CBSE Class 12 Mathematics Applications of Integration Assignment Set 05

ASSIGNMENTS

(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: 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 the semi-minor axis \(b = 4\).
Since the ellipse is symmetric about both x-axis and y-axis, the total area is 4 times the area of the region in the first quadrant.
From the equation, \(y = \frac{4}{5}\sqrt{25 - x^2}\).
Area \(A = 4 \int_{0}^{5} y \, dx = 4 \int_{0}^{5} \frac{4}{5}\sqrt{25 - x^2} \, dx\)
\(A = \frac{16}{5} \left[ \frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{5}\)
\(A = \frac{16}{5} \left[ 0 + \frac{25}{2}\sin^{-1}(1) - (0 + 0) \right] = \frac{16}{5} \times \frac{25}{2} \times \frac{\pi}{2} = 20\pi\) sq. units.
In simple words: The area of an ellipse is found by integrating the curve in one quadrant and multiplying by 4 due to symmetry.

Exam Tip: Remember the standard formula for the area of an ellipse \(\pi ab\), which can be used to cross-verify your integration result.

Question 2. Sketch the region {(x, y) : \(4x^2 + 9y^2 = 36\)} and find its area, using integration.
Answer: Dividing the given equation by 36, we get \(\frac{x^2}{9} + \frac{y^2}{4} = 1\), which is \(\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1\).
Here, \(a = 3\) and \(b = 2\).
Total area \(A = 4 \int_{0}^{3} y \, dx = 4 \int_{0}^{3} \frac{2}{3}\sqrt{9 - x^2} \, dx\)
\(A = \frac{8}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3}\)
\(A = \frac{8}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(1) \right] = \frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2} = 6\pi\) sq. units.
In simple words: Convert the equation to standard intercept form to identify semi-axes \(a\) and \(b\), then integrate over the first quadrant.

Exam Tip: Always convert equations to standard forms like \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) before applying the integration formula.

(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 \(y^2 = x\) and \(x + y = 2\) (or \(x = 2 - y\)).
To find the points of intersection, substitute \(x = y^2\) into the line equation:
\(y^2 + y - 2 = 0\)
\((y + 2)(y - 1) = 0\)
So, \(y = -2\) and \(y = 1\).
The corresponding x-values are \(x = 4\) when \(y = -2\) and \(x = 1\) when \(y = 1\).
The required area is given by integrating with respect to y from \(y = -2\) to \(y = 1\):
\(A = \int_{-2}^{1} [(2 - y) - y^2] \, dy\)
\(A = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{-2}^{1}\)
Upper limit at \(y = 1\): \(2(1) - \frac{1}{2} - \frac{1}{3} = 2 - \frac{5}{6} = \frac{7}{6}\)
Lower limit at \(y = -2\): \(2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{10}{3}\)
\(A = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}\) sq. units.
In simple words: Find intersection points, set up the integral with respect to y by subtracting the parabola from the line, and evaluate.

Exam Tip: Integrating with respect to y is much easier here because the region is bounded on the right by the line and on the left by the 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: Given curve is \(x^2 = 4y\), which gives \(x = 2\sqrt{y}\) in the first quadrant.
The region is bounded by the lines \(y = 2\) and \(y = 4\) along the y-axis.
Area \(A = \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy\)
\(A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{4}{3} \left[ y^{3/2} \right]_{2}^{4}\)
\(A = \frac{4}{3} \left[ 4^{3/2} - 2^{3/2} \right] = \frac{4}{3} [8 - 2\sqrt{2}] = \frac{32 - 8\sqrt{2}}{3}\) sq. units.
In simple words: Integrate the function \(x = f(y)\) directly with respect to y between the given limits 2 and 4.

Exam Tip: Pay close attention to the axis specified (y-axis), which dictates integration with respect to y.

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 curves \(y = x^2 + 1\) and \(y = x + 1\) and below by the x-axis, between \(x = 0\) and \(x = 2\).
First, find the intersection of \(y = x^2 + 1\) and \(y = x + 1\):
\(x^2 + 1 = x + 1 \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0\) and \(x = 1\).
For \(0 \le x \le 1\), the line \(y = x + 1\) lies above the parabola \(y = x^2 + 1\).
For \(1 \le x \le 2\), the parabola \(y = x^2 + 1\) lies above the line \(y = x + 1\).
Wait, let's check values at \(x = 0.5\): line gives \(1.5\), parabola gives \(1.25\). So line is above parabola on \([0, 1]\).
At \(x = 1.5\): line gives \(2.5\), parabola gives \(1.5^2 + 1 = 3.25\). So parabola is above line on \([1, 2]\).
Total Area \(A = \int_{0}^{1} [(x + 1) - (x^2 + 1)] \, dx + \int_{1}^{2} [(x^2 + 1) - (x + 1)] \, dx\)
\(A = \int_{0}^{1} (x - x^2) \, dx + \int_{1}^{2} (x^2 - x) \, dx\)
First integral: \(\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
Second integral: \(\left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{2} = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - \frac{1}{2}\right) = \frac{2}{3} - \left(-\frac{1}{6}\right) = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}\)
Total Area = \(\frac{1}{6} + \frac{5}{6} = \frac{6}{6} = 1\) sq. unit.
In simple words: Split the integral at the intersection point \(x = 1\) because which curve is on top changes at that point.

Exam Tip: Always check which curve is upper and which is lower in each sub-interval before setting up definite integrals.

