Class 12 Maths Chapter Notes: Application of Integrals

Key Concepts & Formulas

• Definite integral represents the algebraic area bounded by a curve, the x-axis (or y-axis), and ordinates (or abscissae).
• Area is always non-negative; if the integral yields a negative value, take its absolute value.
• For area under y = f(x) from x = a to x = b, the formula is Area = ∫_a^b y dx = ∫_a^b f(x) dx.
• For area under x = g(y) from y = c to y = d, the formula is Area = ∫_c^d x dy = ∫_c^d g(y) dy.
• When a curve crosses the axis, the area must be calculated in parts, taking the absolute value of each part.
• Area between two curves, y = f(x) and y = g(x), from x = a to x = b (where f(x) ≥ g(x)), is Area = ∫_a^b [f(x) - g(x)] dx.
• Always sketch the region to correctly identify the upper/lower curves and limits of integration.
• Symmetry can significantly simplify calculations (e.g., area of a circle, ellipse) by calculating a part and multiplying.

Definitions

Definite Integral

A definite integral represents the signed area between the graph of a function and the x-axis, over a specified interval. When interpreted as geometric area, its absolute value is taken.

Area Under a Curve

The area of the region bounded by the graph of a function y = f(x), the x-axis, and the vertical lines x=a and x=b.

Area Between Two Curves

The area of the region enclosed between the graphs of two functions, y = f(x) and y = g(x), over a given interval [a, b].

Ordinate

The y-coordinate of a point. In the context of integration, vertical lines (x=constant) are often referred to as ordinates, defining the boundaries of integration along the x-axis.

Abscissa

The x-coordinate of a point. In the context of integration, horizontal lines (y=constant) are often referred to as abscissae, defining the boundaries of integration along the y-axis.

Upper Curve

In the context of finding the area between two curves, the function whose graph lies above the other function's graph in the specified interval.

Lower Curve

In the context of finding the area between two curves, the function whose graph lies below the other function's graph in the specified interval.

Understanding Area Calculation with Integrals

The Application of Integrals is a powerful branch of calculus that allows us to calculate various geometric quantities, most notably the area of bounded regions. Fundamentally, a definite integral ∫_a^b f(x) dx can be interpreted as the algebraic sum of areas bounded by the curve y = f(x), the x-axis, and the vertical lines (ordinates) x = a and x = b. If the curve lies above the x-axis, the area is positive; if it lies below, the area is negative. Since geometric area is always a non-negative quantity, we usually take the absolute value of the definite integral or split the integration into parts where the function changes sign and then sum their absolute values.

When calculating the area bounded by a curve y = f(x), the x-axis, and the lines x = a and x = b, the formula used is Area = |∫_a^b f(x) dx|. Similarly, if a curve is defined as x = g(y), bounded by the y-axis, and horizontal lines y = c and y = d, the area is Area = |∫_c^d g(y) dy|. The choice between integrating with respect to x or y depends on which approach simplifies the calculation and correctly defines the region. Often, sketching the graph of the function(s) is an essential first step to correctly identify the limits of integration and the 'upper' and 'lower' boundaries of the region. This visual aid helps in identifying points of intersection which are crucial for setting up the limits.

For finding the area between two curves, say y = f(x) and y = g(x), over an interval [a, b], where f(x) ≥ g(x) in that interval, the formula becomes Area = ∫_a^b [f(x) - g(x)] dx. It is crucial to determine the points of intersection of the curves, as these points often define the limits of integration. If the relative positions of the curves change within the interval (i.e., they cross each other), the area must be calculated by splitting the integral into sub-intervals. For each sub-interval, ensure that (upper curve - lower curve) is correctly identified, and then sum the absolute values of these calculated areas. Symmetry plays a significant role in simplifying these calculations. For instance, if a region is symmetric about an axis (like a circle or ellipse), you can calculate the area of one symmetric part and multiply it by the number of symmetric parts to get the total area, thereby reducing the integration complexity.

