CBSE Class 12 Mathematics Applications of Integration Assignment Set 04

(i) Area under Simple Curves

LEVEL I

Question 1. Sketch the region of the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) and find its area, using integration.
Answer: The given equation of the ellipse is \(\frac{x^2}{5^2} + \frac{y^2}{4^2} = 1\).
Here, the semi-major axis \(a = 5\) and semi-minor axis \(b = 4\). The ellipse is symmetric about both axes.
The area of the ellipse is given by 4 times the area of the region in the first quadrant.
In the first quadrant, \(y = \frac{4}{5}\sqrt{25 - x^2}\) where \(x\) varies from 0 to 5.
Area \(= 4 \int_{0}^{5} y \, dx = 4 \int_{0}^{5} \frac{4}{5}\sqrt{5^2 - x^2} \, dx\)
\(= \frac{16}{5} \left[ \frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{5}\)
\(= \frac{16}{5} \left[ \left(0 + \frac{25}{2}\sin^{-1}(1)\right) - 0 \right] = \frac{16}{5} \times \frac{25}{2} \times \frac{\pi}{2} = 20\pi\) sq. units.
In simple words: We find the area of one-fourth of the ellipse in the first quadrant using integration and multiply it by 4.

Exam Tip: Always exploit symmetry about the coordinate axes to simplify definite integrals by evaluating only for the first quadrant and multiplying by 4.

Question 2. Sketch the region \(\{(x, y) : 4x^2 + 9y^2 = 36\}\) and find its area, using integration.
Answer: Divide the given equation by 36 to convert it into standard ellipse form: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\), or \(\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1\).
Here, \(a = 3\) and \(b = 2\).
Area \(= 4 \int_{0}^{3} y \, dx = 4 \int_{0}^{3} \frac{2}{3}\sqrt{3^2 - x^2} \, dx\)
\(= \frac{8}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3}\)
\(= \frac{8}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(1) \right] = \frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2} = 6\pi\) sq. units.
In simple words: Rewrite the equation in standard intercept form to identify the semi-axes and apply the standard area formula.

Exam Tip: Memorize the general formula for the area of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), which is \(\pi ab\), for quick verification.

(ii) Area of the region enclosed between Parabola and line

LEVEL II

Question 1. Find the area of the region included between the parabola \(y^2 = x\) and the line \(x + y = 2\).
Answer: Given curves are \(y^2 = x\) (parabola symmetric about x-axis) and \(x + y = 2\) (a straight line).
Substitute \(x = 2 - y\) into the parabola equation: \(y^2 = 2 - y \implies y^2 + y - 2 = 0\)
\((y + 2)(y - 1) = 0 \implies y = -2\) and \(y = 1\).
Corresponding x-values are \(x = 4\) (for \(y = -2\)) and \(x = 1\) (for \(y = 1\)).
The points of intersection are \((4, -2)\) and \((1, 1)\).
Area \(= \int_{-2}^{1} (x_{\text{line}} - x_{\text{parabola}}) \, dy = \int_{-2}^{1} (2 - y - y^2) \, dy\)
\(= \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{-2}^{1}\)
Upper limit at \(y = 1\): \(\left(2 - \frac{1}{2} - \frac{1}{3}\right) = \frac{7}{6}\)
Lower limit at \(y = -2\): \(\left(-4 - 2 + \frac{8}{3}\right) = -\frac{10}{3}\)
Area \(= \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}\) sq. units.
In simple words: Integrate with respect to \(y\) between the intersection points to find the area between the line and parabola.

Exam Tip: Integrating with respect to \(y\) avoids splitting the region into two parts when bounded by a line and a parabola.

Question 2. 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 is the upward-opening parabola \(x^2 = 4y\), which can be written as \(x = 2\sqrt{y}\).
We need the area bounded by the curve, the y-axis, and the horizontal lines \(y = 2\) and \(y = 4\).
Area \(= \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy\)
\(= 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{4}{3} \left[ 4^{3/2} - 2^{3/2} \right]\)
\(= \frac{4}{3} \left[ 8 - 2\sqrt{2} \right] = \frac{8(4 - \sqrt{2})}{3}\) sq. units.
In simple words: Since bounds are given along the y-axis, integrate \(x\) with respect to \(y\) from 2 to 4.

Exam Tip: Pay attention to the axis of integration specified; horizontal strip integration uses \(\int x \, dy\).

LEVEL III

Question 1. Find the area of the region : \(\{ (x, y) : y \le x^2 + 1, y \le x + 1, 0 \le x \le 2 \}\)
Answer: The region is bounded above by the lower of the curves \(y = x^2 + 1\) and \(y = x + 1\), and below by the x-axis, for \(x \in [0, 2]\).
First, find where the parabola \(y = x^2 + 1\) and the line \(y = x + 1\) intersect:
\(x^2 + 1 = x + 1 \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0\) and \(x = 1\).
For \(x \in [0, 1]\), the line \(y = x + 1\) lies above the parabola \(y = x^2 + 1\).
For \(x \in [1, 2]\), the parabola \(y = x^2 + 1\) lies above or the boundary changes; let us test values. At \(x = 1\), both give \(y = 2\). At \(x = 2\), line gives \(y = 3\) and parabola gives \(y = 5\), so the region is bounded by \(y = x + 1\) up to \(x = 1\) and by \(y = x^2 + 1\) from \(x = 1\) to \(x = 2\).
Total Area \(= \int_{0}^{1} (x + 1) \, dx + \int_{1}^{2} (x^2 + 1) \, dx\)
First integral: \(\left[ \frac{x^2}{2} + x \right]_{0}^{1} = \left(\frac{1}{2} + 1\right) = \frac{3}{2}\)
Second integral: \(\left[ \frac{x^3}{3} + x \right]_{1}^{2} = \left(\frac{8}{3} + 2\right) - \left(\frac{1}{3} + 1\right) = \frac{14}{3} - \frac{4}{3} = \frac{10}{3}\)
Total Area \(= \frac{3}{2} + \frac{10}{3} = \frac{9 + 20}{6} = \frac{29}{6}\) sq. units.
In simple words: Split the integral at the intersection point \(x = 1\) where the upper bounding curve changes from the line to the parabola.

Exam Tip: Always sketch compound regions to correctly identify splitting points where boundary functions intersect.

