Application of Integrals Class 12 NCERT Complete Guide

Fundamental Principles of Area under Curves

To find the area bounded by the curve $y = f(x)$, the x-axis, and the abscissae $x = a$ and $x = b$, we divide the entire region into infinite, vertically oriented elementary strips. Each thin strip of height $y$ and infinitesimal width $dx$ has an area equal to $y \, dx$. Summing these strips from $x = a$ to $x = b$ gives the total area. Mathematically, this is expressed as:

$Area = \int_{a}^{b} y \, dx = \int_{a}^{b} f(x) \, dx$

Similarly, if the curve is bounded by the y-axis, the ordinates $y = c$ and $y = d$, we use horizontal elementary strips of width $dy$ and length $x$. The area is then defined as:

$Area = \int_{c}^{d} x \, dy = \int_{c}^{d} g(y) \, dy$

An important rule to remember is that geometric area is always positive. If a curve lies below the x-axis (where $y < 0$) in a given interval $[a, b]$, the integral will yield a negative value. To find the actual physical area, we must take the absolute value of the integral for that segment: $Area = |\int_{a}^{b} f(x) \, dx|$.

Step-by-Step Methodology to Calculate Area

1. Sketch the Curves — Draw a rough sketch of the given curves, lines, or axes to visualize the bounded region clearly. Identify the exact boundary equations.
2. Find Intersection Points — Solve the equations of the intersecting curves simultaneously to determine the limits of integration (the points where boundaries cross).
3. Determine the Strip Orientation — Choose whether to use vertical strips ($dx$ width, integrating with respect to $x$) or horizontal strips ($dy$ width, integrating with respect to $y$) based on which shape is simpler.
4. Set up the Definite Integral — Express the area equation as an integral. If the region is bounded between two curves $y_1$ and $y_2$, the area is $\int_{a}^{b} (y_{upper} - y_{lower}) \, dx$.
5. Integrate and Evaluate Limits — Find the antiderivative using standard integration formulas, substitute the upper and lower limits, and simplify to get the final area in square units.

Worked Examples with CBSE Marking Pattern

• Example 1: Find the area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Step 1: The ellipse is symmetrical about both axes. Therefore, the total area $A$ is 4 times the area of the region in the first quadrant. Step 2: Express $y$ in terms of $x$ from the equation: $y = b \sqrt{1 - \frac{x^2}{a^2}} = \frac{b}{a} \sqrt{a^2 - x^2}$ for $x \in [0, a]$. Step 3: Set up the integral: $A = 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx$. Step 4: Use the standard formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$. Step 5: Evaluate with limits: $A = \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) \right]_{0}^{a} = \frac{4b}{a} \left( \frac{a^2}{2} \sin^{-1}(1) - 0 \right) = \frac{4b}{a} \cdot \frac{a^2}{2} \cdot \frac{\pi}{2} = \pi a b$ square units.
• Example 2: Find the area of the region bounded by the curve $y^2 = 4x$ and the line $y = 2x$. Step 1: Find the intersection points. Substitute $y = 2x$ into $y^2 = 4x$: $(2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies 4x(x - 1) = 0$. So, intersection points are $(0, 0)$ and $(1, 2)$. Step 2: Identify the upper and lower functions. Over the interval $[0, 1]$, the parabola curve $y_1 = 2\sqrt{x}$ lies above the line $y_2 = 2x$. Step 3: Set up the integral: $A = \int_{0}^{1} (y_1 - y_2) \, dx = \int_{0}^{1} (2\sqrt{x} - 2x) \, dx$. Step 4: Integrate: $A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - x^2 \right]_{0}^{1} = \left[ \frac{4}{3}x^{3/2} - x^2 \right]_{0}^{1} = \left(\frac{4}{3} - 1\right) - 0 = \frac{1}{3}$ square units.