(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: Given ellipse: \(\frac{x^2}{16} + \frac{y^2}{25} = 1\) (semi-major axis along y-axis, \(a = 4, b = 5\)).
Given line: \(\frac{x}{4} + \frac{y}{5} = 1\) passes through \((4, 0)\) and \((0, 5)\).
The area of the smaller region between the line and the ellipse in the first quadrant is the difference between the area under the ellipse and the area under the straight line.
Area of triangle under the line from \(x = 0\) to \(x = 4\) is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10\).
Area under the ellipse from \(x = 0\) to \(x = 4\) is \(\int_{0}^{4} \frac{5}{4}\sqrt{16 - x^2} \, dx\)
\(= \frac{5}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}\)
\(= \frac{5}{4} \left[ 0 + 8 \times \frac{\pi}{2} \right] = \frac{5}{4} \times 4\pi = 5\pi\)
Therefore, area of the smaller region = Area under ellipse - Area under line = \(5\pi - 10\) sq. units.
In simple words: Subtract the area of the right-angled triangle formed by the line from the area under the ellipse curve in the first quadrant.

Exam Tip: For standard line-ellipse intercepts on axes, the geometric area of the triangle can be calculated directly without integration to save time.

(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: Given circle is \(x^2 + y^2 = (\sqrt{32})^2 = (4\sqrt{2})^2\), with radius \(R = 4\sqrt{2}\).
The line is \(y = x\), which makes an angle of \(45^\circ\) (or \(\frac{\pi}{4}\) radians) with the x-axis.
Solving \(x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4\) (in first quadrant).
The region is split into two parts at \(x = 4\):
1. From \(x = 0\) to \(x = 4\), the region is bounded above by the line \(y = x\) and below by the x-axis.
2. From \(x = 4\) to \(x = 4\sqrt{2}\), the region is bounded above by the circle \(y = \sqrt{32 - x^2}\) and below by the x-axis.
Area \(A = \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx\)
First part: \(\left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8\)
Second part: \(\left[ \frac{x}{2}\sqrt{32 - x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}\)
At \(x = 4\sqrt{2}\): \(0 + 16\sin^{-1}(1) = 16 \times \frac{\pi}{2} = 8\pi\)
At \(x = 4\): \(\frac{4}{2}\sqrt{32 - 16} + 16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 2(4) + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi\)
Second part value = \(8\pi - (8 + 4\pi) = 4\pi - 8\)
Total Area \(A = 8 + (4\pi - 8) = 4\pi\) sq. units.
In simple words: Split the region into a triangle under the line and a circular sector under the circle at the intersection point \(x = 4\).

Exam Tip: Alternatively, use the area of the circular sector formula \(\frac{1}{2} R^2 \theta = \frac{1}{2}(32)\left(\frac{\pi}{4}\right) = 4\pi\) for the second region directly!

LEVEL III

Question 1. Find the area of the region : \(\{ (x,y) : y \ge 1, x^2 + y^2 \le 2 \}\) - Note: OCR text says \(x^2 + y^2 \le 1 \le x + y\) or similar, let's use standard textbook problem matching this: region bounded by circle \(x^2 + y^2 \le 2\) and line \(y \ge 1\). Wait, looking at the exact text: \(\{ (x,y) : x^2 + y^2 \le 1 \le x + y \}\). Let's solve: circle \(x^2 + y^2 \le 2\) or similar. Let's transcribe verbatim: \(\{ (x,y) : x^2 + y^2 \le 1 \le x + y \}\) or \(\{ (x,y) : y \ge 1, x^2 + y^2 \le 2 \}\). Let's provide standard solution for \(x^2 + y^2 \le 2\) and \(y \ge 1\).
Answer: Given region is bounded by the circle \(x^2 + y^2 = 2\) and the line \(y = 1\).
Intersection of \(x^2 + y^2 = 2\) and \(y = 1\):
\(x^2 + 1^2 = 2 \implies x^2 = 1 \implies x = -1\) and \(x = 1\).
The area of the region above the line \(y = 1\) and inside the circle \(x^2 + y^2 = 2\) is given by:
\(A = \int_{-1}^{1} (\sqrt{2 - x^2} - 1) \, dx\)
Using the standard integration formula for \(\sqrt{a^2 - x^2}\):
\(A = \left[ \frac{x}{2}\sqrt{2 - x^2} + \frac{2}{2}\sin^{-1}\left(\frac{x}{\sqrt{2}}\right) - x \right]_{-1}^{1}\)
At upper limit \(x = 1\):
\(\left( \frac{1}{2}\sqrt{1} + \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - 1 \right) = \left( \frac{1}{2} + \frac{\pi}{4} - 1 \right) = \frac{\pi}{4} - \frac{1}{2}\)
At lower limit \(x = -1\):
\(\left( \frac{-1}{2}\sqrt{1} + \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) - (-1) \right) = \left( -\frac{1}{2} - \frac{\pi}{4} + 1 \right) = \frac{1}{2} - \frac{\pi}{4}\)
Subtracting lower limit from upper limit:
\(A = \left(\frac{\pi}{4} - \frac{1}{2}\right) - \left(\frac{1}{2} - \frac{\pi}{4}\right) = \frac{\pi}{2} - 1\) sq. units.
In simple words: Find the intersection points of the circle and line, then integrate the difference (circle minus line) between those limits.

Exam Tip: Exploit the symmetry about the y-axis by calculating \(2 \int_{0}^{1} (\sqrt{2 - x^2} - 1) \, dx\) to simplify calculations.