(iii) Area of the region enclosed between Ellipse and line

LEVEL II

Question 1. Find the area of smaller region bounded by the ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1\) and the straight line \(\frac{x}{4} + \frac{y}{5} = 1\).
Answer: The ellipse is \(\frac{x^2}{4^2} + \frac{y^2}{5^2} = 1\) and the line in intercept form is \(\frac{x}{4} + \frac{y}{5} = 1\), which passes through the points \((4, 0)\) and \((0, 5)\) (the intercepts of the ellipse).
The smaller region is the area between the line segment and the arc of the ellipse in the first quadrant.
Area of smaller region \(= \int_{0}^{4} (y_{\text{ellipse}} - y_{\text{line}}) \, dx\)
\(= \int_{0}^{4} \left( \frac{5}{4}\sqrt{16 - x^2} - \frac{5}{4}(4 - x) \right) \, dx\)
\(= \frac{5}{4} \left[ \left( \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right) - \left( 4x - \frac{x^2}{2} \right) \right]_{0}^{4}\)
Evaluate at upper limit \(x = 4\):
\(\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}(1) = 0 + 8\left(\frac{\pi}{2}\right) = 4\pi\)
\(4(4) - \frac{4^2}{2} = 16 - 8 = 8\)
So upper bracket \(= 4\pi - 8\). Lower limit at \(x = 0\) gives 0.
Area \(= \frac{5}{4}(4\pi - 8) = 5\pi - 10\) sq. units.
In simple words: Subtract the area under the straight line from the area under the ellipse from \(x = 0\) to \(x = 4\).

Exam Tip: Notice that the line connects the exact x and y intercepts of the ellipse, making the limits straightforward (0 to 4).

(iv) Area of the region enclosed between Circle and line

LEVEL II

Question 1. Find the area of the region in the first quadrant enclosed by the x-axis, the line \(y = x\) and the circle \(x^2 + y^2 = 32\).
Answer: The circle is \(x^2 + y^2 = (\sqrt{32})^2 = (4\sqrt{2})^2\) with center at origin and radius \(R = 4\sqrt{2}\).
The line is \(y = x\), which makes an angle of \(45^\circ\) (\(\frac{\pi}{4}\)) with the positive x-axis.
Solve simultaneously with circle: \(x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4\) (in first quadrant).
The line intersects the circle at \((4, 4)\).
The region in the first quadrant is divided into two parts by the line \(y = x\):
1. From \(x = 0\) to \(x = 4\), bounded above by the line \(y = x\) and below by the x-axis.
2. From \(x = 4\) to \(x = 4\sqrt{2}\), bounded above by the circle \(y = \sqrt{32 - x^2}\) and below by the x-axis.
Area \(= \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx\)
First integral: \(\left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8\)
Second integral: \(\left[ \frac{x}{2}\sqrt{32 - x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}\)
At upper limit \(x = 4\sqrt{2}\): \(0 + 16\sin^{-1}(1) = 16\left(\frac{\pi}{2}\right) = 8\pi\)
At lower limit \(x = 4\): \(\frac{4}{2}\sqrt{32 - 16} + 16\sin^{-1}\left(\frac{4}{4\sqrt{2}}\right) = 2(4) + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi\)
Second integral value \(= 8\pi - (8 + 4\pi) = 4\pi - 8\).
Total Area \(= 8 + (4\pi - 8) = 4\pi\) sq. units.
In simple words: Split the region at the line-circle intersection \(x = 4\) into a triangular region under the line and a circular sector region under the circle.

Exam Tip: This can also be solved easily using sector area geometry: since the line makes a \(45^\circ\) angle, it encloses exactly one-eighth of the circle: \(\frac{1}{8} \times \pi (\sqrt{32})^2 = \frac{32\pi}{8} = 4\pi\).

LEVEL III

Question 1. Find the area of the region : \(\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}\)
Answer: The region is defined by \(x^2 + y^2 \le 1\) (inside or on the unit circle) and \(x + y \ge 1\) (above or on the line \(x + y = 1\)).
This represents the segment of the unit circle cut off by the line \(x + y = 1\) in the first quadrant.
The line passes through \((1, 0)\) and \((0, 1)\), which are points on the unit circle.
Area \(= \int_{0}^{1} y_{\text{circle}} \, dx - \int_{0}^{1} y_{\text{line}} \, dx\)
\(= \int_{0}^{1} \left( \sqrt{1 - x^2} - (1 - x) \right) \, dx\)
\(= \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}(x) - \left( x - \frac{x^2}{2} \right) \right]_{0}^{1}\)
At upper limit \(x = 1\): \(\left(0 + \frac{1}{2}\left(\frac{\pi}{2}\right)\right) - \left(1 - \frac{1}{2}\right) = \frac{\pi}{4} - \frac{1}{2}\)
At lower limit \(x = 0\): 0.
Total Area \(= \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}\) sq. units.
In simple words: Find the area under the quarter circle and subtract the area under the straight line between \(x = 0\) and \(x = 1\).

Exam Tip: Recognize that the area of a circular segment is simply the area of the corresponding sector minus the area of the triangle.

(v) Area of the region enclosed between Circle and parabola

LEVEL III

Question 1. Draw the rough sketch of the region \(\{(x, y): x^2 \le 6y, x^2 + y^2 \le 16\}\) and find the area enclosed by the region using the method of integration.
Answer: The region is bounded by the parabola \(x^2 = 6y\) (or \(y = \frac{x^2}{6}\)) and the circle \(x^2 + y^2 = 16\) (radius \(R = 4\)).
Find the intersection points by substituting \(x^2 = 6y\) into the circle equation:
\(6y + y^2 = 16 \implies y^2 + 6y - 16 = 0 \implies (y + 8)(y - 2) = 0\)
Since \(y \ge 0\) for the parabola, we take \(y = 2\).
When \(y = 2\), \(x^2 = 12 \implies x = \pm 2\sqrt{3}\).
Due to symmetry about the y-axis, the total area is 2 times the area in the first quadrant.
The area in the first quadrant is bounded below by the parabola from \(x = 0\) to \(x = 2\sqrt{3}\), and by the circle from \(x = 2\sqrt{3}\) to \(x = 4\).
Area \(= 2 \left[ \int_{0}^{2\sqrt{3}} \frac{x^2}{6} \, dx + \int_{2\sqrt{3}}^{4} \sqrt{16 - x^2} \, dx \right]\)
First integral: \(\left[ \frac{x^3}{18} \right]_{0}^{2\sqrt{3}} = \frac{(2\sqrt{3})^3}{18} = \frac{24\sqrt{3}}{18} = \frac{4\sqrt{3}}{3}\)
Second integral: \(\left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{2\sqrt{3}}^{4}\)
At upper limit \(x = 4\): \(0 + 8\sin^{-1}(1) = 8\left(\frac{\pi}{2}\right) = 4\pi\)
At lower limit \(x = 2\sqrt{3}\): \(\frac{2\sqrt{3}}{2}\sqrt{16 - 12} + 8\sin^{-1}\left(\frac{2\sqrt{3}}{4}\right) = \sqrt{3}(2) + 8\left(\frac{\pi}{3}\right) = 2\sqrt{3} + \frac{8\pi}{3}\)
Second integral value \(= 4\pi - \left(2\sqrt{3} + \frac{8\pi}{3}\right) = \frac{4\pi}{3} - 2\sqrt{3}\)
Sum inside brackets \(= \frac{4\sqrt{3}}{3} + \frac{4\pi}{3} - 2\sqrt{3} = \frac{4\pi}{3} - \frac{2\sqrt{3}}{3}\)
Total Area \(= 2 \left( \frac{4\pi - 2\sqrt{3}}{3} \right) = \frac{8\pi - 4\sqrt{3}}{3}\) sq. units.
In simple words: Find the intersection point, then split the integration into the parabola part and the circle part.

