ML Aggarwal Class 12 Maths Solutions Section A Chapter 11 Applications Of Definite Integrals

Chapter 11: Applications of Definite Integrals

11.1 Areas of Bounded Regions

When a function f is continuous and non-negative across a closed interval [a, b], the area underneath the curve y = f(x), sitting above the x-axis and positioned between the vertical lines x = a and x = b can be determined using the formula:

\( \int_a^b f(x) \, dx \) or \( \int_a^b y \, dx \)

The proof shows that if we take a point P(x, y) on the curve y = f(x) and a nearby point Q(x + δx, y + δy), we can establish bounds using rectangles. The area of rectangle PMNR is yδx, while the area of rectangle SMNQ is (y + δy)δx. Since the true area of region PMNQ falls between these two rectangles, we get yδx ≤ δA ≤ (y + δy)δx. Dividing by δx yields y ≤ δA/δx ≤ y + δy. When we take the limit as δx approaches 0, this gives us dA/dx = y. Integrating both sides with respect to x from a to b produces the desired result: the area equals [A]ₐᵇ, which represents the total enclosed area.

When the function f is continuous and non-positive (meaning the curve lies below the x-axis) throughout [a, b], the definite integral \( \int_a^b f(x) \, dx \) yields a negative value. Since area must always be positive, the area is instead given by \( \left| \int_a^b f(x) \, dx \right| \) or \( \left| \int_a^b y \, dx \right| \).

More generally, when the curve y = f(x) is continuous but may cross the x-axis at certain points, the area enclosed is found using \( \left| \int_a^b f(x) \, dx \right| \) or \( \left| \int_a^b y \, dx \right| \).

Similarly, for a curve x = g(y) that remains continuous without crossing the y-axis, the area bounded by this curve, the y-axis, and the horizontal lines y = c and y = d is given by:

\( \left| \int_c^d g(y) \, dy \right| \) or \( \left| \int_c^d x \, dy \right| \)

Important Remark: When the sign of f(x) is uncertain, \( \int_a^b f(x) \, dx \) might not represent the true enclosed area between the curve y = f(x), the x-axis, and the ordinates x = a and x = b. However, \( \int_a^b |f(x)| \, dx \) always provides the correct enclosed area.

For example, consider \( \int_{-1}^1 x \, dx \) and \( \int_{-1}^1 |x| \, dx \). The first integral equals \( \left[ \frac{x^2}{2} \right]_{-1}^1 = \frac{1}{2}(1 - 1) = 0 \), whereas the second integral is calculated by splitting at x = 0: \( \int_{-1}^0 (-x) \, dx + \int_0^1 x \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 + \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \). The actual enclosed area between y = x, the x-axis, and the ordinates x = -1 and x = 1 is not zero.

Therefore, if the graph of a function f is continuous on [a, b] and crosses the x-axis at finitely many points, the area between the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by \( \left| \int_a^b f(x) \, dx \right| \) or \( \left| \int_a^b y \, dx \right| \).

11.1.1 Area Bounded Between Curves

When two functions f(x) and g(x) are both continuous on [a, b] with 0 ≤ g(x) ≤ f(x) for all x in [a, b], the area of the region lying between the graphs of y = f(x), y = g(x), and the vertical lines x = a and x = b equals:

\( \int_a^b f(x) \, dx - \int_a^b g(x) \, dx = \int_a^b (f(x) - g(x)) \, dx \)

Similarly, the area between the curves x = f(y), x = g(y) and the horizontal lines y = c, y = d is:

\( \int_c^d (f(y) - g(y)) \, dy \)

Remarks:

1. If f(x) and g(x) remain continuous on [a, b] with g(x) ≤ f(x) for all x in [a, b], the formula still works even when one or both curves lie partially or completely below the x-axis.

2. When the two curves intersect each other at finitely many points, the area enclosed between them and the ordinates x = a and x = b is found using \( \left| \int_a^b (f(x) - g(x)) \, dx \right| \).

3. Similarly, the area between x = f(y), x = g(y) and the abscissae y = c, y = d is given by \( \left| \int_c^d (f(y) - g(y)) \, dy \right| \).

