CBSE Class 12 Mathematics Application of Integrals Notes Set 02

Important Notes and Formulas

Introduction

We know the methods of evaluating definite integrals. These integrals are used in evaluating certain types of bounded regions. For evaluation of bounded regions defined by given functions, we shall also require to draw a rough sketch of the given function. The process of drawing a rough sketch of a given function is called curve sketching.

Sketching of Some Common Curves

(1) Straight line: The general equation of a straight line is ax + by + c = 0. To draw a straight line, find the points where it meets the coordinate axes by putting y = 0 and x = 0 respectively in its equation. By joining these two points, we get the sketch of the line.

(2) Region represented by a linear inequality: To find the region represented by the linear inequalities ax + by ≤ c and ax + by ≥ c, we proceed as follows.
(i) Convert the inequality into an equality to obtain a linear equation in x, y.
(ii) Draw the straight line represented by it.
(iii) The straight line obtained in (ii) divides the xy-plane in two parts. To determine the region represented by the inequality, choose some convenient point, e.g. the origin or some point on the coordinate axes. If the coordinates of the point satisfy the inequality, then the region containing the point is the required region; otherwise the region not containing the point is the required region.

(3) Circle: The equation of a circle having centre at (0, 0) and radius r is given by \( x^2 + y^2 = r^2 \). The equation of a circle having centre at (h, k) and radius r is given by \( (x - h)^2 + (y - k)^2 = r^2 \). The general equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). This represents the circle whose centre is at (−g, −f) and radius equal to \( \sqrt{g^2 + f^2 - c} \).
The figure of the circle \( x^2 + y^2 = (2)^2 \) is given. Here the centre is (0, 0) and the radius is 2.

(4) Parabola: There are four standard forms of a parabola with vertex at the origin and the axis along either of the coordinate axes.
(i) \( y^2 = \pm 4ax \): For this parabola
(a) Vertex: (0, 0)    (b) Focus: (±a, 0)
(c) Directrix: x ± a = 0    (d) Latus rectum: 4a
(e) Axis: y = 0    (f) Symmetry: It is symmetric about the x-axis.

(ii) \( x^2 = \pm 4ay \): For this parabola
(a) Vertex: (0, 0)    (b) Focus: (0, ±a)
(c) Directrix: y ± a = 0    (d) Latus rectum: 4a
(e) Axis: x = 0    (f) Symmetry: It is symmetric about the y-axis.

(5) Ellipse: The standard equation of the ellipse having its centre at the origin and major and minor axes along the coordinate axes is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here a > b. The figure of the ellipse is given.

Area of Bounded Regions

(1) The area bounded by a cartesian curve y = f(x), the x-axis and the ordinates x = a and x = b is given by
Area \( = \int_{a}^{b} y\,dx = \int_{a}^{b} f(x)\,dx \)

(2) If the curve y = f(x) lies below the x-axis, then the integral of y = f(x) between the ordinates x = a and x = b is negative. So, the area is given by \( \left| \int_{a}^{b} y\,dx \right| \).

(3) The area bounded by a cartesian curve x = f(y), the y-axis and the abscissae y = c and y = d is given by
Area \( = \int_{c}^{d} x\,dy = \int_{c}^{d} f(y)\,dy \)

Sign Convention for Finding the Areas using Integration

While applying the sign convention, we discuss the following three cases.
Case I: In the expression \( \int_{a}^{b} f(x)\,dx \), if b > a and f(x) > 0 for all a ≤ x ≤ b, then this integration gives the area enclosed between the curve f(x), the x-axis and the lines x = a and x = b, which is positive. No modification is needed.
Case II: In the expression \( \int_{a}^{b} f(x)\,dx \), if b > a and f(x) < 0 for all a ≤ x ≤ b, then this integration will be negative. But the numerical or absolute value is taken to mean the area enclosed between the curve y = f(x), the x-axis and the lines x = a and x = b.
Case III: In the expression \( \int_{a}^{b} f(x)\,dx \), where b > a but f(x) changes its sign a number of times in the interval a ≤ x ≤ b, we must divide the region [a, b] in such a way that we clearly get the points lying between [a, b] where f(x) changes its sign. For the region where f(x) > 0 we just integrate to get the area in that region, and then add the absolute value of the integral calculated in the region where f(x) < 0 to get the desired area between the curve y = f(x), the x-axis and the lines x = a and x = b.