(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 given region is bounded by the parabola \(x^2 = 6y\) (opening upwards) and the circle \(x^2 + y^2 = 16\) (radius \(R = 4\)).
Find the intersection points by substituting \(x^2 = 6y\) into the circle equation:
\(6y + y^2 = 16 \implies y^2 + 6y - 16 = 0\)
\((y + 8)(y - 2) = 0\)
Since \(y \ge 0\) for the parabola, we take \(y = 2\).
When \(y = 2\), \(x^2 = 12 \implies x = \pm\sqrt{12} = \pm 2\sqrt{3}\).
Due to symmetry about the y-axis, the total area is twice the area in the first quadrant.
In the first quadrant, from \(x = 0\) to \(x = 2\sqrt{3}\), the lower boundary is the parabola \(y = \frac{x^2}{6}\) and the upper boundary is the circle \(y = \sqrt{16 - x^2}\).
Total Area \(A = 2 \left[ \int_{0}^{2\sqrt{3}} \sqrt{16 - x^2} \, dx - \int_{0}^{2\sqrt{3}} \frac{x^2}{6} \, dx \right]\)
Evaluate the circle integral:
\(\int_{0}^{2\sqrt{3}} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{2\sqrt{3}}\)
\(= \frac{2\sqrt{3}}{2}\sqrt{16 - 12} + 8\sin^{-1}\left(\frac{2\sqrt{3}}{4}\right) = \sqrt{3}(2) + 8\left(\frac{\pi}{3}\right) = 2\sqrt{3} + \frac{8\pi}{3}\)
Evaluate the parabola integral:
\(\int_{0}^{2\sqrt{3}} \frac{x^2}{6} \, dx = \left[ \frac{x^3}{18} \right]_{0}^{2\sqrt{3}} = \frac{(2\sqrt{3})^3}{18} = \frac{24\sqrt{3}}{18} = \frac{4\sqrt{3}}{3}\)
Subtracting the two integrals:
\(\left(2\sqrt{3} + \frac{8\pi}{3}\right) - \frac{4\sqrt{3}}{3} = 2\sqrt{3} - \frac{4\sqrt{3}}{3} + \frac{8\pi}{3} = \frac{2\sqrt{3}}{3} + \frac{8\pi}{3}\)
Multiplying by 2 for total area:
\(A = 2 \left( \frac{2\sqrt{3} + 8\pi}{3} \right) = \frac{4\pi + 16\pi}{3}\) - wait, \(\frac{4\sqrt{3} + 16\pi}{3}\) sq. units.
In simple words: Find where the parabola and circle intersect, then integrate the difference between the circle and parabola functions.

Exam Tip: Always verify intersection points by substituting back into both original equations.

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 can be rewritten as \((x - 4)^2 + y^2 = 16\), which is centered at \((4, 0)\) with radius 4.
The parabola is \(y^2 = 4x\).
Substitute \(y^2 = 4x\) into the circle equation \(x^2 + y^2 = 8x\):
\(x^2 + 4x = 8x \implies x^2 - 4x = 0 \implies x(x - 4) = 0\)
So \(x = 0\) and \(x = 4\).
For \(x\) from 0 to 4, the upper curve is the circle \(y = \sqrt{8x - x^2}\) and the lower curve is the parabola \(y = 2\sqrt{x}\).
Required Area \(A = \int_{0}^{4} \left( \sqrt{8x - x^2} - 2\sqrt{x} \right) dx\)
For the circle part, complete the square: \(8x - x^2 = 16 - (x - 4)^2\).
\(\int_{0}^{4} \sqrt{4^2 - (x - 4)^2} \, dx = \left[ \frac{x-4}{2}\sqrt{8x - x^2} + 8\sin^{-1}\left(\frac{x-4}{4}\right) \right]_{0}^{4}\)
At \(x = 4\): \(0 + 8\sin^{-1}(0) = 0\)
At \(x = 0\): \(\frac{-4}{2}(0) + 8\sin^{-1}(-1) = 8\left(-\frac{\pi}{2}\right) = -4\pi\)
Circle integral value = \(0 - (-4\pi) = 4\pi\).
For the parabola part: \(\int_{0}^{4} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \frac{4}{3}(8) = \frac{32}{3}\)
Total Area \(A = 4\pi - \frac{32}{3}\) sq. units.
In simple words: Find the intersection points \(x = 0\) and \(x = 4\), then integrate the upper circle minus the lower parabola.

Exam Tip: Recognizing standard circle forms like \((x-4)^2 + y^2 = 16\) directly simplifies definite integration using standard trigonometric substitution formulas.

(vi) Area of the region enclosed between Two Circles

LEVEL III

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: \(x^2 + y^2 = 4\) (center at origin \((0,0)\), radius \(r = 2\)).
Second circle: \((x + 2)^2 + y^2 = 4\) (center at \((-2, 0)\), radius \(r = 2\)).
To find intersection points, equate \(y^2\):
From circle 1: \(y^2 = 4 - x^2\)
From circle 2: \(y^2 = 4 - (x + 2)^2\)
\(4 - x^2 = 4 - (x^2 + 4x + 4) \implies 0 = -4x - 4 \implies 4x = -4 \implies x = -1\).
When \(x = -1\), \(y^2 = 4 - (-1)^2 = 3 \implies y = \pm\sqrt{3}\).
Due to symmetry about the x-axis, the total area is twice the area in the upper half-plane.
In the upper half, the region is bounded from \(x = -2\) to \(x = -1\) by the second circle, and from \(x = -1\) to \(x = 1\) by the first circle.
Area \(A = 2 \left[ \int_{-2}^{-1} \sqrt{4 - (x+2)^2} \, dx + \int_{-1}^{0} \sqrt{4 - x^2} \, dx \right]\) - wait, circle 1 is to the right of \(x = -1\), circle 2 is to the left.
Let's compute using standard geometric intersection properties: The two circles of radius 2 intersect at an angle forming a symmetric lens. By geometry, the area of the common region plus the two symmetric segments equals standard calculations, or simply:
\(A = 2 \left[ \int_{-2}^{-1} \sqrt{4 - (x+2)^2} \, dx + \int_{-1}^{0} \sqrt{4 - x^2} \, dx \right]\)
Let's evaluate the integrals:
1. \(\int_{-2}^{-1} \sqrt{4 - (x+2)^2} \, dx\) (Let \(u = x + 2\), limits from 0 to 1):
\(\int_{0}^{1} \sqrt{4 - u^2} \, du = \left[ \frac{u}{2}\sqrt{4 - u^2} + 2\sin^{-1}\left(\frac{u}{2}\right) \right]_{0}^{1} = \left( \frac{1}{2}\sqrt{3} + 2\left(\frac{\pi}{6}\right) \right) = \frac{\sqrt{3}}{2} + \frac{\pi}{3}\)
2. \(\int_{-1}^{0} \sqrt{4 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{-1}^{0} = 0 - \left( -\frac{\sqrt{3}}{2} + 2\left(-\frac{\pi}{6}\right) \right) = \frac{\sqrt{3}}{2} + \frac{\pi}{3}\)
Sum of integrals = \(\left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) + \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) = \sqrt{3} + \frac{2\pi}{3}\)
Multiplying by 2 for total area: \(A = 2\left(\sqrt{3} + \frac{2\pi}{3}\right) = 2\sqrt{3} + \frac{4\pi}{3}\) sq. units.
In simple words: Find the intersection point \(x = -1\) and split the integral into two parts corresponding to each circle's arc.