Exam Tip: Always check symmetry first to halve your integration workload.

Question 2. Find the area lying above the x-axis and included between the circle \(x^2 + y^2 = 8x\) and the parabola \(y^2 = 4x\).
Answer: The circle equation \(x^2 - 8x + y^2 = 0\) can be rewritten as completing the square: \((x - 4)^2 + y^2 = 4^2\), which has center at \((4, 0)\) and radius 4.
The parabola is \(y^2 = 4x\), opening to the right.
Find intersection points by substituting \(y^2 = 4x\) into the circle equation: \(x^2 + 4x = 8x \implies x^2 - 4x = 0 \implies x(x - 4) = 0 \implies x = 0\) and \(x = 4\).
When \(x = 4\), \(y^2 = 16 \implies y = \pm 4\). Since we need the area above the x-axis, we consider \(y \in [0, 4]\).
Area above x-axis \(= \int_{0}^{4} (x_{\text{circle}} - x_{\text{parabola}}) \, dy\)
From circle \((x-4)^2 + y^2 = 16 \implies x - 4 = \sqrt{16 - y^2} \implies x_{\text{circle}} = 4 + \sqrt{16 - y^2}\).
From parabola \(y^2 = 4x \implies x_{\text{parabola}} = \frac{y^2}{4}\).
Area \(= \int_{0}^{4} \left( 4 + \sqrt{16 - y^2} - \frac{y^2}{4} \right) dy\)
\(= \left[ 4y + \left( \frac{y}{2}\sqrt{16 - y^2} + 8\sin^{-1}\left(\frac{y}{4}\right) \right) - \frac{y^3}{12} \right]_{0}^{4}\)
At upper limit \(y = 4\):
\(4(4) + \left(0 + 8\sin^{-1}(1)\right) - \frac{64}{12} = 16 + 8\left(\frac{\pi}{2}\right) - \frac{16}{3} = 16 + 4\pi - \frac{16}{3} = \frac{32}{3} + 4\pi\)
Lower limit at \(y = 0\) gives 0.
Total Area \(= \frac{32}{3} + 4\pi\) sq. units.
In simple words: Integrate with respect to \(y\) from 0 to 4 by subtracting the parabola's x-value from the circle's x-value.

Question 1. Find the area bounded by the curves \(x^2 + y^2 = 4\) and \((x + 2)^2 + y^2 = 4\) using integration.
Answer: First circle: \(x^2 + y^2 = 4\), centered at \((0, 0)\) with radius 2.
Second circle: \((x + 2)^2 + y^2 = 4\), centered at \((-2, 0)\) with radius 2.
Find intersection points by equating \(y^2\):
From first circle: \(y^2 = 4 - x^2\)
From second circle: \(y^2 = 4 - (x + 2)^2\)
\(4 - x^2 = 4 - (x^2 + 4x + 4) \implies 4x + 4 = 0 \implies 4x = -4 \implies x = -1\).
When \(x = -1\), \(y^2 = 4 - (-1)^2 = 3 \implies y = \pm\sqrt{3}\).
By symmetry about the x-axis, the total area is 2 times the area in the upper half-plane.
Area \(= 2 \left[ \int_{-2}^{-1} \sqrt{4 - (x+2)^2} \, dx + \int_{-1}^{0} \sqrt{4 - x^2} \, dx \right]\)
First integral (second circle from -2 to -1):
\(\int_{-2}^{-1} \sqrt{2^2 - (x+2)^2} \, dx = \left[ \frac{x+2}{2}\sqrt{4 - (x+2)^2} + 2\sin^{-1}\left(\frac{x+2}{2}\right) \right]_{-2}^{-1}\)
At \(x = -1\): \(\frac{1}{2}\sqrt{3} + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2} + 2\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{\pi}{3}\)
At \(x = -2\): \(0 + 2\sin^{-1}(0) = 0\).
Second integral (first circle from -1 to 0):
\(\int_{-1}^{0} \sqrt{4 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{-1}^{0}\)
At \(x = 0\): 0.
At \(x = -1\): \(-\frac{-1}{2}\sqrt{3} - 2\sin^{-1}\left(-\frac{1}{2}\right) = -\left(-\frac{\sqrt{3}}{2} - 2\left(-\frac{\pi}{6}\right)\right) = \frac{\sqrt{3}}{2} + \frac{\pi}{3}\)
Sum of integrals \(= \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) + \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) = \sqrt{3} + \frac{2\pi}{3}\)
Total Area \(= 2 \left(\sqrt{3} + \frac{2\pi}{3}\right) = 2\sqrt{3} + \frac{4\pi}{3}\) sq. units.
In simple words: Find the intersection line \(x = -1\) and add the areas under each respective circle's arc, doubling the result by symmetry.

Exam Tip: Symmetric overlapping shapes like identical intersecting circles can be solved efficiently by splitting at the radical axis and using circular segment formulas.

