NCERT Solutions & Concepts: Application of Integrals Ex 8.3 Class 12

Understanding Bounded Areas & the Geometry of Integration

To find the area enclosed between two intersecting curves, say $y = f(x)$ and $y = g(x)$, from $x = a$ to $x = b$ (where $f(x) \ge g(x)$), we divide the region into thin vertical strips of width $dx$ and height $[f(x) - g(x)]$. Summing these strips over the interval yields the definite integral:

$\text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx$

Here, $f(x)$ is the upper curve and $g(x)$ is the lower curve. Alternatively, if the region is bounded by curves defined as functions of $y$, say $x = f(y)$ and $x = g(y)$ from $y = c$ to $y = d$ (where $f(y) \ge g(y)$), we use horizontal strips of thickness $dy$:

$\text{Area} = \int_{c}^{d} [f(y) - g(y)] \, dy$

In this setup, $f(y)$ is the rightmost curve and $g(y)$ is the leftmost curve. The critical first step is always sketching the curves. Without a clear sketch, it is incredibly easy to mistake the upper and lower boundaries or to miss finding the correct points of intersection, which serve as your integration limits.

Step-by-Step Process to Solve Ex 8.3 Problems

1. Sketch the Curves — Identify the equations given (e.g., parabolas like $y^2 = 4ax$, circles like $x^2 + y^2 = r^2$, or lines). Sketch them on a Cartesian plane to visualize the bounded region.
2. Find the Intersection Points — Solve the equations of the bounding curves simultaneously. The $x$-coordinates (or $y$-coordinates) of these intersection points will define your limits of integration ($a$ and $b$).
3. Set up the Integral — Determine if vertical strips ($dx$) or horizontal strips ($dy$) are more convenient. Identify which function is 'upper' and which is 'lower' over the intervals, and construct the integral: $\int (y_{\text{upper}} - y_{\text{lower}}) \, dx$.
4. Integrate and Evaluate — Perform the definite integration using standard formulas (especially $\int \sqrt{a^2 - x^2} \, dx$). Substitute the upper and lower limits carefully to calculate the final enclosed area.

Fully Worked CBSE Board Examples

• Example 1: Area of a Triangle given Vertices Find the area of the region bounded by the triangle whose vertices are $A(2, 0)$, $B(4, 5)$, and $C(6, 3)$. Step 1: Find the equations of the sides. - Equation of AB: $y - 0 = \frac{5 - 0}{4 - 2}(x - 2) \implies y = \frac{5}{2}(x - 2)$ - Equation of BC: $y - 5 = \frac{3 - 5}{6 - 4}(x - 4) \implies y - 5 = -1(x - 4) \implies y = 9 - x$ - Equation of AC: $y - 0 = \frac{3 - 0}{6 - 2}(x - 2) \implies y = \frac{3}{4}(x - 2)$ Step 2: Set up the area integral. The area of $\triangle ABC$ is the area under $AB$ (from $x=2$ to $4$) plus the area under $BC$ (from $x=4$ to $6$), minus the area under $AC$ (from $x=2$ to $6$). $\text{Area} = \int_{2}^{4} \frac{5}{2}(x-2) \, dx + \int_{4}^{6} (9-x) \, dx - \int_{2}^{6} \frac{3}{4}(x-2) \, dx$ Step 3: Integrate each term. - Term 1: $\frac{5}{2} \left[ \frac{(x-2)^2}{2} \right]_{2}^{4} = \frac{5}{4} [4 - 0] = 5$ - Term 2: $\left[ 9x - \frac{x^2}{2} \right]_{4}^{6} = (54 - 18) - (36 - 8) = 36 - 28 = 8$ - Term 3: $\frac{3}{4} \left[ \frac{(x-2)^2}{2} \right]_{2}^{6} = \frac{3}{8} [16 - 0] = 6$ Step 4: Combine the values. $\text{Total Area} = 5 + 8 - 6 = 7 \text{ sq. units.}$
• Example 2: Area bounded by a Parabola and a Line Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = x$. Step 1: Find points of intersection. Substitute $y=x$ into $y^2 = 4x$: $x^2 = 4x \implies x(x-4) = 0 \implies x=0, x=4$ The intersection points are $(0,0)$ and $(4,4)$. Step 2: Set up the integral. Over $[0, 4]$, the curve $y = 2\sqrt{x}$ lies above the line $y = x$. $\text{Area} = \int_{0}^{4} (2\sqrt{x} - x) \, dx$ Step 3: Integrate and evaluate. $\text{Area} = \left[ 2 \cdot \frac{2}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{4} = \left[ \frac{4}{3}(8) - 8 \right] - [0]$ $\text{Area} = \frac{32}{3} - 8 = \frac{8}{3} \text{ sq. units.}$

Exam Traps & Board Presentation Secrets

1. Symmetry Rules: Always look for symmetry. If a region is symmetric about the x-axis or y-axis, integrate for one half/quadrant and multiply by 2 or 4. State this clearly in your steps to earn presentation marks.
2. The Integral Formula $\sqrt{a^2-x^2}$: This is the most common integration formula used in circle and ellipse area problems:

$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$
Memorize this perfectly! A single wrong sign here can cost you 3 to 4 marks.