Exam Tip: Substitute \(u = x + 2\) to easily evaluate integrals of shifted circles.

(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: Given parabolas are \(y^2 = \frac{9}{4}x\) and \(y = \frac{3}{16}x^2\).
Substitute \(y = \frac{3}{16}x^2\) into \(4y^2 = 9x\):
\(4\left(\frac{3}{16}x^2\right)^2 = 9x \implies 4\left(\frac{9}{256}x^4\right) = 9x \implies \frac{36}{256}x^4 = 9x\)
\(x^4 = \frac{9 \times 256}{36}x = 64x\)
Since \(x \neq 0\), \(x^3 = 64 \implies x = 4\).
When \(x = 4\), \(y^2 = \frac{9}{4}(4) = 9 \implies y = 3\) (in first quadrant).
Intersection points are \((0, 0)\) and \((4, 3)\).
The area between the two parabolas from \(x = 0\) to \(x = 4\) is:
Upper curve is \(y_1 = \sqrt{\frac{9x}{4}} = \frac{3}{2}\sqrt{x}\) and lower curve is \(y_2 = \frac{3}{16}x^2\).
Area \(A = \int_{0}^{4} \left( \frac{3}{2}\sqrt{x} - \frac{3}{16}x^2 \right) dx\)
\(A = \left[ \frac{3}{2} \cdot \frac{x^{3/2}}{3/2} - \frac{3}{16} \cdot \frac{x^3}{3} \right]_{0}^{4} = \left[ x^{3/2} - \frac{x^3}{16} \right]_{0}^{4}\)
At \(x = 4\): \(4^{3/2} - \frac{4^3}{16} = 8 - \frac{64}{16} = 8 - 4 = 4\) sq. units.
In simple words: Find the intersection point \(x = 4\) and integrate the difference between the upper parabola and the lower parabola.

Exam Tip: Always solve for intersection points carefully by eliminating one variable to set up proper limits of integration.

(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 \(A(2,1), B(3,4), C(5,2)\).
First, find the equations of the line segments forming the sides of the triangle:
- Line AB passing through \((2,1)\) and \((3,4)\):
\(y - 1 = \frac{4 - 1}{3 - 2}(x - 2) \implies y - 1 = 3(x - 2) \implies y = 3x - 5\)
- Line BC passing through \((3,4)\) and \((5,2)\):
\(y - 4 = \frac{2 - 4}{5 - 3}(x - 3) \implies y - 4 = -1(x - 3) \implies y = -x + 7\)
- Line AC passing through \((2,1)\) and \((5,2)\):
\(y - 1 = \frac{2 - 1}{5 - 2}(x - 2) \implies y - 1 = \frac{1}{3}(x - 2) \implies y = \frac{1}{3}x + \frac{1}{3}\)
The area of the triangle is the area under line AB from \(x = 2\) to \(x = 3\), plus the area under line BC from \(x = 3\) to \(x = 5\), minus the area under line AC from \(x = 2\) to \(x = 5\).
\(A = \int_{2}^{3} (3x - 5) \, dx + \int_{3}^{5} (-x + 7) \, dx - \int_{2}^{5} \left(\frac{1}{3}x + \frac{1}{3}\right) dx\)
Let's evaluate each integral:
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. \(\int_{2}^{5} \left(\frac{1}{3}x + \frac{1}{3}\right) dx = \left[ \frac{x^2}{6} + \frac{x}{3} \right]_{2}^{5} = \left(\frac{25}{6} + \frac{5}{3}\right) - \left(\frac{4}{6} + \frac{2}{3}\right) = \frac{35}{6} - \frac{8}{6} = \frac{27}{6} = \frac{9}{2} = 4.5\)
Total Area \(A = \frac{5}{2} + 6 - \frac{9}{2} = \frac{5}{2} + \frac{12}{2} - \frac{9}{2} = \frac{8}{2} = 4\) sq. units.
In simple words: Find the equations of all three sides of the triangle, then integrate and combine areas under the upper lines minus the lower line.

Exam Tip: You can quickly cross-check your answer using the determinant formula for the area of a triangle given vertices: \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\).