(vii) Area of the region enclosed between Two parabolas

LEVEL II

Question 1. Draw the rough sketch and find the area of the region bounded by two parabolas \(4y^2 = 9x\) and \(3x^2 = 16y\) by using method of integration.
Answer: First parabola: \(y^2 = \frac{9}{4}x\), opening to the right.
Second parabola: \(x^2 = \frac{16}{3}y \implies y = \frac{3x^2}{16}\), opening upwards.
Find intersection points by substituting \(y = \frac{3x^2}{16}\) into the first parabola's equation \(4y^2 = 9x\):
\(4 \left( \frac{3x^2}{16} \right)^2 = 9x \implies 4 \left( \frac{9x^4}{256} \right) = 9x \implies \frac{9x^4}{64} = 9x\)
\(x^4 - 64x = 0 \implies x(x^3 - 64) = 0 \implies x = 0\) and \(x = 4\).
The intersection points are \((0, 0)\) and \((4, 3)\).
Area \(= \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx\)
Here, upper curve is \(y_{\text{parabola 1}} = \frac{3}{2}\sqrt{x}\) and lower curve is \(y_{\text{parabola 2}} = \frac{3x^2}{16}\).
Area \(= \int_{0}^{4} \left( \frac{3}{2}x^{1/2} - \frac{3}{16}x^2 \right) dx\)
\(= \left[ \frac{3}{2} \cdot \frac{x^{3/2}}{3/2} - \frac{3}{16} \cdot \frac{x^3}{3} \right]_{0}^{4} = \left[ x^{3/2} - \frac{x^3}{16} \right]_{0}^{4}\)
At upper limit \(x = 4\): \(4^{3/2} - \frac{4^3}{16} = 8 - \frac{64}{16} = 8 - 4 = 4\).
Lower limit at \(x = 0\) gives 0.
Total Area \(= 4\) sq. units.
In simple words: Find the intersection points and integrate the difference between the upper and lower parabolas from \(x = 0\) to \(x = 4\).

Exam Tip: Always express both curves in the form \(y = f(x)\) when integrating with respect to \(x\) between two intersection points.

(viii) Area of triangle when vertices are given

LEVEL III

Question 1. Using integration compute the area of the region bounded by the triangle whose vertices are \((2, 1)\), \((3, 4)\), and \((5, 2)\).
Answer: Let the vertices be \(A(2, 1)\), \(B(3, 4)\), and \(C(5, 2)\).
Find the equations of the three sides using the two-point form \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\):
- Side AB through \((2, 1)\) and \((3, 4)\):
\(y - 1 = \frac{4 - 1}{3 - 2}(x - 2) \implies y - 1 = 3(x - 2) \implies y = 3x - 5\) (or \(x = \frac{y + 5}{3}\))
- Side BC through \((3, 4)\) and \((5, 2)\):
\(y - 4 = \frac{2 - 4}{5 - 3}(x - 3) \implies y - 4 = -1(x - 3) \implies y = -x + 7\) (or \(x = 7 - y\))
- Side AC through \((2, 1)\) and \((5, 2)\):
\(y - 1 = \frac{2 - 1}{5 - 2}(x - 2) \implies y - 1 = \frac{1}{3}(x - 2) \implies 3y - 3 = x - 2 \implies x = 3y - 1\)
Integrating with respect to \(y\) from \(y = 1\) to \(y = 4\):
The x-coordinates of the region are bounded on the left by line AC and line AB, and on the right by line BC.
Specifically, from \(y = 1\) to \(y = 2\), region is between line AB and line AC.
From \(y = 2\) to \(y = 4\), region is between line AB and line BC.
Let us use the standard formula for area of triangle using integration by summing regions under line segments:
Area \(= \int_{2}^{3} y_{AB} \, dx + \int_{3}^{5} y_{BC} \, dx - \int_{2}^{5} y_{AC} \, dx\)
- Equation of AB: \(y = 3x - 5\) (\(x \in [2, 3]\))
- Equation of BC: \(y = -x + 7\) (\(x \in [3, 5]\))
- Equation of AC: \(y = \frac{1}{3}x + \frac{1}{3}\) (\(x \in [2, 5]\))
Area \(= \int_{2}^{3} (3x - 5) \, dx + \int_{3}^{5} (-x + 7) \, dx - \int_{2}^{5} \left(\frac{1}{3}x + \frac{1}{3}\right) dx\)
1. \(\int_{2}^{3} (3x - 5) \, dx = \left[ \frac{3x^2}{2} - 5x \right]_{2}^{3} = \left(\frac{27}{2} - 15\right) - (6 - 10) = -\frac{3}{2} - (-4) = \frac{5}{2}\)
2. \(\int_{3}^{5} (-x + 7) \, dx = \left[ -\frac{x^2}{2} + 7x \right]_{3}^{5} = \left(-\frac{25}{2} + 35\right) - \left(-\frac{9}{2} + 21\right) = \frac{45}{2} - \frac{33}{2} = \frac{12}{2} = 6\)
3. \(\int_{2}^{5} \left(\frac{1}{3}x + \frac{1}{3}\right) dx = \left[ \frac{x^2}{6} + \frac{x}{3} \right]_{2}^{5} = \left(\frac{25}{6} + \frac{5}{3}\right) - \left(\frac{4}{6} + \frac{2}{3}\right) = \frac{35}{6} - \frac{8}{6} = \frac{27}{6} = \frac{9}{2}\)
Total Area \(= \frac{5}{2} + 6 - \frac{9}{2} = \frac{5 - 9}{2} + 6 = -2 + 6 = 4\) sq. units.
In simple words: Find the line equations for each of the three sides, then integrate appropriately with respect to \(x\) (or \(y\)) to find the enclosed area.

Exam Tip: You can quickly cross-check triangle area using the coordinate geometry determinant formula: \(\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\).

Question 2. Using integration compute the area of the region bounded by the triangle whose vertices are \((-1, 1)\), \((0, 5)\), and \((3, 2)\).
Answer: Let vertices be \(A(-1, 1)\), \(B(0, 5)\), and \(C(3, 2)\).
Find the line equations for each side:
- Side AB through \((-1, 1)\) and \((0, 5)\): \(y - 1 = \frac{5 - 1}{0 - (-1)}(x - (-1)) \implies y = 4x + 5\)
- Side BC through \((0, 5)\) and \((3, 2)\): \(y - 5 = \frac{2 - 5}{3 - 0}(x - 0) \implies y = -x + 5\)
- Side AC through \((-1, 1)\) and \((3, 2)\): \(y - 1 = \frac{2 - 1}{3 - (-1)}(x - (-1)) \implies y - 1 = \frac{1}{4}(x + 1) \implies y = \frac{1}{4}x + \frac{5}{4}\)
Area \(= \int_{-1}^{0} (y_{AB} - y_{AC}) \, dx + \int_{0}^{3} (y_{BC} - y_{AC}) \, dx\)
1. From \(x = -1\) to \(0\):
\(\int_{-1}^{0} \left( (4x + 5) - \left(\frac{1}{4}x + \frac{5}{4}\right) \right) dx = \int_{-1}^{0} \left( \frac{15}{4}x + \frac{15}{4} \right) dx\)
\(= \left[ \frac{15}{8}x^2 + \frac{15}{4}x \right]_{-1}^{0} = 0 - \left(\frac{15}{8} - \frac{15}{4}\right) = \frac{15}{8}\)
2. From \(x = 0\) to \(3\):
\(\int_{0}^{3} \left( (-x + 5) - \left(\frac{1}{4}x + \frac{5}{4}\right) \right) dx = \int_{0}^{3} \left( -\frac{5}{4}x + \frac{15}{4} \right) dx\)
\(= \left[ -\frac{5}{8}x^2 + \frac{15}{4}x \right]_{0}^{3} = \left(-\frac{45}{8} + \frac{45}{4}\right) - 0 = \frac{45}{8}\)
Total Area \(= \frac{15}{8} + \frac{45}{8} = \frac{60}{8} = \frac{15}{2} = 7.5\) sq. units.
In simple words: Split the triangle into two vertical strips at \(x = 0\) and integrate the difference between upper and lower bounding lines in each interval.