1. Label Your Sketches: In the CBSE paper, drawing a clean, labeled graph of the curves with shaded regions and points of intersection carries 1 to 1.5 marks of the total question weightage.

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: Analyze the curve and line. The curve $y = x^2$ is an upward-opening parabola symmetric about the y-axis. The line is $y = 4$. Step 2: Find points of intersection. $x^2 = 4 \implies x = -2, 2$. The points of intersection are $(-2, 4)$ and $(2, 4)$. Step 3: Set up the integral using symmetry. The area is symmetric with respect to the y-axis. We integrate from $x = 0$ to $x = 2$ and multiply by 2. The upper boundary is $y = 4$, and the lower boundary is $y = x^2$. $\text{Area} = 2 \int_{0}^{2} (4 - x^2) \, dx$ Step 4: Integrate and evaluate. $\text{Area} = 2 \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = 2 \left[ \left(8 - \frac{8}{3}\right) - 0 \right] = 2 \left( \frac{16}{3} \right) = \frac{32}{3}$ Final answer: The area of the region is $\frac{32}{3}$ sq. units.
• Q: Find the area of the region bounded by the curves $y = x^2 + 2$, $y = x$, $x = 0$, and $x = 3$. A: Step 1: Identify the upper and lower curves. For any $x \in [0, 3]$, $x^2 + 2 > x$. Thus, $y = x^2 + 2$ is the upper curve and $y = x$ is the lower curve. Step 2: Set up the integral. The limits of integration are explicitly given as $x = 0$ and $x = 3$. $\text{Area} = \int_{0}^{3} [(x^2 + 2) - x] \, dx$ Step 3: Perform integration. $\text{Area} = \left[ \frac{x^3}{3} + 2x - \frac{x^2}{2} \right]_{0}^{3}$ $\text{Area} = \left( \frac{27}{3} + 6 - \frac{9}{2} \right) - [0] = 9 + 6 - 4.5 = 10.5 = \frac{21}{2}$ Final answer: The area of the region is $\frac{21}{2}$ sq. units.
• Q: Find the area of the region enclosed between the circle $x^2 + y^2 = 16$ and the line $y = x$ in the first quadrant. A: Step 1: Find the intersection point in the first quadrant. Substitute $y = x$ into $x^2 + y^2 = 16$: $x^2 + x^2 = 16 \implies 2x^2 = 16 \implies x^2 = 8 \implies x = 2\sqrt{2} \quad (\text{since } x > 0)$ Step 2: Split the region. From $x = 0$ to $2\sqrt{2}$, the upper boundary is the line $y = x$. From $x = 2\sqrt{2}$ to $4$ (the radius of the circle), the upper boundary is the circle $y = \sqrt{16 - x^2}$. Step 3: Set up the total area integral. $\text{Area} = \int_{0}^{2\sqrt{2}} x \, dx + \int_{2\sqrt{2}}^{4} \sqrt{16 - x^2} \, dx$ Step 4: Compute the integrals. First part: $\int_{0}^{2\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2\sqrt{2}} = \frac{8}{2} = 4$ Second part: $\int \sqrt{16-x^2}\,dx = \left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{2\sqrt{2}}^{4}$ Substitute $x = 4$: $0 + 8\sin^{-1}(1) = 8\left(\frac{\pi}{2}\right) = 4\pi$ Substitute $x = 2\sqrt{2}$: $\frac{2\sqrt{2}}{2}\sqrt{8} + 8\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 4 + 8\left(\frac{\pi}{4}\right) = 4 + 2\pi$ Subtracting lower limit from upper limit: $4\pi - (4 + 2\pi) = 2\pi - 4$ Step 5: Sum the parts. $\text{Total Area} = 4 + (2\pi - 4) = 2\pi$ Final answer: The area is $2\pi$ sq. units.
• Q: Find the area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$. A: Step 1: Analyze the function behavior. The curve $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$. Since area cannot be negative, we must integrate the absolute value. Step 2: Set up the absolute value integration. $\text{Area} = \int_{0}^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right|$ Step 3: Evaluate the integrals. - Part 1: $[\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1$ - Part 2: $[\sin x]_{\pi/2}^{\pi} = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1$ Step 4: Sum the absolute values. $\text{Total Area} = 1 + |-1| = 2$ Final answer: The area of the region is $2$ sq. units.

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