Hence, if f(x) is as in the above figure, the area enclosed by y = f(x), the x-axis and the lines x = a and x = b is given by
\( A = \int_{a}^{c} f(x)\,dx + \left| \int_{c}^{d} f(x)\,dx \right| + \int_{d}^{e} f(x)\,dx + \left| \int_{e}^{f} f(x)\,dx \right| + \int_{f}^{b} f(x)\,dx \)

Symmetrical Area

If the curve is symmetrical about a coordinate axis (or a line or the origin), then we find the area of one symmetrical portion and multiply it by the number of symmetrical portions to get the required area.

Positive and negative area: Area is always taken as positive. If some part of the area lies above the x-axis and some part lies below the x-axis, then the areas of the two parts should be calculated separately and then their numerical values added to get the desired area.

Important Tips

(i) The area of the region bounded by \( y^2 = 4ax \) and its latus rectum is \( \frac{8a^2}{3} \) square units.
(ii) The area of the region bounded by one arch of sin (ax) or cos (ax) and the x-axis is \( \frac{2}{a} \) sq. units.
(iii) The area of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is πab sq. units.
(iv) The area of the region bounded by the curve y = sin x, the x-axis and the lines x = 0 and x = 2π is 4 sq. units.

Solved Examples

Example 1. Sketch the region bounded by 3x + 4y ≤ 12.
Answer:
Converting the inequality into an equation, we get 3x + 4y = 12. This line meets the x-axis at (4, 0) and the y-axis at (0, 3). Joining these two points, we obtain the straight line represented by 3x + 4y = 12.
This straight line divides the plane in two parts. One part contains the origin, the other does not contain the origin. Clearly, (0, 0) satisfies the inequality 3x + 4y ≤ 12. So, the region represented by 3x + 4y ≤ 12 is the region containing the origin, as shown in the figure.

Example 2. Sketch the parabola \( y^2 = 8x \).
Answer:
\( y^2 = 8x \)
⇒ \( y^2 = 4(2)x \)
Here the vertex is (0, 0) and the focus is (2, 0).

Example 3. The area (in square units) enclosed by the curve \( x^2y = 36 \), the x-axis and the lines x = 6 and x = 9 is [Kerala (Engg.) 2000]
(a) 2    (b) 1    (c) 4    (d) 3
Answer:
Ans. (a) 2
Required area \( = \int_{6}^{9} y\,dx = \int_{6}^{9} \frac{36}{x^2}\,dx \)   [Given \( x^2y = 36 \Rightarrow y = \frac{36}{x^2} \)]
\( = \left[ -\frac{36}{x} \right]_{6}^{9} = -\left[ \frac{36}{9} - \frac{36}{6} \right] = -[4 - 6] = 2 \).

Example 4. The area bounded by the x-axis, the curve y = f(x) and the lines x = 1, x = b is equal to \( \sqrt{b^2 + 1} - \sqrt{2} \) for all b > 1, then f(x) is [MP PET 2000]
(a) \( \sqrt{x - 1} \)    (b) \( \sqrt{x + 1} \)    (c) \( \sqrt{x^2 - 1} \)    (d) \( \frac{x}{\sqrt{1 + x^2}} \)
Answer:
Ans. (d) \( \frac{x}{\sqrt{1 + x^2}} \)
\( \int_{1}^{b} f(x)\,dx = \sqrt{b^2 + 1} - \sqrt{2} = \left[ \sqrt{x^2 + 1} \right]_{1}^{b} \)
⇒ \( f(x) = \frac{1}{2} \cdot \frac{2x}{\sqrt{x^2 + 1}} \)
Hence \( f(x) = \frac{x}{\sqrt{1 + x^2}} \).

