)). Always ensure the function \( x=f(y) \) is correctly derived for the given quadrant.

Question 4. Find the area of the region bounded by the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).
Answer: The ellipse's equation is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \). This ellipse is symmetrical about both axes because it only has even powers of \( y \) and \( x \). The semi-major axis is \( a=4 \) (along the x-axis) and the semi-minor axis is \( b=3 \) (along the y-axis). Due to symmetry, the total area of the ellipse is four times the area in the first quadrant.
From the equation, \( \frac{y^2}{9} = 1 - \frac{x^2}{16} \)
\( y^2 = 9 \left( 1 - \frac{x^2}{16} \right) \)
\( y^2 = \frac{9}{16} (16 - x^2) \)
\( y = \pm \frac{3}{4} \sqrt{16 - x^2} \)
For the first quadrant, \( y = \frac{3}{4} \sqrt{16 - x^2} \).
Area of ellipse \( = 4 \times \text{(Area in first quadrant)} \)
\( = 4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2}\,dx \)
\( = 3 \int_{0}^{4} \sqrt{16 - x^2}\,dx \)
Let \( x = 4 \sin \theta \), so \( dx = 4 \cos \theta\,d\theta \).
When \( x = 0 \), \( 4 \sin \theta = 0 \implies \theta = 0 \).
When \( x = 4 \), \( 4 \sin \theta = 4 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
The integral becomes:
\( = 3 \int_{0}^{\frac{\pi}{2}} \sqrt{16 - (4 \sin \theta)^2} (4 \cos \theta)\,d\theta \)
\( = 3 \int_{0}^{\frac{\pi}{2}} \sqrt{16 - 16 \sin^2 \theta} (4 \cos \theta)\,d\theta \)
\( = 3 \int_{0}^{\frac{\pi}{2}} \sqrt{16(1 - \sin^2 \theta)} (4 \cos \theta)\,d\theta \)
\( = 3 \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2 \theta} (4 \cos \theta)\,d\theta \)
\( = 3 \int_{0}^{\frac{\pi}{2}} (4 \cos \theta) (4 \cos \theta)\,d\theta \)
\( = 3 \int_{0}^{\frac{\pi}{2}} 16 \cos^2 \theta\,d\theta \)
\( = 48 \int_{0}^{\frac{\pi}{2}} \cos^2 \theta\,d\theta \)
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):
\( = 48 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2}\,d\theta \)
\( = 24 \int_{0}^{\frac{\pi}{2}} (1 + \cos 2\theta)\,d\theta \)
\( = 24 \left[ \theta + \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{2}} \)
\( = 24 \left[ \left( \frac{\pi}{2} + \frac{\sin (2 \cdot \frac{\pi}{2})}{2} \right) - \left( 0 + \frac{\sin (2 \cdot 0)}{2} \right) \right] \)
\( = 24 \left[ \left( \frac{\pi}{2} + \frac{\sin \pi}{2} \right) - \left( 0 + \frac{\sin 0}{2} \right) \right] \)
\( = 24 \left[ \left( \frac{\pi}{2} + 0 \right) - (0 + 0) \right] \)
\( = 24 \times \frac{\pi}{2} \)
\( = 12\pi \) sq. units.
In simple words: To find the area of the ellipse, we multiply the area of its first quadrant by four due to its symmetry. We express \( y \) in terms of \( x \) and use a trigonometric substitution to simplify the integration. After evaluating the integral, the total area of the ellipse is \( 12\pi \) square units.

Exam Tip: For ellipses, using the formula \( \text{Area} = \pi ab \) can save time if allowed, where \( a \) and \( b \) are the semi-major and semi-minor axes. If integration is required, remember to use symmetry and the trigonometric substitution \( x = a \sin \theta \).