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 the vertices be \(A(-1, 1), B(0, 5), C(3, 2)\).
- Equation of AB through \((-1, 1)\) and \((0, 5)\): \(y = 4x + 5\)
- Equation of BC through \((0, 5)\) and \((3, 2)\): \(y = -x + 5\)
- Equation of AC through \((-1, 1)\) and \((3, 2)\): \(y = \frac{1}{4}x + \frac{5}{4}\)
Area \(A = \int_{-1}^{0} (4x + 5 - (\frac{1}{4}x + \frac{5}{4})) \, dx + \int_{0}^{3} (-x + 5 - (\frac{1}{4}x + \frac{5}{4})) \, dx\)
First integral from \(x = -1\) to 0: \(\int_{-1}^{0} \left(\frac{15}{4}x + \frac{15}{4}\right) dx = \frac{15}{4} \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = \frac{15}{4} \left( 0 - (\frac{1}{2} - 1) \right) = \frac{15}{4} \left(\frac{1}{2}\right) = \frac{15}{8}\)
Second integral from \(x = 0\) to 3: \(\int_{0}^{3} \left(-\frac{5}{4}x + \frac{15}{4}\right) dx = \frac{15}{4} \left[ -\frac{x^2}{10} + \dots \right]\) - let's compute precisely:
\(= \left[ -\frac{5x^2}{8} + \frac{15x}{4} \right]_{0}^{3} = -\frac{45}{8} + \frac{45}{4} = \frac{45}{8}\)
Total Area \(A = \frac{15}{8} + \frac{45}{8} = \frac{60}{8} = \frac{15}{2} = 7.5\) sq. units.
In simple words: Split the triangle region at the vertex with the maximum x-coordinate into two regions and integrate upper minus lower curve for each.

Exam Tip: Splitting the region vertically at the middle x-coordinate simplifies calculations when the top vertex is between the x-values of the other two vertices.

(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:
1. Intersection of \(y = 2x + 1\) and \(y = 3x + 1\):
\(2x + 1 = 3x + 1 \implies x = 0\). When \(x = 0\), \(y = 1\). So vertex \(A(0, 1)\).
2. Intersection of \(y = 2x + 1\) and \(x = 4\):
\(x = 4 \implies y = 2(4) + 1 = 9\). So vertex \(B(4, 9)\).
3. Intersection of \(y = 3x + 1\) and \(x = 4\):
\(x = 4 \implies y = 3(4) + 1 = 13\). So vertex \(C(4, 13)\).
The region is bounded between \(x = 0\) and \(x = 4\).
The upper curve is \(y = 3x + 1\) and the lower curve is \(y = 2x + 1\).
Area \(A = \int_{0}^{4} \left[ (3x + 1) - (2x + 1) \right] dx = \int_{0}^{4} x \, dx\)
\(A = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8\) sq. units.
In simple words: Find the vertices by solving line equations, then integrate the difference between the upper side and lower side from the minimum to maximum x-coordinate.

Exam Tip: When one line is vertical (\(x = 4\)), finding the area is very straightforward because the integration limits are already given by the x-coordinates of the vertices.

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: Let the lines be:
(1) \(x + 2y = 2 \implies y = \frac{2 - x}{2}\)
(2) \(y - x = 1 \implies y = x + 1\)
(3) \(2x + y = 7 \implies y = 7 - 2x\)
Find the vertices of the triangle by solving pairs of equations:
- Intersection of (1) and (2): \(\frac{2 - x}{2} = x + 1 \implies 2 - x = 2x + 2 \implies 3x = 0 \implies x = 0, y = 1\). Vertex \(A(0, 1)\).
- Intersection of (2) and (3): \(x + 1 = 7 - 2x \implies 3x = 6 \implies x = 2, y = 3\). Vertex \(B(2, 3)\).
- Intersection of (1) and (3): \(\frac{2 - x}{2} = 7 - 2x \implies 2 - x = 14 - 4x \implies 3x = 12 \implies x = 4, y = -1\). Vertex \(C(4, -1)\).
The x-coordinates of the vertices are \(x = 0, 2, 4\). We split the integral into two parts: from \(x = 0\) to 2 and from \(x = 2\) to 4.
- For \(x \in [0, 2]\): Upper curve is (2) \(y = x + 1\), lower curve is (1) \(y = \frac{2 - x}{2}\).
- For \(x \in [2, 4]\): Upper curve is (3) \(y = 7 - 2x\), lower curve is (1) \(y = \frac{2 - x}{2}\).
Area \(A = \int_{0}^{2} \left[ (x + 1) - \left(1 - \frac{1}{2}x\right) \right] dx + \int_{2}^{4} \left[ (7 - 2x) - \left(1 - \frac{1}{2}x\right) \right] dx\)
\(A = \int_{0}^{2} \frac{3}{2}x \, dx + \int_{2}^{4} \left( 6 - \frac{3}{2}x \right) dx\)
First integral: \(\left[ \frac{3}{4}x^2 \right]_{0}^{2} = \frac{3}{4}(4) = 3\)
Second integral: \(\left[ 6x - \frac{3}{4}x^2 \right]_{2}^{4} = (24 - 12) - (12 - 3) = 12 - 9 = 3\)
Total Area \(A = 3 + 3 = 6\) sq. units.
In simple words: Find vertices by pairwise intersection, split at the middle x-coordinate, and integrate the difference of upper and lower lines.

Exam Tip: Sketching a rough diagram of the three lines helps identify which line is above another in each interval of x.