CBSE Board Exam Tips & Common Mistakes

1. Forget Symmetry Multiplication: When calculating the area of a circle or ellipse, students often integrate only in the first quadrant $[0, a]$ and forget to multiply the result by 4. Always look for horizontal or vertical symmetry!
2. Negative Area Pitfall: Remember that integrating $y = \sin x$ from $0$ to $2\pi$ directly yields $0$ because the positive and negative halves cancel out. For area, you must split it: $\int_{0}^{\pi} \sin x \, dx + |\int_{\pi}^{2\pi} \sin x \, dx|$.
3. Incorrect Intersection Limits: Solve equations systematically. Do not guess the limits from a rough diagram; algebra is your safest bet.
4. Write Units: Always append 'sq. units' or 'square units' to your final numerical answer. CBSE markers can deduct 0.5 marks for missing units.

Practice Questions with Solutions

• Q: Find the area of the region bounded by the curve $y = x^2$ and the line $y = 4$. A: Step 1: The curve $y = x^2$ is a parabola opening upwards, symmetric about the y-axis. The line $y = 4$ is parallel to the x-axis. Step 2: Find the intersection points. Setting $x^2 = 4 \implies x = \pm 2$. Thus, the bounds are from $x = -2$ to $x = 2$. Step 3: Since the region is symmetric, we can integrate from $x = 0$ to $x = 2$ and double the result. The upper function is $y = 4$ and the lower function is $y = x^2$. Step 4: Set up the integral: $Area = 2 \int_{0}^{2} (4 - x^2) \, dx$. Step 5: Compute the integral: $Area = 2 \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = 2 \left[ \left(8 - \frac{8}{3}\right) - 0 \right] = 2 \left( \frac{16}{3} \right) = \frac{32}{3}$. Final answer: The area of the region is $\frac{32}{3}$ square units.
• Q: Find the area bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$. A: Step 1: Analyze the curve $y = \cos x$. It is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$. Step 2: To find the total bounded physical area, we must split the integration interval at the point where the curve crosses the x-axis ($x = \pi/2$). Step 3: Set up the absolute value integrals: $Area = \int_{0}^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right|$. Step 4: Evaluate each part: First part: $[\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1$. Second part: $[\sin x]_{\pi/2}^{\pi} = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1$. Step 5: Sum the absolute values: $Area = 1 + |-1| = 1 + 1 = 2$. Final answer: The area is 2 square units.
• Q: Find the area of the region bounded by the parabola $y^2 = 16x$ and its latus rectum. A: Step 1: Identify the standard equation parameters. The parabola is $y^2 = 4ax \implies 4a = 16 \implies a = 4$. Step 2: The latus rectum is the line perpendicular to the axis of symmetry passing through the focus $(a, 0)$, which is the line $x = 4$. Step 3: The region is bounded by $y^2 = 16x$ and the line $x = 4$. By symmetry across the x-axis, the total area is twice the area of the top half. Step 4: Write the integral: $Area = 2 \int_{0}^{4} y \, dx = 2 \int_{0}^{4} 4\sqrt{x} \, dx = 8 \int_{0}^{4} x^{1/2} \, dx$. Step 5: Integrate: $Area = 8 \left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{16}{3} \left( 4^{3/2} - 0 \right) = \frac{16}{3} \cdot 8 = \frac{128}{3}$. Final answer: The area of the region is $\frac{128}{3}$ square units.
• Q: Find the area of the region in the first quadrant enclosed by the x-axis, the line $y = x$, and the circle $x^2 + y^2 = 32$. A: Step 1: Find the intersection point of the line $y = x$ and the circle $x^2 + y^2 = 32$. Substituting $y = x$ gives $x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4$ (since we are in the first quadrant). Step 2: Identify the limits. The region starts at $x = 0$ and is bounded by the line $y = x$ up to $x = 4$. From $x = 4$ to $x = \sqrt{32} = 4\sqrt{2}$, it is bounded by the circle $y = \sqrt{32 - x^2}$. Step 3: Split the area into two parts: $Area = \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx$. Step 4: Compute the first integral: $\int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = 8$. Step 5: Compute the second integral: $\int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{32 - x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}$ At $4\sqrt{2}$: $0 + 16\sin^{-1}(1) = 16 \cdot \frac{\pi}{2} = 8\pi$. At $4$: $\frac{4}{2}\sqrt{16} + 16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 2(4) + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi$. So, second part = $8\pi - (8 + 4\pi) = 4\pi - 8$. Step 6: Combine both parts: $Total \, Area = 8 + (4\pi - 8) = 4\pi$. Final answer: The area is $4\pi$ square units.

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