Question 6. Find the area of the region in the first quadrant enclosed by the x-axis, the line \( x = \sqrt{3}y \) and the circle \( x^2 + y^2 = 4 \).
Answer: Let's consider these two equations. First, \( x^2 + y^2 = 4 \) is equation (1). And second, \( x = \sqrt{3}y \), which can also be written as \( y = \frac{1}{\sqrt{3}}x \), is equation (2). The equation \( x^2 + y^2 = 4 \) shows a circle that has its center at O (0, 0) and a radius of 2. The line \( y = \frac{1}{\sqrt{3}}x \) is a straight line that passes through (0, 0) and crosses the circle at point B \( (\sqrt{3}, 1) \). To find the intersection point: Substitute \( x = \sqrt{3}y \) into \( x^2+y^2=4 \):
\( (\sqrt{3}y)^2 + y^2 = 4 \)
\( 3y^2 + y^2 = 4 \)
\( 4y^2 = 4 \)
\( y^2 = 1 \)
\( y = 1 \) (since in first quadrant)
When \( y=1 \), \( x = \sqrt{3}(1) = \sqrt{3} \). So the intersection point is \( (\sqrt{3}, 1) \).
The required area is the shaded region in the figure, which can be divided into two parts: the area under the line from \( x=0 \) to \( x=\sqrt{3} \), and the area under the circle from \( x=\sqrt{3} \) to \( x=2 \).
\( \text{Required Area} = \text{Area OBL} + \text{Area LBA} \)
\( = \int_{0}^{\sqrt{3}} \frac{1}{\sqrt{3}}x\,dx + \int_{\sqrt{3}}^{2} \sqrt{4-x^2}\,dx \)
For the first integral:
\( \int_{0}^{\sqrt{3}} \frac{1}{\sqrt{3}}x\,dx = \frac{1}{\sqrt{3}} \left[ \frac{x^2}{2} \right]_{0}^{\sqrt{3}} = \frac{1}{\sqrt{3}} \left[ \frac{(\sqrt{3})^2}{2} - 0 \right] = \frac{1}{\sqrt{3}} \cdot \frac{3}{2} = \frac{\sqrt{3}}{2} \).
For the second integral, using the formula \( \int \sqrt{a^2-x^2}\,dx = \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \):
\( \int_{\sqrt{3}}^{2} \sqrt{4-x^2}\,dx = \left[ \frac{x\sqrt{4-x^2}}{2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{\sqrt{3}}^{2} \)
\( = \left( \frac{2\sqrt{4-2^2}}{2} + 2\sin^{-1}\frac{2}{2} \right) - \left( \frac{\sqrt{3}\sqrt{4-(\sqrt{3})^2}}{2} + 2\sin^{-1}\frac{\sqrt{3}}{2} \right) \)
\( = \left( 0 + 2\sin^{-1}1 \right) - \left( \frac{\sqrt{3}\sqrt{4-3}}{2} + 2\sin^{-1}\frac{\sqrt{3}}{2} \right) \)
\( = \left( 2 \cdot \frac{\pi}{2} \right) - \left( \frac{\sqrt{3}}{2} + 2 \cdot \frac{\pi}{3} \right) \)
\( = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} \)
Now, sum the two parts:
\( \text{Total Area} = \frac{\sqrt{3}}{2} + \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} \)
\( = \pi - \frac{2\pi}{3} \)
\( = \frac{3\pi - 2\pi}{3} = \frac{\pi}{3} \) sq. units.
In simple words: To find the area, we split the region into two parts: one under the straight line and one under the curve of the circle. We integrate each part separately, from \( x=0 \) to \( x=\sqrt{3} \) for the line, and from \( x=\sqrt{3} \) to \( x=2 \) for the circle. Adding these two results gives us the total area, which is \( \frac{\pi}{3} \) square units.

Exam Tip: For areas bounded by multiple curves, always sketch the region first to identify intersection points and determine if the area needs to be split into multiple integrals. Choose between \( \int y\,dx \) or \( \int x\,dy \) based on which simplifies the integration limits and functions.