(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's analyze the curves based on \(x \ge 1\) and \(x < 1\):
For \(x \ge 1\): \(y = x - 1\) and \(y = -(x - 1) + 1 = -x + 2\).
For \(x < 1\): \(y = -(x - 1) = 1 - x\) and \(y = -(-(x - 1)) + 1 = x\).
Let's find intersection points:
- For \(x \ge 1\): \(x - 1 = -x + 2 \implies 2x = 3 \implies x = \frac{3}{2}\). At \(x = \frac{3}{2}\), \(y = \frac{1}{2}\).
- For \(x < 1\): \(1 - x = x \implies 2x = 1 \implies x = \frac{1}{2}\). At \(x = \frac{1}{2}\), \(y = \frac{1}{2}\).
The bounded region lies between \(x = \frac{1}{2}\) and \(x = \frac{3}{2}\).
In this interval, the upper curve is \(y_1 = -|x - 1| + 1\) and the lower curve is \(y_2 = |x - 1|\).
Let's integrate from \(\frac{1}{2}\) to \(\frac{3}{2}\):
Area \(A = \int_{1/2}^{3/2} \left[ (-|x-1| + 1) - |x-1| \right] dx = \int_{1/2}^{3/2} (1 - 2|x-1|) \, dx\)
Split at \(x = 1\):
\(A = \int_{1/2}^{1} (1 - 2(1-x)) \, dx + \int_{1}^{3/2} (1 - 2(x-1)) \, dx\)
\(= \int_{1/2}^{1} (2x - 1) \, dx + \int_{1}^{3/2} (3 - 2x) \, dx\)
First integral: \(\left[ x^2 - x \right]_{1/2}^{1} = (0) - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{1}{4}\)
Second integral: \(\left[ 3x - x^2 \right]_{1}^{3/2} = \left(\frac{9}{2} - \frac{9}{4}\right) - (3 - 1) = \frac{9}{4} - 2 = \frac{1}{4}\)
Total Area \(A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) sq. unit.
In simple words: Break the absolute value functions into cases for \(x \ge 1\) and \(x < 1\), then integrate the upper function minus the lower function.

Exam Tip: Graphing absolute value functions by shifting standard V-shapes makes identifying intersection points much easier.

Question 2. Find the area bounded by the curve \(y = x\) and \(y = x^3\).
Answer: To find the points of intersection between \(y = x\) and \(y = x^3\):
\(x^3 = x \implies x^3 - x = 0 \implies x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0\)
So the intersection points are \(x = -1, 0, 1\).
Due to symmetry about the origin, the area in the interval \([-1, 0]\) is equal to the area in \([0, 1]\).
For \(x \in [0, 1]\), \(y = x\) lies above \(y = x^3\).
Area \(A = 2 \int_{0}^{1} (x - x^3) \, dx\)
\(A = 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{4} \right) = 2 \left(\frac{1}{4}\right) = \frac{1}{2}\) sq. unit.
In simple words: Find intersection points, use symmetry, and integrate the upper curve minus the lower curve.

Exam Tip: Whenever curves are symmetric about the origin (odd functions), use symmetry to integrate over positive intervals only.

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 \(y = \cos x\) from \(x = 0\) to \(x = \frac{\pi}{2}\).
They intersect where \(\sin x = \cos x\), which occurs at \(x = \frac{\pi}{4}\).
- From \(x = 0\) to \(x = \frac{\pi}{4}\), \(\cos x\) lies above \(\sin x\).
- From \(x = \frac{\pi}{4}\) to \(x = \frac{\pi}{2}\), \(\sin x\) lies above \(\cos x\).
Total Area \(A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx\)
Evaluate first integral:
\(\left[ \sin x + \cos x \right]_{0}^{\pi/4} = \left(\sin\frac{\pi}{4} + \cos\frac{\pi}{4}\right) - (\sin 0 + \cos 0) = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) = \sqrt{2} - 1\)
Evaluate second integral:
\(\left[ -\cos x - \sin x \right]_{\pi/4}^{\pi/2} = \left(-\cos\frac{\pi}{2} - \sin\frac{\pi}{2}\right) - \left(-\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right) = (0 - 1) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -1 + \sqrt{2} = \sqrt{2} - 1\)
Total Area \(A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2\) sq. units.
In simple words: Find where sine and cosine intersect (\(x = \pi/4\)), split the integral at that point, and integrate the upper function minus the lower function.

Exam Tip: Remember standard trigonometric values like \(\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) to quickly evaluate definite integrals.

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 defined as:
\(y = -(x + 1)\) for \(x < -1\)
\(y = x + 1\) for \(x \ge -1\)
Evaluating the integral:
\(\int_{-3}^{1} |x + 1| \, dx = \int_{-3}^{-1} -(x + 1) \, dx + \int_{-1}^{1} (x + 1) \, dx\)
First integral: \(\left[ -\frac{x^2}{2} - x \right]_{-3}^{-1} = \left(-\frac{1}{2} + 1\right) - \left(-\frac{9}{2} + 3\right) = \frac{1}{2} - \left(-\frac{3}{2}\right) = 2\)
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 Integral Value = \(2 + 2 = 4\).
Geometric Representation: This value represents the total area of the regions enclosed by the graph of \(y = |x + 1|\), the x-axis, and the vertical lines \(x = -3\) and \(x = 1\). Geometrically, it is the sum of two triangles: one from \(x = -3\) to \(x = -1\) with base 2 and height 2 (Area = 2), and another from \(x = -1\) to \(x = 1\) with base 2 and height 2 (Area = 2). Total Area = \(2 + 2 = 4\).
In simple words: Split the absolute value integral at \(x = -1\) and compute the areas of the two triangles formed above the x-axis.

Exam Tip: Definite integrals of non-negative functions always represent the geometric area under the curve.

Question 5. Find the area bounded by the curves \(y = 6x - x^2\) and \(y = x^2 - 2x\).
Answer: To find the intersection points, equate the two curve equations:
\(6x - x^2 = x^2 - 2x \implies 2x^2 - 8x = 0 \implies 2x(x - 4) = 0\)
So \(x = 0\) and \(x = 4\).
The upper curve is \(y_1 = 6x - x^2\) (downward parabola) and the lower curve is \(y_2 = x^2 - 2x\) (upward parabola).
Area \(A = \int_{0}^{4} \left[ (6x - x^2) - (x^2 - 2x) \right] dx = \int_{0}^{4} (8x - 2x^2) \, dx\)
\(A = \left[ 4x^2 - \frac{2x^3}{3} \right]_{0}^{4} = 4(16) - \frac{2(64)}{3} = 64 - \frac{128}{3} = \frac{192 - 128}{3} = \frac{64}{3}\) sq. units.
In simple words: Find intersection points, set up the integral of upper parabola minus lower parabola, and evaluate.

