Application of Integrals — Class 12 Maths Notes

Topics covered in Application of Integrals

1. 1.Introduction to the Application of Integrals

   Brief Overview:
   

   The Application of Integrals in mathematics is a fascinating journey from learning how to calculate the areas of simple geometric shapes like triangles and rectangles, to diving into the complex world of curves using Integral Calculus. While basic geometry allows us to handle straightforward shapes, it's through Integral Calculus that we tackle the more challenging task of finding areas enclosed by curves. This chapter takes you through the adventure of calculating areas under simple curves, between various lines, and around the arcs of circles, parabolas, and ellipses using integrals. This not only deepens our understanding of geometry but also opens up new vistas for solving real-life problems.

   In Detail: Remember the time you were trying to find out how much paint you would need to color a uniquely shaped art project? Or perhaps, calculating the area of your school's football field that's not a perfect rectangle? These are instances where the basic formulas of geometry fall short. Here's where the Application of Integrals becomes your superhero.

   In the previous chapter, you've learned about finding the area bounded by a curve =y=f(x), the x-coordinates (ordinates) =x=a and =x=b, and the x-axis, through calculating definite integrals as the limit of a sum. This was like laying down the foundation. Now, we're going to build upon it by exploring how to find areas under simple curves like lines, circles, parabolas, and ellipses (but we'll stick to their standard forms only). This means we're not just looking at flat, 2-dimensional shapes anymore; we're stepping into a world where we can measure the space taken by curves.

   Imagine drawing a curve, say a simple hill-like shape (which in mathematics could be represented by a parabola), and wondering how much area is under that curve. Using the Application of Integrals, you can find that out! This has vast applications in the real world - from architects designing curved structures to meteorologists predicting the amount of rainfall over a region based on cloud curves. It's a tool that lets us quantify the space curves occupy, helping in fields as diverse as engineering, physics, and even economics.

   This chapter is not just about theory; it's about seeing the world through the lens of mathematics and realizing how much of it can be understood, predicted, and created with the knowledge of integrals.

   Real-Life Applications and Careers:

   • Engineering: Designing bridges, roads, and tunnels with curved paths requires knowledge of integrals to calculate areas and volumes.
   • Meteorology: Predicting weather patterns and the amount of rainfall by analyzing the area under curves on weather charts.
   • Economics: Economists use integrals to determine the consumer surplus and producer surplus in the market, which are areas under certain curves in their graphs.
   • Environmental Science: Calculating the rate of pollution spread or the area affected by a certain pollutant over time.
   • Architecture: Designing buildings and landscapes that include curved shapes, requiring precise calculations of areas.

2. 2.Area under Simple Curves Explained

   Brief Overview:
   

   Finding the area under simple curves is a captivating application of integrals in calculus. It's like piecing together a giant jigsaw puzzle where each piece is infinitesimally small. This puzzle, once completed, shows us the area beneath the curve, above the x-axis, and between the vertical lines =x=a and =x=b.

   In Detail: This Diagram represents the area under the curve =y=f(x), above the x-axis, and between the vertical lines =x=a and =x=b. This area can be thought of as composed of a large number of very thin vertical strips.

   Let's consider an arbitrary strip. It has a height y, which is the value of our function f(x) at that particular point, and a width dx, which is an infinitesimally small change in x. The area of this tiny strip, dA, is ⋅y⋅dx, or ⋅f(x)⋅dx.

   When we sum up the areas of all these thin strips from =x=a to =x=b, we get the total area A under the curve. Mathematically, this is represented by the definite integral:

   =∫ A=∫ab​f(x)dx

   This integral sums up all the infinitesimal areas dA across the region PQRSP (as labeled in your figure).

   Now, if we wanted to find the area bounded by a curve =x=g(y), where the function is in terms of y, and the horizontal lines =y=c and =y=d, we would look at horizontal strips. The area A would then be given by the integral:

   =∫ A=∫cd​g(y)dy

   Here, xdy represents the area of a thin horizontal strip, and we integrate these to find the total area.

   Real-World Application: This mathematical method is not just for solving textbook problems; it has real-world applications. For instance, architects use this concept to calculate the materials needed for curved surfaces, and economists use it to determine the consumer surplus in a market graph, which is the area under a demand curve and above the equilibrium price.

   Careers Involved:

   • Civil Engineering: Calculating the area under roadway curves for construction purposes.
   • Economics: Estimating areas under curves that represent economic data.
   • Environmental Science: Modeling the spread of a pollutant over time in a specific area.
   • Architecture: Designing spaces with complex shapes requires understanding the areas those shapes occupy.

3. 3.Exercise Questions - 1, 2

   1. 1. Find the area of the region bounded by the ellipse 216+29=116x2​+9y2​=1.
      

   2. Solution for the Ellipse 216+29=116x2​+9y2​=1:

      Step 1: Identify the Axes Lengths From the equation of the ellipse, we see the denominators under 2x2 and 2y2 which are the squares of the lengths of the semi-major axis (a) and semi-minor axis (b), respectively.

      • For 2x2, the denominator is 16, which means 2=16a2=16, so =4a=4.
      • For 2y2, the denominator is 9, which means 2=9b2=9, so =3b=3.

      Step 2: Use the Area Formula for an Ellipse The area A of an ellipse is given by the formula =A=πab.

      • Plugging in the values of a and b we get =⋅4⋅3A=π⋅4⋅3.

      Step 3: Perform the Calculation

      • Multiplying these together, we get =⋅12A=π⋅12.

      Final Answer:

      • The area A is 37.7037.70 square units (using the value of π as approximately 3.14159).
   3. 2. Find the area of the region bounded by the ellipse 24+29=14x2​+9y2​=1.

      To find the area of the region bounded by the ellipse 24+29=14x2​+9y2​=1, we will follow a similar process as before, using the formula for the area of an ellipse =A=πab, where a and b are the semi-major and semi-minor axes of the ellipse, respectively.

      For the given equation 24+29=14x2​+9y2​=1, we identify that:

      • 2=4a2=4, hence =2a=2,
      • 2=9b2=9, hence =3b=3.

      Using these values, we calculate the area A of the ellipse as follows:

      =⋅⋅A=π⋅a⋅b =⋅2⋅3A=π⋅2⋅3 =6A=6π

      The area of the ellipse is 66π square units. Using ≈3.14159π≈3.14159, this is approximately 6×3.14159≈18.856×3.14159≈18.85 square units.