Application of Integrals Ex 8.1 Class 12 NCERT Solutions

Understanding Area Under Simple Curves

Finding the area bounded by a curve is one of the foundational geometric applications of definite integration. Consider a continuous curve $y = f(x)$ defined on the closed interval $[a, b]$.

To find the area bounded by this curve, the x-axis, and the vertical lines $x = a$ and $x = b$, we divide the entire region into infinitely thin vertical strips of width $dx$ and height $y$. The area of one such elemental strip is $y \cdot dx$. By summing (integrating) these elemental strips from the lower limit $x = a$ to the upper limit $x = b$, we get the total area formula:
$\text{Area} = \int_{a}^{b} y \, dx = \int_{a}^{b} f(x) \, dx$

Conversely, if the curve is represented as $x = g(y)$ and we wish to find the area bounded by the curve, the y-axis, and the horizontal lines $y = c$ and $y = d$, we use horizontal elemental strips of thickness $dy$ and length $x$. The area of this strip is $x \cdot dy$, leading to the integral:
$\text{Area} = \int_{c}^{d} x \, dy = \int_{c}^{d} g(y) \, dy$

Symmetry plays a crucial role in simplifying these computations. For symmetrical shapes like circles ($x^2 + y^2 = r^2$) and ellipses ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$), calculating the area in the first quadrant and multiplying it by 4 yields the total area. This prevents complex integral limits and cuts down calculation time significantly.

The Step-by-Step Method to Solve Exercise 8.1 Problems

1. Sketch the Curve — Always draw a rough sketch of the given curves, lines, and coordinate axes. Identify the exact region whose area needs to be calculated. Label key boundary points and axes.
2. Identify Boundaries and Limits — Determine the limits of integration by finding the points of intersection of the curves or utilizing the given equations of lines (e.g., $x=a$ or $y=c$).
3. Choose the Strip Direction — Decide whether to use vertical strips ($y \, dx$) or horizontal strips ($x \, dy$). Use vertical strips if the top and bottom boundaries of the region are easily defined as functions of $x$. Use horizontal strips if the left and right boundaries are simpler to express as functions of $y$.
4. Formulate and Solve the Integral — Set up the definite integral with the correct limits. Substitute the value of $y$ (or $x$) in terms of the variable of integration, and integrate using standard integration formulas. Remember to write the final answer in 'square units'.

Avoid These Common Board Exam Mistakes

1. Ignoring Quadrant Restrictions: Pay close attention to terms like 'in the first quadrant'. If the question specifies a quadrant, do not calculate the entire symmetric area.
2. Sign of Definite Integrals: Area is a physical quantity and must always be positive. If a curve falls below the x-axis, the integral evaluates to a negative value. You must take the absolute value (modulus) of that integral segment to represent the physical area: $\text{Area} = \left| \int y \, dx \right|$.
3. Incorrect Standard Integration Formulas: Many problems in Ex 8.1 involve circles or ellipses, requiring the standard integration formula:

$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$
Memorize this formula perfectly, as it is used in almost every circle and ellipse problem in CBSE board papers!

Practice Questions with Solutions

• Q: Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$, $x = 4$ and the x-axis in the first quadrant. A: Step 1: Identify the given equations. The curve is a right-opening parabola $y^2 = x$. Since we are looking for the area in the first quadrant, we take $y = \sqrt{x}$. Step 2: Determine limits. The region is bounded by vertical lines $x = 1$ and $x = 4$, and the x-axis ($y = 0$). Step 3: Formulate the integral using vertical strips. $\text{Area} = \int_{1}^{4} y \, dx = \int_{1}^{4} \sqrt{x} \, dx$ Step 4: Integrate and evaluate. $\text{Area} = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \frac{2}{3} \left[ 4^{3/2} - 1^{3/2} \right]$ Since $4^{3/2} = (2^2)^{3/2} = 8$, we get: $\text{Area} = \frac{2}{3} (8 - 1) = \frac{14}{3}\text{ square units}$ Final answer: The area of the bounded region is $\frac{14}{3}$ square units.
• Q: Find the area of the region bounded by the ellipse $\frac{x^2}{16} + rac{y^2}{9} = 1$. A: Step 1: The given curve is an ellipse centered at the origin with semi-major axis $a = 4$ and semi-minor axis $b = 3$. Step 2: By symmetry, the total area is $4$ times the area of the region in the first quadrant. $\text{Total Area} = 4 \times \int_{0}^{4} y \, dx$ From the ellipse equation, $y^2 = 9\left(1 - \frac{x^2}{16}\right) \implies y = \frac{3}{4}\sqrt{16 - x^2}$ in the first quadrant. Step 3: Substitute $y$ into the integral. $\text{Total Area} = 4 \times \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx$ Step 4: Apply the integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$. $\text{Total Area} = 3 \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$ Evaluating at upper limit $x = 4$: $3 \left[ 0 + 8\sin^{-1}(1) \right] = 3 \left[ 8 \cdot \frac{\pi}{2} \right] = 12\pi$ Evaluating at lower limit $x = 0$ gives $0$. $\text{Total Area} = 12\pi - 0 = 12\pi\text{ square units}$ Final answer: The area of the ellipse is $12\pi$ square units.
• Q: Find the area of the region bounded by the curve $x^2 = 4y$, $y = 2$, $y = 4$ and the y-axis in the first quadrant. A: Step 1: The curve is an upward-opening parabola $x^2 = 4y$. In the first quadrant, $x = 2\sqrt{y}$. Step 2: Since boundaries are given along the y-axis ($y = 2$ to $y = 4$), we use horizontal strips ($x \, dy$). Step 3: Setup the definite integral. $\text{Area} = \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy$ Step 4: Evaluate the integral. $\text{Area} = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{4}{3} \left[ 4^{3/2} - 2^{3/2} \right]$ Since $4^{3/2} = 8$ and $2^{3/2} = 2\sqrt{2}$: $\text{Area} = \frac{4}{3} (8 - 2\sqrt{2}) = \frac{32 - 8\sqrt{2}}{3}\text{ square units}$ Final answer: The area of the region is $\frac{32 - 8\sqrt{2}}{3}$ 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 $x^2 + y^2 = 32$. $x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4\text{ (since first quadrant)}$ At $x = 4$, $y = 4$. The intersection point is $(4, 4)$. The circle's radius is $\sqrt{32} = 4\sqrt{2}$. Step 2: The region under the curve is split into two parts relative to the x-axis: - From $x=0$ to $x=4$, bounded by the line $y = x$. - From $x=4$ to $x=4\sqrt{2}$, bounded by the circle $y = \sqrt{32 - x^2}$. Step 3: Set up the sum of two integrals. $\text{Area} = \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx$ Step 4: Evaluate each integral. First part: $\int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = 8$ Second part: $\left[ \frac{x}{2}\sqrt{32-x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}$ At $x = 4\sqrt{2}$: $0 + 16\sin^{-1}(1) = 16 \cdot \frac{\pi}{2} = 8\pi$ At $x = 4$: $\frac{4}{2}\sqrt{32-16} + 16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 2(4) + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi$ Subtracting lower limit from upper limit: $8\pi - (8 + 4\pi) = 4\pi - 8$ Total Area = $8 + (4\pi - 8) = 4\pi$ square units. Final answer: The area of the enclosed region is $4\pi$ square units.

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