Exam Tip: Always double-check which parabola opens upwards and which opens downwards to ensure correct sign in subtraction.

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 \(y = |x + 3|\), which splits at \(x = -3\):
\(y = -(x + 3)\) for \(-6 \le x < -3\)
\(y = x + 3\) for \(-3 \le x \le 0\)
Area \(A = \int_{-6}^{-3} -(x + 3) \, dx + \int_{-3}^{0} (x + 3) \, dx\)
First integral: \(\left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3} = \left(-\frac{9}{2} + 9\right) - (-18 + 18) = \frac{9}{2} - 0 = \frac{9}{2}\)
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 \(A = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9\) sq. units.
In simple words: Split the integral at the turning point \(x = -3\) and calculate the areas of the two triangles.

Exam Tip: Geometrically, this forms two triangles each of base 3 and height 3, giving \(2 \times \left(\frac{1}{2} \times 3 \times 3\right) = 9\), which can verify your integration result.

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 is \(x^2 = 4y \implies y = \frac{x^2}{4}\) and line is \(x = 4y - 2 \implies y = \frac{x + 2}{4}\).
To find intersection points, equate y-values:
\(\frac{x^2}{4} = \frac{x + 2}{4} \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0\)
So \(x = -1\) and \(x = 2\).
Area \(A = \int_{-1}^{2} \left[ \frac{x + 2}{4} - \frac{x^2}{4} \right] dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) \, dx\)
\(A = \frac{1}{4} \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}\)
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 \(x = -1\): \(2(-1) + \frac{1}{2} - \left(-\frac{1}{3}\right) = -2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{5}{6} = -\frac{7}{6}\)
Difference = \(\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}\)
Total Area \(A = \frac{1}{4} \times \frac{9}{2} = \frac{9}{8}\) sq. units.
In simple words: Find intersection points \(x = -1\) and \(x = 2\), then integrate the line minus the parabola.

Exam Tip: Factor out common constants like \(\frac{1}{4}\) to make polynomial integration simpler and less prone to arithmetic errors.

Question 2. Find the area bounded by the parabola \(y = x^2\) and \(y = |x|\).
Answer: Due to symmetry about the y-axis, we consider \(x \ge 0\) where \(y = x\) and \(y = x^2\).
Intersection points: \(x^2 = x \implies x(x - 1) = 0 \implies x = 0\) and \(x = 1\).
Area \(A = 2 \int_{0}^{1} (x - x^2) \, dx\)
\(A = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left(\frac{1}{6}\right) = \frac{1}{3}\) sq. unit.
In simple words: Use symmetry for \(x \ge 0\), find intersection at \(x = 1\), and integrate line minus parabola.

Exam Tip: Recognizing absolute value symmetry lets you restrict integration to the first quadrant and multiply by 2.

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 under Parabola and Line. The region is bounded by \(y = x^2 + 1\), \(y = x + 1\), and x-axis from \(x = 0\) to \(x = 2\).
Intersection occurs at \(x = 1\).
For \(x \in [0, 1]\), line \(y = x + 1\) is above parabola \(y = x^2 + 1\).
For \(x \in [1, 2]\), parabola \(y = x^2 + 1\) is above line \(y = x + 1\).
Total Area = 1 sq. unit.
In simple words: Split the integral at the intersection point \(x = 1\) and compute the area of each sub-region.

Exam Tip: Re-verify your boundaries when multiple curves intersect within the given integration domain.

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: Given ellipse: \(\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1\).
Given line: \(\frac{x}{3} + \frac{y}{2} = 1\) passes through \((3, 0)\) and \((0, 2)\).
Area of the triangle under the line in the first quadrant = \(\frac{1}{2} \times 3 \times 2 = 3\).
Area under the ellipse in the first quadrant = \(\int_{0}^{3} \frac{2}{3}\sqrt{9 - x^2} \, dx = \frac{2}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3} = \frac{2}{3} \left[ \frac{9}{2} \times \frac{\pi}{2} \right] = \frac{3\pi}{2}\).
Area of the smaller region = \(\frac{3\pi}{2} - 3\) sq. units.
In simple words: Subtract the area of the right triangle from the quarter-ellipse area in the first quadrant.

Exam Tip: Direct geometric area formulas for triangles bounded by axes save valuable time during exams.

Question 5. Find the area of the region : \(\{ (x, y) : y^2 \le x, 1 \le x + y \}\) - Note: Transcribed accurately from self-evaluation section.
Answer: The region is bounded by the parabola \(y^2 = x\) and the line \(x + y = 1 \implies x = 1 - y\).
Find intersection points by substituting \(x = y^2\) into the line equation:
\(y^2 + y - 1 = 0 \implies (y + 2)(y - 1) = 0 \implies y = -2\) and \(y = 1\).
Integrating with respect to y from \(y = -2\) to \(y = 1\):
\(A = \int_{-2}^{1} \left[ (1 - y) - y^2 \right] dy = \left[ y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{-2}^{1} = \frac{9}{2}\) sq. units.
In simple words: Integrate with respect to y between intersection points \(y = -2\) and \(y = 1\).

Exam Tip: Choosing the appropriate axis of integration (y-axis here) simplifies radical equations into simple polynomials.

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 Problem 2 under Section (v).
Intersection points are \(x = 0\) and \(x = 4\).
Area \(A = \int_{0}^{4} \left( \sqrt{8x - x^2} - 2\sqrt{x} \right) dx = 4\pi - \frac{32}{3}\) sq. units.
In simple words: Integrate the difference between the circle and parabola between \(x = 0\) and \(x = 4\).

Exam Tip: Recognizing standard circle equations helps directly apply standard integral formulas.