Question 7. Find the area of the smaller part of the circle \( x^2 + y^2 = a^2 \) cut off by the line \( x = \frac{a}{\sqrt{2}} \).
Answer: The equations for the given curves are these: First, \( x^2 + y^2 = a^2 \) is equation (1). And second, \( x = \frac{a}{\sqrt{2}} \) is equation (2). Evidently, (1) depicts a circle with radius \( a \) and center at (0,0), and (2) is a straight line equation running parallel to the y-axis, located at a distance of \( \frac{a}{\sqrt{2}} \) units to the y-axis's right. The line \( x = \frac{a}{\sqrt{2}} \) cuts the circle, and we need to find the area of the smaller segment created. Due to symmetry, we can find the area in the first quadrant and multiply by two.
By solving (1) and (2) together, we get the intersection points. Substitute \( x = \frac{a}{\sqrt{2}} \) into \( x^2 + y^2 = a^2 \):
\( \left( \frac{a}{\sqrt{2}} \right)^2 + y^2 = a^2 \)
\( \frac{a^2}{2} + y^2 = a^2 \)
\( y^2 = a^2 - \frac{a^2}{2} \)
\( y^2 = \frac{a^2}{2} \)
\( y = \pm \frac{a}{\sqrt{2}} \).
So the intersection points are \( \left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right) \) and \( \left( \frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}} \right) \).
The smaller area formed by these two curves is the shaded part that you can see in the diagram. We integrate \( y\,dx \) from \( x = \frac{a}{\sqrt{2}} \) to \( x = a \).
From \( x^2 + y^2 = a^2 \), we have \( y = \sqrt{a^2 - x^2} \).
\( \text{Required Area} = 2 \int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^2 - x^2}\,dx \)
Using the formula \( \int \sqrt{a^2-x^2}\,dx = \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \):
\( = 2 \left[ \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \right]_{\frac{a}{\sqrt{2}}}^{a} \)
\( = 2 \left[ \left( \frac{a\sqrt{a^2-a^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{a}{a} \right) - \left( \frac{\frac{a}{\sqrt{2}}\sqrt{a^2-\left(\frac{a}{\sqrt{2}}\right)^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{\frac{a}{\sqrt{2}}}{a} \right) \right] \)
\( = 2 \left[ \left( 0 + \frac{a^2}{2}\sin^{-1}1 \right) - \left( \frac{\frac{a}{\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}}}{2} + \frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt{2}} \right) \right] \)
\( = 2 \left[ \frac{a^2}{2} \cdot \frac{\pi}{2} - \left( \frac{\frac{a}{\sqrt{2}}\sqrt{\frac{a^2}{2}}}{2} + \frac{a^2}{2} \cdot \frac{\pi}{4} \right) \right] \)
\( = 2 \left[ \frac{\pi a^2}{4} - \left( \frac{\frac{a}{\sqrt{2}} \cdot \frac{a}{\sqrt{2}}}{2} + \frac{\pi a^2}{8} \right) \right] \)
\( = 2 \left[ \frac{\pi a^2}{4} - \left( \frac{a^2/2}{2} + \frac{\pi a^2}{8} \right) \right] \)
\( = 2 \left[ \frac{\pi a^2}{4} - \left( \frac{a^2}{4} + \frac{\pi a^2}{8} \right) \right] \)
\( = 2 \left[ \frac{\pi a^2}{4} - \frac{a^2}{4} - \frac{\pi a^2}{8} \right] \)
\( = \frac{\pi a^2}{2} - \frac{a^2}{2} - \frac{\pi a^2}{4} \)
\( = \frac{2\pi a^2 - 2a^2 - \pi a^2}{4} \)
\( = \frac{\pi a^2 - 2a^2}{4} = \frac{a^2}{4}(\pi - 2) \) sq. units.
In simple words: We're finding the area of a smaller circular piece cut by a vertical line. Because the circle is symmetrical, we calculate the area in one half (from \( x=\frac{a}{\sqrt{2}} \) to \( x=a \)) and then double it. The formula for integrating \( \sqrt{a^2-x^2} \) is used, and after applying the limits, the total area is \( \frac{a^2}{4}(\pi-2) \) square units.

Exam Tip: For circular segments, recognize that the integral \( \int \sqrt{a^2-x^2}\,dx \) is a standard form. Remember to use symmetry to simplify calculations and choose the correct limits of integration based on the segment being calculated.