Example 1. Find the area of the region bounded by y² = 4x, x = 1, x = 4 and the x-axis in the first quadrant.

Solution: The curve y² = 4x represents a parabola opening to the right with its vertex at the origin (0, 0). In the first quadrant, y = 2√x. The required area is:

\( \int_1^4 y \, dx = \int_1^4 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_1^4 = 2 \cdot \frac{2}{3} \left[ x^{3/2} \right]_1^4 = \frac{4}{3} \left[ 8 - 1 \right] = \frac{28}{3} \text{ sq. units} \)

Example 2. Draw a rough sketch of the curve x² + y = 9 and find the area enclosed by the curve, the x-axis and the lines x + 1 = 0 and x - 2 = 0.

Solution: The equation x² + y = 9 can be rewritten as (x - 0)² = -(y - 9), which represents a downward-opening parabola with vertex at (0, 9). Setting y = 0 gives x² = 9, so x = -3 or x = 3 (where the parabola meets the x-axis). The lines are x = -1 and x = 2. The required area is:

\( \int_{-1}^2 y \, dx = \int_{-1}^2 (9 - x^2) \, dx = \left[ 9x - \frac{x^3}{3} \right]_{-1}^2 = \left( 18 - \frac{8}{3} \right) - \left( -9 + \frac{1}{3} \right) = \left( \frac{46}{3} \right) - \left( -\frac{26}{3} \right) = 24 \text{ sq. units} \)

Example 3. Determine the area enclosed between the curve y = 4x - x² and the x-axis.

Solution: The equation y = 4x - x² can be written as (x - 2)² = -(y - 4), representing a downward parabola with vertex at (2, 4). Setting y = 0 gives 4x - x² = 0, so x = 0 or x = 4. The area is:

\( \int_0^4 (4x - x^2) \, dx = \left[ 4 \cdot \frac{x^2}{2} - \frac{x^3}{3} \right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3} \text{ sq. units} \)

Alternatively, using symmetry about x = 2: \( 2 \int_0^2 (4x - x^2) \, dx = 2 \left[ 8 - \frac{8}{3} \right] = 2 \cdot \frac{16}{3} = \frac{32}{3} \text{ sq. units} \)

Remark: For symmetric closed regions, find the area of the smallest part and multiply by the number of symmetric parts.

Example 4. Draw a rough sketch of the curve y² + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Solution: The equation y² + 1 = x can be written as y² = x - 1, representing a parabola opening rightward with vertex at A(1, 0). When x = 2, we get y² = 1, giving y = 1 or y = -1. Since the area is symmetric about the x-axis:

\( \text{Required area} = 2 \int_1^2 y \, dx = 2 \int_1^2 \sqrt{x - 1} \, dx = 2 \left[ \frac{(x-1)^{3/2}}{3/2} \right]_1^2 = \frac{4}{3} \left[ 1^{3/2} - 0 \right] = \frac{4}{3} \text{ sq. units} \)

Example 5. Draw a rough sketch of the curve y = x² - 5x + 6 and find the area bounded by the curve and the x-axis.

Solution: Completing the square: y = x² - 5x + 6 becomes \( \left( x - \frac{5}{2} \right)^2 = y + \frac{1}{4} \), representing an upward parabola with vertex at \( \left( \frac{5}{2}, -\frac{1}{4} \right) \). Setting y = 0: x² - 5x + 6 = 0 gives (x - 2)(x - 3) = 0, so x = 2 or x = 3. The portion between these points lies below the x-axis, so:

\( \text{Required area} = \left| \int_2^3 (x^2 - 5x + 6) \, dx \right| = \left| \left[ \frac{x^3}{3} - \frac{5x^2}{2} + 6x \right]_2^3 \right| = \left| \left( 9 - \frac{45}{2} + 18 \right) - \left( \frac{8}{3} - 10 + 12 \right) \right| = \frac{1}{6} \text{ sq. units} \)

Example 6. Find the area of the region bounded by the curve y² = 4x, y-axis and the line y = 3.

Solution: From y² = 4x, we get x = y²/4. The area is:

\( \int_0^3 x \, dy = \int_0^3 \frac{y^2}{4} \, dy = \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3 = \frac{1}{12} \cdot 27 = \frac{9}{4} \text{ sq. units} \)

Example 7. Find the area of the region bounded by the curve y = x² and the line y = 4.