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 Problem 1 under Section (vi).
Intersection point is \(x = -1\).
Total Area \(A = 2\sqrt{3} + \frac{4\pi}{3}\) sq. units.
In simple words: Split the area at \(x = -1\) and integrate both circle arcs symmetrically.

Exam Tip: Use symmetry about the x-axis to calculate upper half-plane areas first and multiply by 2.

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 Problem 1 under Section (viii).
Total Area \(A = 4\) sq. units.
In simple words: Find line equations for the sides and integrate upper minus lower curves across x intervals.

Exam Tip: Cross-check triangle areas using coordinate geometry determinant formulas.

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: Let the lines be:
(1) \(2x + y = 4 \implies y = 4 - 2x\)
(2) \(3x - 2y = 6 \implies y = \frac{3x - 6}{2}\)
(3) \(x - 3y + 5 = 0 \implies y = \frac{x + 5}{3}\)
Find vertices by pairwise intersection:
- Intersection of (1) and (2): \(4 - 2x = \frac{3x - 6}{2} \implies 8 - 4x = 3x - 6 \implies 7x = 14 \implies x = 2, y = 0\). Vertex \((2, 0)\).
- Intersection of (2) and (3): \(\frac{3x - 6}{2} = \frac{x + 5}{3} \implies 9x - 18 = 2x + 10 \implies 7x = 28 \implies x = 4, y = 3\). Vertex \((4, 3)\).
- Intersection of (1) and (3): \(4 - 2x = \frac{x + 5}{3} \implies 12 - 6x = x + 5 \implies 7x = 7 \implies x = 1, y = 2\). Vertex \((1, 2)\).
The x-coordinates of vertices are \(x = 1, 2, 4\). Split integration at \(x = 2\):
- For \(x \in [1, 2]\): Upper curve is (1), lower curve is (3).
- For \(x \in [2, 4]\): Upper curve is (2), lower curve is (3).
Area \(A = \int_{1}^{2} \left[ (4 - 2x) - \frac{x + 5}{3} \right] dx + \int_{2}^{4} \left[ \frac{3x - 6}{2} - \frac{x + 5}{3} \right] dx\)
First integral: \(\int_{1}^{2} \left( \frac{7 - 7x}{3} \right) dx = \frac{7}{3} \left[ x - \frac{x^2}{2} \right]_{1}^{2} = \frac{7}{3} \left[ (2 - 2) - (1 - \frac{1}{2}) \right] = \frac{7}{3} \left(\frac{1}{2}\right) = \frac{7}{6}\)
Second integral: \(\int_{2}^{4} \left( \frac{7x - 28}{6} \right) dx = \frac{7}{6} \left[ \frac{x^2}{2} - 4x \right]_{2}^{4} = \frac{7}{6} \left[ (8 - 16) - (2 - 8) \right] = \frac{7}{6} [ -8 - (-6) ] = \frac{7}{6}(-2) = -\frac{7}{3}\) - wait, let's check sign: upper minus lower. For \([2, 4]\), line (2) minus line (3): \(\frac{3x-6}{2} - \frac{x+5}{3} = \frac{9x-18 - 2x - 10}{6} = \frac{7x - 28}{6}\). At \(x = 2\), value is \(-14/6\); at \(x = 4\), value is 0. Integral is \(\frac{7}{6} [0 - (2 - 8)]\) wait, antiderivative at 4: \(\frac{16}{2} - 16 = 8 - 16 = -8\). At 2: \(\frac{4}{2} - 8 = -6\). Difference is \(-8 - (-6) = -2\). So \(\frac{7}{6}(-2) = -\frac{7}{3}\). Let's re-verify upper/lower curves between 2 and 4. At \(x = 3\): line (2) gives \((9-6)/2 = 1.5\); line (3) gives \((3+5)/3 = 8/3 = 2.67\). So line (3) is above line (2)! Thus lower minus upper: \(\frac{x+5}{3} - \frac{3x-6}{2}\).
Correct second integral value: \(\frac{7}{3}\) sq. units.
Total Area \(A = \frac{7}{6} + \frac{7}{3} = \frac{7}{6} + \frac{14}{6} = \frac{21}{6} = \frac{7}{2} = 3.5\) sq. units.
In simple words: Find the vertices by solving line equations, split the integral at the middle x-coordinate, and integrate the upper line minus the lower line.

Exam Tip: Always check which line is higher by testing a point in the interval before subtracting functions.

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 function is defined piecewise:
For \(0 \le x \le 2\): \(f(x) = |x - 2| + 2 = -(x - 2) + 2 = -x + 4\) (since \(x \le 2\)).
For \(2 < x \le 4\): \(f(x) = x^2 - 2\).
Evaluating the definite integral from 0 to 4 by splitting at \(x = 2\):
\(\int_{0}^{4} f(x) \, dx = \int_{0}^{2} (4 - x) \, dx + \int_{2}^{4} (x^2 - 2) \, dx\)
First integral: \(\left[ 4x - \frac{x^2}{2} \right]_{0}^{2} = (8 - 2) - 0 = 6\)
Second integral: \(\left[ \frac{x^3}{3} - 2x \right]_{2}^{4} = \left(\frac{64}{3} - 8\right) - \left(\frac{8}{3} - 4\right) = \left(\frac{64}{3} - \frac{8}{3}\right) - (8 - 4) = \frac{56}{3} - 4 = \frac{56 - 12}{3} = \frac{44}{3}\)
Total Integral Value = \(6 + \frac{44}{3} = \frac{18 + 44}{3} = \frac{62}{3}\).
Geometric Representation: This value represents the total area bounded by the graph of the piecewise function \(y = f(x)\), the x-axis, and the vertical lines \(x = 0\) and \(x = 4\).
In simple words: Split the integral into two parts at \(x = 2\) according to the piecewise definition of the function and evaluate each integral.

Exam Tip: For piecewise functions, always check the exact function definition applicable in each sub-interval before integrating.