Question 8. The area between \( x = y^2 \) and \( x = 4 \) is divided into two equal parts by the line \( x = a \). Find the value of \( a \).
Answer: The curve \( x = y^2 \) forms a parabola, which is displayed in the figure. Its vertex is at point O, and its axis lies along the x-axis. QR represents the ordinate that goes along \( x = 4 \). The total area of the region bounded by \( x = y^2 \) and \( x = 4 \) is \( A_1 \). This area is symmetrical about the x-axis, so we can integrate \( y\,dx \) from \( x=0 \) to \( x=4 \) and multiply by 2.
From \( x = y^2 \), we get \( y = \sqrt{x} \) (for the upper part).
\( A_1 = 2 \int_{0}^{4} \sqrt{x}\,dx \)
\( = 2 \int_{0}^{4} x^{\frac{1}{2}}\,dx \)
\( = 2 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{4} \)
\( = 2 \cdot \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{0}^{4} \)
\( = \frac{4}{3} [ 4^{\frac{3}{2}} - 0^{\frac{3}{2}} ] \)
\( = \frac{4}{3} [ 8 - 0 ] \)
\( = \frac{32}{3} \) sq. units.
The line \( x = a \) divides this area into two equal parts. So, the area from \( x=0 \) to \( x=a \) must be half of \( A_1 \). Let \( A_2 \) be the area from \( x=0 \) to \( x=a \).
\( A_2 = 2 \int_{0}^{a} \sqrt{x}\,dx \)
\( = 2 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{a} \)
\( = \frac{4}{3} [ a^{\frac{3}{2}} - 0 ] \)
\( = \frac{4}{3} a^{\frac{3}{2}} \)
Given that \( A_2 = \frac{A_1}{2} \):
\( \frac{4}{3} a^{\frac{3}{2}} = \frac{1}{2} \cdot \frac{32}{3} \)
\( \frac{4}{3} a^{\frac{3}{2}} = \frac{16}{3} \)
\( 4 a^{\frac{3}{2}} = 16 \)
\( a^{\frac{3}{2}} = 4 \)
To find \( a \), raise both sides to the power of \( \frac{2}{3} \):
\( a = 4^{\frac{2}{3}} \) units.
In simple words: We find the total area under the parabola \( x=y^2 \) up to \( x=4 \). Then, we set up a new integral for the area from \( x=0 \) to \( x=a \) and make it equal to half the total area. Solving this equation helps us determine the value of \( a \).

Exam Tip: When a region is divided into equal parts, calculate the total area first. Then, set up an integral for one of the divided parts (with an unknown limit like \( a \)) and equate it to the appropriate fraction of the total area. This approach helps in finding the unknown parameter.

Question 9. Find the area of the region bounded by the parabola \( y = x^2 \) and lines \( y = |x| \).
Answer: It is clear that \( x^2 = y \) depicts a parabola, whose vertex is at (0, 0), with the positive y-axis as its axis, and it opens in an upward direction. The equation \( y = |x| \) means two lines: \( y = x \) (for \( x \ge 0 \)) and \( y = -x \) (for \( x < 0 \)). Both of these lines pass through the origin and create angles of \( 45^\circ \) and \( 135^\circ \) with the positive x-axis, respectively. The required region is the shaded region shown in the figure. Both curves are symmetrical around the y-axis, so we can find the area in the first quadrant and multiply by 2.
First, find the intersection points of \( y = x^2 \) and \( y = x \) (in the first quadrant):
\( x^2 = x \)
\( x^2 - x = 0 \)
\( x(x-1) = 0 \)
So, \( x = 0 \) or \( x = 1 \).
The intersection points are \( (0,0) \) and \( (1,1) \).
The area in the first quadrant is the integral of the upper curve minus the lower curve, from \( x=0 \) to \( x=1 \). Here, \( y=|x| \) (which is \( y=x \)) is the upper curve and \( y=x^2 \) is the lower curve.
\( \text{Area in first quadrant} = \int_{0}^{1} (x - x^2)\,dx \)
\( = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \)
\( = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \)
\( = \frac{1}{2} - \frac{1}{3} \)
\( = \frac{3-2}{6} = \frac{1}{6} \) sq. units.
Due to symmetry, the total required area is twice the area in the first quadrant:
\( \text{Total Area} = 2 \times \frac{1}{6} = \frac{1}{3} \) sq. units.
In simple words: We need to find the area between a parabola and two lines that form a 'V' shape. Since the shape is symmetrical, we just find the area in the first quarter of the graph by subtracting the parabola's function from the line's function and then doubling the result to get the full area. This gives us \( \frac{1}{3} \) square units.

Exam Tip: For problems involving absolute value functions, always split the integral into parts based on the definition of the absolute value (e.g., \( |x| = x \) for \( x \ge 0 \) and \( |x| = -x \) for \( x < 0 \)). Leverage symmetry to simplify calculations if the region is symmetrical.