Steps to Find Area of a Region Using Integrals

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Worked Examples

• {"title":"Area under a parabola","bodyMarkdown":"Find the area of the region bounded by $y = x^2$, the x-axis, and the lines $x = 1$ and $x = 2$.\n\nSolution: The region is above the x-axis. \n$Area = \\int_1^2 x^2 dx = [\\frac{x^3}{3}]_1^2 = (\\frac{2^3}{3}) - (\\frac{1^3}{3}) = \\frac{8}{3} - \\frac{1}{3} = \\frac{7}{3}$ square units."}
• {"title":"Area of an ellipse","bodyMarkdown":"Find the area of the ellipse $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$.\n\nSolution: Due to symmetry, we can find the area of one quadrant and multiply by 4. From $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$, we get $y = \\frac{b}{a}\\sqrt{a^2 - x^2}$.\n$Area = 4 \\cdot \\int_0^a \\frac{b}{a}\\sqrt{a^2 - x^2} dx$. Using the formula $\\int\\sqrt{a^2 - x^2} dx = \\frac{x}{2}\\sqrt{a^2 - x^2} + \\frac{a^2}{2}\\sin^{-1}(\\frac{x}{a})$,\n$Area = 4 \\cdot (\\frac{b}{a}) \\cdot [\\frac{x}{2}\\sqrt{a^2 - x^2} + \\frac{a^2}{2}\\sin^{-1}(\\frac{x}{a})]_0^a$\n$Area = 4 \\cdot (\\frac{b}{a}) \\cdot [(0 + \\frac{a^2}{2}\\sin^{-1}(1)) - (0 + 0)]$\n$Area = 4 \\cdot (\\frac{b}{a}) \\cdot (\\frac{a^2}{2}) \\cdot (\\frac{\\pi}{2}) = \\pi ab$ square units."}
• {"title":"Area between two curves","bodyMarkdown":"Find the area enclosed by the parabola $y = x^2$ and the line $y = x$.\n\nSolution: First, find intersection points: $x^2 = x \\Rightarrow x^2 - x = 0 \\Rightarrow x(x-1) = 0$. So, $x=0$ and $x=1$.\nIn the interval $[0, 1]$, the line $y=x$ is above the parabola $y=x^2$.\n$Area = \\int_0^1 (x - x^2) dx = [\\frac{x^2}{2} - \\frac{x^3}{3}]_0^1 = (\\frac{1}{2} - \\frac{1}{3}) - (0 - 0) = \\frac{3}{6} - \\frac{2}{6} = \\frac{1}{6}$ square units."}

YoLearn Exam Tip

When solving problems on Application of Integrals, always draw a neat sketch of the region. This simple step helps you correctly identify the curves, their intersection points, and which curve is 'upper' or 'lower' in different intervals. Pay close attention to limits of integration; they are often determined by intersection points or given lines. Remember that area is always positive; if your integral yields a negative value, it implies the region is below the axis or the 'upper-lower' order was swapped. Take the absolute value for the final answer. For symmetrical figures, use symmetry to simplify calculations, but ensure you multiply by the correct factor (e.g., 2, 4). Common traps include incorrect limits and failing to split the integral when curves intersect multiple times or cross an axis.

Practice Questions with Solutions

• Q: What is the area of the region bounded by $y = \sin(x)$, x-axis, $x = 0$, and $x = \pi$? A: $\int_0^\pi \sin(x) dx = [-\cos(x)]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2$ square units.
• Q: How would you find the area bounded by $x = y^2$ and the line $x = 4$? A: Integrate with respect to y. Intersection points are $y^2 = 4 \Rightarrow y = \pm2$. $x=y^2$ is symmetric about the x-axis. Area = $\int_{-2}^2 (4 - y^2) dy = [4y - \frac{y^3}{3}]_{-2}^2 = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = 16 - \frac{16}{3} = \frac{32}{3}$ square units.
• Q: If a curve $y = f(x)$ lies partly above and partly below the x-axis, how do you calculate the total area bounded by the curve, x-axis, and limits x=a to x=b? A: Find the points where $f(x) = 0$ within $[a, b]$. Split the integral into sub-intervals based on these points. Calculate $|\int f(x) dx|$ for each sub-interval and sum them up. For example, if $f(x)$ is negative from a to c and positive from c to b, the area is $|\int_a^c f(x) dx| + \int_c^b f(x) dx$.
• Q: What is the significance of plotting the graph in application of integrals? A: Plotting the graph helps visualize the region, identify points of intersection, correctly determine which function is the 'upper' or 'lower' curve, and choose appropriate limits of integration, thereby preventing errors in setting up the integral.

Frequently Asked Questions