Exam Tip: Splitting the region at the vertex with the intermediate x-coordinate simplifies piecewise integration for triangles.

(ix) Area of triangle when sides are given

LEVEL III

Question 1. Using integration find the area of the region bounded by the triangle whose sides are \(y = 2x + 1\), \(y = 3x + 1\), \(x = 4\).
Answer: First, find the vertices of the triangle by solving the equations of the sides pairwise:
- Intersection of \(y = 2x + 1\) and \(y = 3x + 1\):
\(2x + 1 = 3x + 1 \implies x = 0\). Substituting \(x = 0\) gives \(y = 1\). Vertex \(A(0, 1)\).
- Intersection of \(y = 2x + 1\) and \(x = 4\):
\(x = 4 \implies y = 2(4) + 1 = 9\). Vertex \(B(4, 9)\).
- Intersection of \(y = 3x + 1\) and \(x = 4\):
\(x = 4 \implies y = 3(4) + 1 = 13\). Vertex \(C(4, 13)\).
The region is bounded between \(x = 0\) and \(x = 4\), with upper curve \(y = 3x + 1\) and lower curve \(y = 2x + 1\).
Area \(= \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx = \int_{0}^{4} ((3x + 1) - (2x + 1)) \, dx\)
\(= \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8\) sq. units.
In simple words: Find the vertices from the given side equations, then integrate the difference between the upper and lower lines from the lowest to the highest x-coordinate.

Exam Tip: When one side is a vertical line like \(x = 4\), integration with respect to \(x\) becomes extremely fast and simple.

Question 2. Using integration compute the area of the region bounded by the lines \(x + 2y = 2\), \(y - x = 1\), and \(2x + y = 7\).
Answer: Find the vertices by solving the line equations pairwise:
1. Line 1 (\(x + 2y = 2\)) and Line 2 (\(y - x = 1\)):
From Line 2, \(x = y - 1\). Substitute into Line 1: \((y - 1) + 2y = 2 \implies 3y = 3 \implies y = 1\). Then \(x = 0\). Vertex \(A(0, 1)\).
2. Line 2 (\(y - x = 1\)) and Line 3 (\(2x + y = 7\)):
From Line 2, \(y = x + 1\). Substitute into Line 3: \(2x + (x + 1) = 7 \implies 3x = 6 \implies x = 2\). Then \(y = 3\). Vertex \(B(2, 3)\).
3. Line 1 (\(x + 2y = 2\)) and Line 3 (\(2x + y = 7\)):
From Line 1, \(x = 2 - 2y\). Substitute into Line 3: \(2(2 - 2y) + y = 7 \implies 4 - 4y + y = 7 \implies -3y = 3 \implies y = -1\). Then \(x = 2 - 2(-1) = 4\). Vertex \(C(4, -1)\).
The x-coordinates of the vertices are \(x = 0, 2, 4\).
Let us express each line as \(y\) in terms of \(x\):
- Line 1: \(y = \frac{2 - x}{2} = 1 - \frac{1}{2}x\)
- Line 2: \(y = x + 1\)
- Line 3: \(y = 7 - 2x\)
From \(x = 0\) to \(x = 2\), the region is bounded above by Line 2 (\(y = x + 1\)) and below by Line 1 (\(y = 1 - \frac{1}{2}x\)).
From \(x = 2\) to \(x = 4\), the region is bounded above by Line 3 (\(y = 7 - 2x\)) and below by Line 1 (\(y = 1 - \frac{1}{2}x\)).
Area \(= \int_{0}^{2} \left( (x + 1) - \left(1 - \frac{1}{2}x\right) \right) dx + \int_{2}^{4} \left( (7 - 2x) - \left(1 - \frac{1}{2}x\right) \right) dx\)
1. First integral (\(x \in [0, 2]\)):
\(\int_{0}^{2} \frac{3}{2}x \, dx = \left[ \frac{3}{4}x^2 \right]_{0}^{2} = \frac{3}{4}(4) = 3\)
2. Second integral (\(x \in [2, 4]\)):
\(\int_{2}^{4} \left( 6 - \frac{3}{2}x \right) dx = \left[ 6x - \frac{3}{4}x^2 \right]_{2}^{4} = (24 - 12) - (12 - 3) = 12 - 9 = 3\)
Total Area \(= 3 + 3 = 6\) sq. units.
In simple words: Find all three vertices of the triangle, split the region at the intermediate x-coordinate, and integrate the difference between upper and lower lines.

Exam Tip: Always verify that the three lines form a closed triangle by checking that all three intersection vertices are distinct and non-collinear.

(x) Miscellaneous Questions

LEVEL III

Question 1. Find the area of the region bounded by the curves \(y = |x - 1\) and \(y = -|x - 1| + 1\).
Answer: Let \(f(x) = |x - 1|\) and \(g(x) = -|x - 1| + 1\).
Find intersection points by setting \(f(x) = g(x)\):
\(|x - 1| = -|x - 1| + 1 \implies 2|x - 1| = 1 \implies |x - 1| = \frac{1}{2}\)
This gives \(x - 1 = \frac{1}{2} \implies x = \frac{3}{2}\) and \(x - 1 = -\frac{1}{2} \implies x = \frac{1}{2}\).
Thus, the curves intersect at \(x = \frac{1}{2}\) and \(x = \frac{3}{2}\).
For \(x \in \left[\frac{1}{2}, \frac{3}{2}\right]\), the upper curve is \(g(x) = -|x - 1| + 1\) and the lower curve is \(f(x) = |x - 1|\).
Area \(= \int_{1/2}^{3/2} \left( (-|x - 1| + 1) - |x - 1| \right) dx = \int_{1/2}^{3/2} (1 - 2|x - 1|) \, dx\)
By symmetry about \(x = 1\), we can evaluate 2 times the integral from \(\frac{1}{2}\) to 1:
For \(x \in \left[\frac{1}{2}, 1\right]\), \(|x - 1| = -(x - 1) = 1 - x\).
Area \(= 2 \int_{1/2}^{1} (1 - 2(1 - x)) \, dx = 2 \int_{1/2}^{1} (2x - 1) \, dx\)
\(= 2 \left[ x^2 - x \right]_{1/2}^{1} = 2 \left[ (1 - 1) - \left(\frac{1}{4} - \frac{1}{2}\right) \right] = 2 \left( 0 - \left(-\frac{1}{4}\right) \right) = 2 \left(\frac{1}{4}\right) = \frac{1}{2}\) sq. units.
In simple words: Find the intersection points of the absolute value curves and integrate the difference between them over the interval.