Question 10. Find the area bounded by the curve \( x^2 = 4y \) and the straight line \( x = 4y - 2 \).
Answer: The given curve is \( x^2 = 4y \) (1), which is an upward-opening parabola with its vertex at (0, 0) and symmetry about the y-axis. The line's equation is \( x = 4y - 2 \) (2). By solving (1) and (2) at the same time, we get their intersection points. Substitute \( 4y = x^2 \) from (1) into (2):
\( x = x^2 - 2 \)
\( x^2 - x - 2 = 0 \)
\( (x-2)(x+1) = 0 \)
This gives \( x = 2 \) or \( x = -1 \).
Using equation (2), when \( y = \frac{1}{4} \), the value of \( x \) is \( 1 - 2 = -1 \). When \( y = 1 \), the value of \( x \) becomes \( 4 - 2 = 2 \). So, the intersection points are \( (-1, \frac{1}{4}) \) and \( (2, 1) \). The area of the region bounded by these curves is found by integrating the difference between the upper function (the line) and the lower function (the parabola) with respect to \( x \), from \( x=-1 \) to \( x=2 \).
From (1), \( y_{parabola} = \frac{x^2}{4} \).
From (2), \( y_{line} = \frac{x+2}{4} \).
\( \text{Required Area} = \int_{-1}^{2} (y_{line} - y_{parabola})\,dx \)
\( = \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^2}{4} \right)\,dx \)
\( = \frac{1}{4} \int_{-1}^{2} (x+2 - x^2)\,dx \)
\( = \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} \)
\( = \frac{1}{4} \left[ \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right) \right] \)
\( = \frac{1}{4} \left[ \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \right] \)
\( = \frac{1}{4} \left[ \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{3-12+2}{6} \right) \right] \)
\( = \frac{1}{4} \left[ \left( 6 - \frac{8}{3} \right) - \left( -\frac{7}{6} \right) \right] \)
\( = \frac{1}{4} \left[ \left( \frac{18-8}{3} \right) + \frac{7}{6} \right] \)
\( = \frac{1}{4} \left[ \frac{10}{3} + \frac{7}{6} \right] \)
\( = \frac{1}{4} \left[ \frac{20+7}{6} \right] \)
\( = \frac{1}{4} \left[ \frac{27}{6} \right] \)
\( = \frac{1}{4} \times \frac{9}{2} = \frac{9}{8} \) sq. units.
In simple words: To find the area, we first determine where the line and parabola cross each other. Then, we integrate the difference between the line's equation and the parabola's equation, from the leftmost crossing point to the rightmost. This calculation gives us the area of \( \frac{9}{8} \) square units.

Exam Tip: Always sketch the curves to correctly identify which function is above the other. This prevents errors in setting up the difference \( y_{upper} - y_{lower} \). Also, accurately find all intersection points to establish the correct limits of integration.

Question 11. Find the area of the region bounded by the curve \( y^2 = 4x \) and the line \( x = 3 \).
Answer: The curve \( y^2 = 4x \) is a parabola, as depicted in the diagram. This parabola opens to the right, and its axis is the x-axis. The area enclosed by the curve \( y^2 = 4x \) and the line \( x = 3 \) is denoted as A. Since the region is symmetrical about the x-axis, we can calculate the area of the upper half (where \( y = \sqrt{4x} = 2\sqrt{x} \)) from \( x=0 \) to \( x=3 \) and then double it to get the total area.
\( \text{Area} = 2 \int_{0}^{3} y\,dx \)
\( = 2 \int_{0}^{3} 2\sqrt{x}\,dx \)
\( = 4 \int_{0}^{3} x^{\frac{1}{2}}\,dx \)
\( = 4 \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{3} \)
\( = 4 \cdot \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{0}^{3} \)
\( = \frac{8}{3} [ 3^{\frac{3}{2}} - 0^{\frac{3}{2}} ] \)
\( = \frac{8}{3} [ 3\sqrt{3} ] \)
\( = 8\sqrt{3} \) sq. units.
In simple words: To find the area, we use the fact that the region is symmetrical. We integrate the top half of the parabola \( y=2\sqrt{x} \) from \( x=0 \) to \( x=3 \) and then double the result. This calculation shows the total area is \( 8\sqrt{3} \) square units.