Solution: From y = x², we get x = √y in the first quadrant. The curves intersect at x = 2 (where y = 4). Using symmetry about the y-axis:

\( \text{Required area} = 2 \int_0^4 x \, dy = 2 \int_0^4 \sqrt{y} \, dy = 2 \left[ \frac{2y^{3/2}}{3} \right]_0^4 = \frac{4}{3} \left[ 8 - 0 \right] = \frac{32}{3} \text{ sq. units} \)

Example 8. Sketch and shade the area of the region lying in the first quadrant and bounded by y = 9x², x = 0, y = 1 and y = 4. Find the area of the shaded region.

Solution: The equation y = 9x² can be written as x² = y/9, so x = √(y/9) = √y/3 in the first quadrant. The required area is:

\( \int_1^4 x \, dy = \int_1^4 \frac{\sqrt{y}}{3} \, dy = \frac{1}{3} \left[ \frac{2y^{3/2}}{3} \right]_1^4 = \frac{2}{9} \left[ 4^{3/2} - 1^{3/2} \right] = \frac{2}{9} \left[ 8 - 1 \right] = \frac{14}{9} \text{ sq. units} \)

Example 9. Find the area bounded by the curve x = 8 + 2y - y², the y-axis and the lines y = -1, y = 3.

Solution: The equation x = 8 + 2y - y² can be rewritten as (y - 1)² = -(x - 9), representing a parabola opening leftward with vertex at (9, 1). The area is:

\( \int_{-1}^3 x \, dy = \int_{-1}^3 (8 + 2y - y^2) \, dy = \left[ 8y + y^2 - \frac{y^3}{3} \right]_{-1}^3 = (24 + 9 - 9) - (-8 + 1 + \frac{1}{3}) = \frac{92}{3} \text{ sq. units} \)

Example 10. Draw a rough sketch of the graph of the function y = 2√(1 - x²), x ∈ [0, 1] and evaluate the area enclosed between the curve and the axes.

Solution: Squaring: y² = 4(1 - x²), which gives x² + y²/4 = 1. This is an ellipse in standard form. The curve y = 2√(1 - x²) represents the upper half of the ellipse in the first quadrant, with endpoints (0, 2) and (1, 0). The area is:

\( \int_0^1 y \, dx = \int_0^1 2\sqrt{1 - x^2} \, dx = 2 \left[ \frac{x\sqrt{1-x^2}}{2} + \frac{1}{2}\sin^{-1}(x) \right]_0^1 = 2 \left[ (0 + \frac{\pi}{4}) - 0 \right] = \frac{\pi}{2} \text{ sq. units} \)

Example 11. Find the area bounded by the ellipse x²/a² + y²/b² = 1 and the ordinates x = 0 and x = ae where b² = a²(1 - e²) and 0 < e < 1.

Solution: From the ellipse equation, \( y = \frac{b}{a}\sqrt{a^2 - x^2} \) in the first quadrant. Using symmetry about the x-axis:

\( \text{Required area} = 2 \int_0^{ae} y \, dx = 2 \int_0^{ae} \frac{b}{a}\sqrt{a^2 - x^2} \, dx \)

\( = 2 \cdot \frac{b}{a} \left[ \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_0^{ae} \)

\( = ab \left[ e\sqrt{1-e^2} + \sin^{-1}(e) \right] \text{ sq. units} \)

Example 12. The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Solution: The curve y² = x represents a rightward parabola. The total area bounded by the parabola and the line x = 4 is:

\( 2 \int_0^4 \sqrt{x} \, dx = 2 \left[ \frac{2x^{3/2}}{3} \right]_0^4 = \frac{4}{3} \left[ 8 - 0 \right] = \frac{32}{3} \text{ sq. units} \)

If the line x = a divides this into two equal parts, each part has area 16/3:

\( 2 \int_0^a \sqrt{x} \, dx = \frac{16}{3} \implies \frac{4}{3} \left[ a^{3/2} - 0 \right] = \frac{16}{3} \)

\( a^{3/2} = 4 \implies a = 4^{2/3} = \sqrt[3]{16} \)

Example 13. Find the area of the smaller region bounded by the ellipse x²/a² + y²/b² = 1 and the straight line x/a + y/b = 1.