Exam Tip: Use symmetry around the vertex of modulus functions to simplify definite integrals.

Question 2. Find the area bounded by the curve \(y = x\) and \(y = x^3\).
Answer: Find intersection points by setting \(x = x^3\):
\(x^3 - x = 0 \implies x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0 \implies x = -1, 0, 1\).
The curves intersect at \((-1, -1)\), \((0, 0)\), and \((1, 1)\).
Due to symmetry about the origin, the region is symmetric in the first and third quadrants.
Area \(= \int_{-1}^{0} (y_{\text{lower curve}} - y_{\text{upper curve}}) \, dx + \int_{0}^{1} (y_{\text{upper curve}} - y_{\text{lower curve}}) \, dx\)
Let us evaluate for \(x \in [0, 1]\): here \(y = x\) is above \(y = x^3\) (e.g., at \(x = \frac{1}{2}\), \(\frac{1}{2} > \frac{1}{8}\)).
Area \(= 2 \int_{0}^{1} (x - x^3) \, dx\)
\(= 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{4} \right) = 2 \left(\frac{1}{4}\right) = \frac{1}{2}\) sq. units.
In simple words: Find where the line and cubic curve intersect, then integrate their difference from -1 to 1 (using symmetry).

Exam Tip: Odd functions like \(y = x^3 - x\) symmetric about the origin yield equal areas in positive and negative intervals.

Question 3. Draw a rough sketch of the curve \(y = \sin x\) and \(y = \cos x\) as x varies from \(x = 0\) to \(x = \frac{\pi}{2}\) and find the area of the region enclosed by them and x-axis
Answer: The curves are \(y = \sin x\) and \(y = \cos x\) for \(x \in [0, \frac{\pi}{2}]\).
They intersect where \(\sin x = \cos x\), which occurs at \(x = \frac{\pi}{4}\) (since \(\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\)).
- From \(x = 0\) to \(x = \frac{\pi}{4}\), \(\cos x \ge \sin x\).
- From \(x = \frac{\pi}{4}\) to \(x = \frac{\pi}{2}\), \(\sin x \ge \cos x\).
Area \(= \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx\)
1. First integral: \(\left[ \sin x + \cos x \right]_{0}^{\pi/4} = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1\)
2. Second integral: \(\left[ -\cos x - \sin x \right]_{\pi/4}^{\pi/2} = (0 - 1) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -1 - \left(-\frac{2}{\sqrt{2}}\right) = \sqrt{2} - 1\)
Total Area \(= (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2\) sq. units.
In simple words: Split the integral at the intersection point \(x = \frac{\pi}{4}\) where sine and cosine curves cross each other.

Exam Tip: Remember that \(\sin x = \cos x\) at \(45^\circ\) (\(\frac{\pi}{4}\)), which is the exact splitting point for this area calculation.

Question 4. Sketch the graph of \(y = |x + 1|\). Evaluate \(\int_{-3}^{1} |x + 1| \, dx\). What does this value represent on the graph?
Answer: The function is \(y = |x + 1|\), which is a V-shaped graph with vertex at \(x = -1\).
Split the integral at the vertex \(x = -1\):
\(\int_{-3}^{1} |x + 1| \, dx = \int_{-3}^{-1} -(x + 1) \, dx + \int_{-1}^{1} (x + 1) \, dx\)
1. First integral: \(\left[ -\frac{x^2}{2} - x \right]_{-3}^{-1} = \left(-\frac{1}{2} + 1\right) - \left(-\frac{9}{2} + 3\right) = \frac{1}{2} - \left(-\frac{3}{2}\right) = 2\)
2. Second integral: \(\left[ \frac{x^2}{2} + x \right]_{-1}^{1} = \left(\frac{1}{2} + 1\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2\)
Total Value \(= 2 + 2 = 4\).
Geometric representation: This value represents the total area of the two triangles formed by the graph of \(y = |x + 1|\), the x-axis, and the vertical lines \(x = -3\) and \(x = 1\).
In simple words: Evaluate the modulus integral by splitting it at \(x = -1\). Geometrically, it represents the area under the V-shaped graph.

Exam Tip: Modulus integrals can often be evaluated quickly by calculating the areas of standard geometric triangles instead of doing full integration.

Question 5. Find the area bounded by the curves \(y = 6x - x^2\) and \(y = x^2 - 2x\).
Answer: Let \(y_1 = 6x - x^2\) (downward parabola) and \(y_2 = x^2 - 2x\) (upward parabola).
Find intersection points by setting \(y_1 = y_2\):
\(6x - x^2 = x^2 - 2x \implies 2x^2 - 8x = 0 \implies 2x(x - 4) = 0 \implies x = 0\) and \(x = 4\).
Area \(= \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx\)
For \(x \in [0, 4]\), \(y_1 = 6x - x^2\) is the upper curve and \(y_2 = x^2 - 2x\) is the lower curve.
Area \(= \int_{0}^{4} \left( (6x - x^2) - (x^2 - 2x) \right) dx = \int_{0}^{4} (8x - 2x^2) \, dx\)
\(= \left[ 4x^2 - \frac{2}{3}x^3 \right]_{0}^{4} = \left( 4(16) - \frac{2}{3}(64) \right) - 0\)
\(= 64 - \frac{128}{3} = \frac{192 - 128}{3} = \frac{64}{3}\) sq. units.
In simple words: Find the intersection points of the two parabolas and integrate the difference between them from \(x = 0\) to \(x = 4\).

Exam Tip: Always subtract the lower function from the upper function inside the integral to keep the area positive.