Exam Tip: For parabolas opening along the x-axis (like \( y^2=kx \)), when bounded by a vertical line \( x=c \), it's often easiest to integrate with respect to \( x \). Remember to use \( y = \pm \sqrt{kx} \) and account for symmetry by multiplying the integral for the upper half by two.

Choose the correct answers in the following questions 12 and 13:

Question 12. Area lying in the first quadrant and bounded by the circle \( x^2 + y^2 = 4 \) and the lines \( x = 0 \) and \( x = 2 \) is
(a) \( \pi \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{4} \)
Answer: (a) \( \pi \)
We draw a circle with a radius of 2 units. The region is in the first quadrant, bounded by the circle, the y-axis (where \( x = 0 \)), and the line \( x = 2 \). This specific region is precisely a quarter of the entire circle. The area A of the needed region is found by integrating \( y\,dx \) from 0 to 2. The circle's equation is \( x^2 + y^2 = 4 \), so \( y = \sqrt{4 - x^2} \).
\( \text{Area} = \int_{0}^{2} \sqrt{4 - x^2}\,dx \)
Using the formula \( \int \sqrt{a^2-x^2}\,dx = \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \), with \( a=2 \):
\( = \left[ \frac{x\sqrt{4-x^2}}{2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2} \)
\( = \left( \frac{2\sqrt{4-2^2}}{2} + 2\sin^{-1}\frac{2}{2} \right) - \left( \frac{0\sqrt{4-0^2}}{2} + 2\sin^{-1}\frac{0}{2} \right) \)
\( = \left( 0 + 2\sin^{-1}1 \right) - \left( 0 + 2\sin^{-1}0 \right) \)
\( = \left( 2 \cdot \frac{\pi}{2} \right) - (0) \)
\( = \pi \) sq. units.
This corresponds to option (a).
In simple words: The region described is a quarter circle of radius 2. We calculate its area using integration from \( x=0 \) to \( x=2 \). The result is \( \pi \), which matches option (a).

Exam Tip: For regions defined by a circle in a specific quadrant and bounded by axes, it often forms a simple fraction of the total circle's area. Recognize that \( \int_0^a \sqrt{a^2-x^2}\,dx \) represents the area of a quarter circle with radius \( a \), which is \( \frac{\pi a^2}{4} \). In this case, \( \frac{\pi (2^2)}{4} = \pi \).

Question 13. Area of the region bounded by the curve \( y^2 = 4x \), y-axis and the line \( y = 3 \) is
(a) \( \pi \)
(b) \( \frac{9}{4} \)
(c) \( \frac{9}{3} \)
(d) \( \frac{9}{2} \)
Answer: (b) \( \frac{9}{4} \)
The curve \( y^2 = 4x \) forms a parabola, and the line \( y = 3 \) is a horizontal line. The area of the region bounded by the parabola \( y^2 = 4x \), the y-axis, and the line \( y = 3 \) is calculated as follows. Since the region is bounded by the y-axis and a horizontal line, it's easier to integrate with respect to \( y \).
From \( y^2 = 4x \), we get \( x = \frac{y^2}{4} \).
The limits of integration for \( y \) are from 0 to 3.
\( \text{Area} = \int_{0}^{3} x\,dy \)
\( = \int_{0}^{3} \frac{y^2}{4}\,dy \)
\( = \frac{1}{4} \int_{0}^{3} y^2\,dy \)
\( = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3} \)
\( = \frac{1}{4} \left[ \frac{3^3}{3} - \frac{0^3}{3} \right] \)
\( = \frac{1}{4} \left[ \frac{27}{3} - 0 \right] \)
\( = \frac{1}{4} \times 9 \)
\( = \frac{9}{4} \) sq. units.
This corresponds to option (b).
In simple words: We're finding the area bordered by a parabola, the y-axis, and a horizontal line. We calculate this area by integrating the parabola's equation (rearranged for \( x \) in terms of \( y \)) from \( y=0 \) to \( y=3 \). The final area is \( \frac{9}{4} \) square units, which matches option (b).

Exam Tip: When the region is bounded by the y-axis (or a vertical line) and a horizontal line, consider integrating with respect to \( y \) by expressing \( x \) as a function of \( y \) (i.e., \( \int x\,dy \)). This simplifies the setup significantly.