Example 5. The area of the region bounded by the curve \( y = x - x^2 \) between x = 0 and x = 1 is [Pb. CET 1994, 89]
(a) \( \frac{1}{6} \)    (b) \( \frac{1}{3} \)    (c) \( \frac{1}{2} \)    (d) \( \frac{5}{6} \)
Answer:
Ans. (a) \( \frac{1}{6} \)
Required area \( = \int_{0}^{1} (x - x^2)\,dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \).

Example 6. Find the area bounded between the curve \( y^2 = 2y - x \) and the y-axis.
(a) \( \frac{4}{3} \)    (b) \( \frac{2}{3} \)    (c) \( \frac{1}{3} \)    (d) 5
Answer:
Ans. (a) \( \frac{4}{3} \)
The area between the given curve \( x = 2y - y^2 \) and the y-axis will be as shown in the figure.

∴ Required area \( = \int_{0}^{2} (2y - y^2)\,dy = \left[ y^2 - \frac{y^3}{3} \right]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3} \)

Example 7. Find the area bounded by the curve x = a cos t, y = b sin t in the first quadrant.
(a) \( \frac{\pi ab}{4} \)    (b) \( \frac{\pi a^2 b}{4} \)    (c) \( \frac{\pi ab^2}{4} \)    (d) None of these
Answer:
Ans. (a) \( \frac{\pi ab}{4} \)
Clearly the given equations are the parametric equations of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The curve meets the x-axis in the first quadrant at (a, 0).
In the first quadrant, \( y = \frac{b}{a}\sqrt{a^2 - x^2} \).
∴ Required area \( = \int_{0}^{a} y\,dx = \frac{b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx = \frac{b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \right]_{0}^{a} \)
\( = \frac{b}{a} \times \frac{a^2}{2} \times \frac{\pi}{2} = \frac{\pi ab}{4} \)

Example 8. Find the whole area of the circle \( x^2 + y^2 = a^2 \).
(a) π    (b) \( \pi a^2 \)    (c) \( \pi a^3 \)    (d) \( a^2 \)
Answer:
Ans. (b) \( \pi a^2 \)
The required area is symmetric about both the axes, as shown in the figure.

∴ Required area \( = 4\int_{0}^{a} \sqrt{a^2 - x^2}\,dx = 4\left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \right]_{0}^{a} \)
\( = 4\left[ \frac{\pi}{2} \times \frac{a^2}{2} \right] = \pi a^2 \)

Example 9. Find the area bounded by the parabola \( y^2 = 4x \) and its latus rectum. [Rajasthan PET 1997, 94, 92, 84]
(a) \( \frac{8}{3} \)    (b) \( \frac{4}{3} \)    (c) \( \frac{16}{3} \)    (d) None of these
Answer:
Ans. (a) \( \frac{8}{3} \)
Since the curve is symmetrical about the x-axis, the required area

\( = 2\int_{0}^{1} y\,dx = 2\int_{0}^{1} \sqrt{4x}\,dx = 4 \cdot \frac{2}{3}\left[ x^{3/2} \right]_{0}^{1} = \frac{8}{3} \)

Example 10. The area of the bounded region by the curve y = sin x, the x-axis and the lines x = 0 and x = π is [Rajasthan PET 1989, 92]
(a) 4    (b) 2    (c) 0    (d) None of these
Answer:
Ans. (b) 2

Required area \( = \int_{0}^{\pi} \sin x\,dx = 2\int_{0}^{\pi/2} \sin x\,dx = 2\left[ -\cos x \right]_{0}^{\pi/2} = 2\left[ \left(-\cos\frac{\pi}{2}\right) - (-\cos 0) \right] = 2(1) \) = 2 square units.
Trick: For the curve y = sin x or cos x, the areas are \( \int_{0}^{\pi/2} |\sin x|\,dx = 1, \int_{0}^{\pi} |\sin x|\,dx = 2, \int_{0}^{3\pi/2} |\sin x|\,dx = 3, \int_{0}^{2\pi} |\sin x|\,dx = 4 \) and so on.