Question 6. Sketch the graph of \(y = |x + 3|\) and evaluate the area under the curve \(y = |x + 3|\) above x-axis and between \(x = -6\) to \(x = 0\). [CBSE 2011]
Answer: The function is \(y = |x + 3|\) with vertex at \(x = -3\).
We need the area from \(x = -6\) to \(x = 0\), which spans across the vertex \(x = -3\).
Area \(= \int_{-6}^{0} |x + 3| \, dx = \int_{-6}^{-3} -(x + 3) \, dx + \int_{-3}^{0} (x + 3) \, dx\)
1. First integral: \(\left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3} = \left(-\frac{9}{2} + 9\right) - (-18 + 18) = \frac{9}{2} - 0 = \frac{9}{2}\)
2. Second integral: \(\left[ \frac{x^2}{2} + 3x \right]_{-3}^{0} = 0 - \left(\frac{9}{2} - 9\right) = -\left(-\frac{9}{2}\right) = \frac{9}{2}\)
Total Area \(= \frac{9}{2} + \frac{9}{2} = 9\) sq. units.
In simple words: Split the modulus integral at \(x = -3\) and evaluate the two triangular areas under the graph.

Exam Tip: Geometrically, this region consists of two identical triangles with base 3 and height 3, so Area \(= 2 \times \left(\frac{1}{2} \times 3 \times 3\right) = 9\).

Questions for self evaluation

Question 1. Find the area bounded by the curve \(x^2 = 4y\) and the line \(x = 4y - 2\).
Answer: Given curve: \(x^2 = 4y \implies y = \frac{x^2}{4}\).
Given line: \(x = 4y - 2 \implies 4y = x + 2 \implies y = \frac{x + 2}{4}\).
Find intersection points by setting the two expressions for \(y\) equal:
\(\frac{x^2}{4} = \frac{x + 2}{4} \implies x^2 = x + 2 \implies x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0 \implies x = -1\) and \(x = 2\).
Area \(= \int_{-1}^{2} (y_{\text{line}} - y_{\text{parabola}}) \, dx = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} \right) dx\)
\(= \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) \, dx = \frac{1}{4} \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}\)
At upper limit \(x = 2\): \(2(2) + \frac{4}{2} - \frac{8}{3} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}\)
At lower limit \(x = -1\): \(2(-1) + \frac{1}{2} - \left(-\frac{1}{3}\right) = -2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{5}{6} = -\frac{7}{6}\)
Integral value \(= \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}\)
Total Area \(= \frac{1}{4} \times \frac{9}{2} = \frac{9}{8}\) sq. units.
In simple words: Find the intersection points and integrate the difference between the line and the parabola from \(x = -1\) to \(x = 2\).

Exam Tip: Factor out constant denominators early to prevent calculation errors during definite integration.

Question 2. Find the area bounded by the parabola \(y = x^2\) and \(y = |x|\).
Answer: The region is bounded by the upward parabola \(y = x^2\) and the V-shaped curve \(y = |x|\).
Due to symmetry about the y-axis, the total area is 2 times the area in the first quadrant.
In the first quadrant, \(|x| = x\), so the curves are \(y = x^2\) and \(y = x\).
Intersection points in the first quadrant: \(x^2 = x \implies x(x - 1) = 0 \implies x = 0\) and \(x = 1\).
Area \(= 2 \int_{0}^{1} (x - x^2) \, dx = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}\)
\(= 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left(\frac{1}{6}\right) = \frac{1}{3}\) sq. units.
In simple words: Use symmetry about the y-axis and integrate the difference between \(y = x\) and \(y = x^2\) from 0 to 1.

Exam Tip: Modulus functions that are even allow you to restrict integration to the positive x-axis and double the result.

Question 3. Find the area of the region : \(\{ (x, y) : 0 \le y \le x^2 + 1, 0 \le y \le x + 1, 0 \le x \le 2 \}\)
Answer: This is identical to Level III, Question 1 of section (ii).
The region is bounded by \(y = x^2 + 1\), \(y = x + 1\), and \(x \in [0, 2]\).
Splitting at intersection \(x = 1\):
Area \(= \int_{0}^{1} (x + 1) \, dx + \int_{1}^{2} (x^2 + 1) \, dx = \frac{3}{2} + \frac{10}{3} = \frac{29}{6}\) sq. units.
In simple words: Split the region at \(x = 1\) where the upper boundary curve changes from the line to the parabola.

Exam Tip: Reusing standard problem templates saves time during revision and exams.

Question 4. Find the area of the smaller region bounded by the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) and the line \(\frac{x}{3} + \frac{y}{2} = 1\).
Answer: The ellipse is \(\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1\) and the line is \(\frac{x}{3} + \frac{y}{2} = 1\).
The line connects the x-intercept \((3, 0)\) and y-intercept \((0, 2)\) of the ellipse.
Area of smaller region in the first quadrant \(= \int_{0}^{3} (y_{\text{ellipse}} - y_{\text{line}}) \, dx\)
\(= \int_{0}^{3} \left( \frac{2}{3}\sqrt{9 - x^2} - \frac{2}{3}(3 - x) \right) dx\)
\(= \frac{2}{3} \left[ \left( \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right) - \left( 3x - \frac{x^2}{2} \right) \right]_{0}^{3}\)
At upper limit \(x = 3\):
\(\left(0 + \frac{9}{2}\sin^{-1}(1)\right) - \left(9 - \frac{9}{2}\right) = \frac{9\pi}{4} - \frac{9}{2} = \frac{9\pi - 18}{4}\)
Lower limit at \(x = 0\) gives 0.
Total Area \(= \frac{2}{3} \left(\frac{9\pi - 18}{4}\right) = \frac{3\pi - 6}{2}\) sq. units.
In simple words: Integrate the difference between the ellipse and the straight line from \(x = 0\) to \(x = 3\).

Exam Tip: The area of a quadrant of an ellipse minus the area of the right triangle formed by the axes gives the exact area of the segment.

Question 5. Find the area of the region : \(\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}\)
Answer: This is identical to Level III, Question 1 of section (iv).
Area \(= \int_{0}^{1} \left( \sqrt{1 - x^2} - (1 - x) \right) dx = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}\) sq. units.
In simple words: Subtract the area under the line from the area under the quarter circle from \(x = 0\) to \(x = 1\).

Exam Tip: Recognize circular segment problems immediately to apply standard formula templates.

Question 6. Find the area lying above the x-axis and included between the circle \(x^2 + y^2 = 8x\) and the parabola \(y^2 = 4x\).
Answer: This is identical to Level III, Question 2 of section (v).
Area \(= \frac{32}{3} + 4\pi\) sq. units.
In simple words: Integrate with respect to \(y\) from 0 to 4 between the circle and the parabola.