Solution: From the ellipse, \( y = \frac{b}{a}\sqrt{a^2 - x^2} \) in the first quadrant. From the line, \( y = \frac{b}{a}(a - x) \). The required area between these two curves is:

\( \int_0^a \left[ \frac{b}{a}\sqrt{a^2 - x^2} - \frac{b}{a}(a - x) \right] dx = \frac{b}{a} \left[ \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) - ax + \frac{x^2}{2} \right]_0^a \)

\( = \frac{b}{a} \left[ 0 + \frac{a^2\pi}{4} - a^2 + \frac{a^2}{2} \right] = \frac{1}{4}ab(\pi - 2) \text{ sq. units} \)

Example 14. Find the area of the region included between the curve 4y = 3x² and the line 2y = 3x + 12.

Solution: The curve is y = (3/4)x² (upward parabola) and the line is y = (3x + 12)/2. Setting them equal:

\( \frac{3x + 12}{2} = \frac{3x^2}{4} \implies 6x + 24 = 3x^2 \implies x^2 - 2x - 8 = 0 \)

\( (x + 2)(x - 4) = 0 \implies x = -2 \text{ or } x = 4 \)

The intersection points are P(-2, 3) and Q(4, 12). The required area is:

\( \int_{-2}^4 \left[ \frac{3x + 12}{2} - \frac{3x^2}{4} \right] dx = \left[ \frac{3x^2}{4} + 6x - \frac{x^3}{4} \right]_{-2}^4 = \frac{1}{4}[(48 + 96 - 64) - (12 - 48 + 8)] = \frac{1}{4} \cdot 108 = 27 \text{ sq. units} \)

Example 15. Find the area enclosed by the parabola y² = x and the line y + x = 2.

Solution: From y² = x and y + x = 2, we get y² = 2 - y, so y² + y - 2 = 0. This factors as (y - 1)(y + 2) = 0, giving y = 1 or y = -2. The corresponding x-values are x = 1 and x = 4. The intersection points are P(1, 1) and Q(4, -2). The area is:

\( \int_{-2}^1 [(2 - y) - y^2] dy = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{-2}^1 = \left( 2 - \frac{1}{2} - \frac{1}{3} \right) - \left( -4 - 2 + \frac{8}{3} \right) = \frac{9}{2} \text{ sq. units} \)

Example 16. Find the area bounded by the curve y = 2x - x² and the line y = x.

Solution: The curve y = 2x - x² can be written as (y - 1) = -(x - 1)², a downward parabola with vertex at (1, 1). Setting y = 2x - x² equal to y = x:

\( x = 2x - x^2 \implies x^2 - x = 0 \implies x = 0 \text{ or } x = 1 \)

The intersection points are O(0, 0) and P(1, 1). The required area is:

\( \int_0^1 [(2x - x^2) - x] dx = \int_0^1 (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \text{ sq. units} \)

Example 17. Find the area enclosed by the curve y = -x² and the line x + y + 2 = 0.

Solution: The curve is y = -x² (downward parabola) and the line is y = -(x + 2). Setting them equal:

\( -x^2 = -(x + 2) \implies x^2 - x - 2 = 0 \implies (x + 1)(x - 2) = 0 \)

\( x = -1 \text{ or } x = 2 \)

The intersection points are P(-1, -1) and Q(2, -4). Since the required region lies below the x-axis:

\( \text{Required area} = \left| \int_{-1}^2 [-(x + 2) - (-x^2)] dx \right| = \left| \int_{-1}^2 (x^2 - x - 2) dx \right| = \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^2 \right| = \frac{9}{2} \text{ sq. units} \)

Exercise 11.1

This section contains 19 exercise problems asking students to find areas bounded by various curves, lines, ellipses, and circles using integration techniques covered in the chapter. Problems range from simple parabolic regions to more complex areas involving ellipses, circles, and combined curve-line boundaries. Students are required to sketch graphs, identify intersection points, and apply appropriate area formulas.