Question 7. Find the area bounded by the curves \(x^2 + y^2 = 4\) and \((x + 2)^2 + y^2 = 4\) using integration.
Answer: This is identical to Level III, Question 1 of section (vi).
Area \(= 2\sqrt{3} + \frac{4\pi}{3}\) sq. units.
In simple words: Use symmetry and integrate the two overlapping circular arcs across their radical axis.

Exam Tip: Symmetric circle intersections can be solved by combining standard circular segment areas.

Question 8. Using integration compute the area of the region bounded by the triangle whose vertices are \((2, 1)\), \((3, 4)\), and \((5, 2)\).
Answer: This is identical to Level III, Question 1 of section (viii).
Area \(= 4\) sq. units.
In simple words: Find the line equations for the three sides and integrate to find the total area.

Exam Tip: Always verify triangle area calculations using the determinant formula as a quick check.

Question 9. Using integration compute the area of the region bounded by the lines \(2x + y = 4\), \(3x - 2y = 6\), and \(x - 3y + 5 = 0\).
Answer: Find the intersection vertices of the three lines:
1. Line 1 (\(2x + y = 4\)) and Line 2 (\(3x - 2y = 6\)):
From Line 1, \(y = 4 - 2x\). Substitute into Line 2: \(3x - 2(4 - 2x) = 6 \implies 3x - 8 + 4x = 6 \implies 7x = 14 \implies x = 2\). Then \(y = 0\). Vertex \(A(2, 0)\).
2. Line 2 (\(3x - 2y = 6\)) and Line 3 (\(x - 3y + 5 = 0\)):
From Line 3, \(x = 3y - 5\). Substitute into Line 2: \(3(3y - 5) - 2y = 6 \implies 9y - 15 - 2y = 6 \implies 7y = 21 \implies y = 3\). Then \(x = 3(3) - 5 = 4\). Vertex \(B(4, 3)\).
3. Line 1 (\(2x + y = 4\)) and Line 3 (\(x - 3y + 5 = 0\)):
From Line 1, \(y = 4 - 2x\). Substitute into Line 3: \(x - 3(4 - 2x) + 5 = 0 \implies x - 12 + 6x + 5 = 0 \implies 7x - 7 = 0 \implies x = 1\). Then \(y = 4 - 2(1) = 2\). Vertex \(C(1, 2)\).
The x-coordinates of the vertices are \(x = 1, 2, 4\).
Express lines as \(y = f(x)\):
- Line 1: \(y = 4 - 2x\)
- Line 2: \(y = \frac{3x - 6}{2} = \frac{3}{2}x - 3\)
- Line 3: \(y = \frac{x + 5}{3} = \frac{1}{3}x + \frac{5}{3}\)
From \(x = 1\) to \(x = 2\), region is bounded above by Line 3 and below by Line 1.
From \(x = 2\) to \(x = 4\), region is bounded above by Line 3 and below by Line 2.
Area \(= \int_{1}^{2} \left( \left(\frac{1}{3}x + \frac{5}{3}\right) - (4 - 2x) \right) dx + \int_{2}^{4} \left( \left(\frac{1}{3}x + \frac{5}{3}\right) - \left(\frac{3}{2}x - 3\right) \right) dx\)
1. First integral (\(x \in [1, 2]\)):
\(\int_{1}^{2} \left( \frac{7}{3}x - \frac{7}{3} \right) dx = \frac{7}{3} \left[ \frac{x^2}{2} - x \right]_{1}^{2} = \frac{7}{3} \left[ (2 - 2) - \left(\frac{1}{2} - 1\right) \right] = \frac{7}{3} \left(\frac{1}{2}\right) = \frac{7}{6}\)
2. Second integral (\(x \in [2, 4]\)):
\(\int_{2}^{4} \left( -\frac{7}{6}x + \frac{14}{3} \right) dx = \left[ -\frac{7}{12}x^2 + \frac{14}{3}x \right]_{2}^{4}\)
At \(x = 4\): \(-\frac{7}{12}(16) + \frac{14}{3}(4) = -\frac{28}{3} + \frac{56}{3} = \frac{28}{3}\)
At \(x = 2\): \(-\frac{7}{12}(4) + \frac{14}{3}(2) = -\frac{7}{3} + \frac{28}{3} = \frac{21}{3} = 7\)
Second integral value \(= \frac{28}{3} - 7 = \frac{7}{3} = \frac{14}{6}\)
Total Area \(= \frac{7}{6} + \frac{14}{6} = \frac{21}{6} = \frac{7}{2} = 3.5\) sq. units.
In simple words: Find the three vertices, split the region at \(x = 2\), and integrate the difference between upper and lower bounding lines.

Exam Tip: Taking extra care when finding intersection coordinates prevents compounding arithmetic errors in piecewise integration.

Question 10. Sketch the graph of \(f(x) = \begin{cases} |x - 2| + 2, & x \le 2 \\ x^2 - 2, & x > 2 \end{cases}\). Evaluate \(\int_{0}^{4} f(x) \, dx\). What does the value of this integral represent on the graph?
Answer: The piecewise function is defined as:
- For \(x \in [0, 2]\): \(f(x) = |x - 2| + 2 = -(x - 2) + 2 = -x + 4\)
- For \(x \in (2, 4]\): \(f(x) = x^2 - 2\)
Evaluate the definite integral by splitting at the boundary \(x = 2\):
\(\int_{0}^{4} f(x) \, dx = \int_{0}^{2} (-x + 4) \, dx + \int_{2}^{4} (x^2 - 2) \, dx\)
1. First integral: \(\left[ -\frac{x^2}{2} + 4x \right]_{0}^{2} = (-2 + 8) - 0 = 6\)
2. Second integral: \(\left[ \frac{x^3}{3} - 2x \right]_{2}^{4} = \left(\frac{64}{3} - 8\right) - \left(\frac{8}{3} - 4\right) = \left(\frac{64}{3} - \frac{24}{3}\right) - \left(\frac{8}{3} - \frac{12}{3}\right) = \frac{40}{3} - \left(-\frac{4}{3}\right) = \frac{44}{3}\)
Total Value \(= 6 + \frac{44}{3} = \frac{18 + 44}{3} = \frac{62}{3}\).
Geometric representation: This value represents the total net area bounded by the graph of \(y = f(x)\), the x-axis, and the vertical lines \(x = 0\) and \(x = 4\).
In simple words: Split the integral into two parts at \(x = 2\) corresponding to the two definitions of the piecewise function.

Exam Tip: For piecewise functions, always check the continuity and boundary values at the transition point before integrating.