Chapter 8: Application of Integrals

Dear Class 12 Student! You have learned the mechanical process of finding anti-derivatives. Now we apply it to find the exact geometric area enclosed by complex curves. This chapter is highly scoring, but R.D. Sharma warns: If you cannot accurately trace the curve, you cannot find the area. Mastering the sketches of parabolas, ellipses, and circles is your first, non-negotiable step. Let's trace and integrate!

1. Prerequisites: Curve Tracing (Crucial 1st Step)

Before setting up any integral, you must perfectly sketch the bounded region. Memorize these standard forms:

Standard Curves Dictionary 1. Straight Lines: $ax + by + c = 0$. Easiest plotted using $x$ and $y$ intercepts.

2. Circles:
- Center $(0,0)$: $x^2 + y^2 = a^2$ (Radius is $a$).
- Center $(h,k)$: $(x-h)^2 + (y-k)^2 = r^2$.

3. Parabolas (The 4 Standard Forms):
- $y^2 = 4ax \implies$ Opens to the Right (Symmetric about X-axis).
- $y^2 = -4ax \implies$ Opens to the Left.
- $x^2 = 4ay \implies$ Opens Upwards (Symmetric about Y-axis).
- $x^2 = -4ay \implies$ Opens Downwards.
Shifted Parabola: e.g., $(y-k)^2 = 4a(x-h)$ has its vertex at $(h,k)$.

4. Ellipses: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The major axis is along $x$ if $a > b$.

5. Modulus Function: $y = |x|$ (V-shape at origin). $y = |x-a|$ (V-shape shifted to $x=a$).

Practice Problem 1: Intersection & Tracing Question: Find the points of intersection of the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x$.

Solution:
1. Analyze the circle: $x^2 - 8x + y^2 = 0$. Complete the square: $(x^2 - 8x + 16) - 16 + y^2 = 0 \implies (x-4)^2 + y^2 = 4^2$. It's a circle centered at $(4,0)$ with radius $4$.
2. Analyze the parabola: $y^2 = 4x$ opens to the right from the origin.
3. Find Intersections: Substitute $y^2 = 4x$ into the circle's equation:
$x^2 + (4x) = 8x \implies x^2 - 4x = 0 \implies x(x - 4) = 0$.
4. The x-coordinates are $x = 0$ and $x = 4$.
5. Find y-coordinates using $y^2 = 4x$:
- If $x=0$, $y^2 = 0 \implies y=0$. Point: $(0,0)$.
- If $x=4$, $y^2 = 16 \implies y = \pm 4$. Points: $(4,4)$ and $(4,-4)$.
6. Points of intersection: $(0,0)$, $(4,4)$, and $(4,-4)$.

2. Area Under Simple Curves

Vertical Strips: Area Bounded by Curve and X-axis

We imagine dividing the area into infinite vertical rectangles of infinitesimally small width $dx$ and height $y=f(x)$. The total area is the sum (integral) of these strips from $x=a$ to $x=b$.

The Two Fundamental Formulas 1. Using Vertical Strips (Bounded by X-axis):
If the region is bounded by the curve $y = f(x)$, the x-axis, and the ordinates $x=a$ and $x=b$:
$$\text{Area} = \int_a^b y \, dx = \int_a^b f(x) \, dx$$

2. Using Horizontal Strips (Bounded by Y-axis):
If the region is bounded by the curve $x = g(y)$, the y-axis, and the abscissae $y=c$ and $y=d$:
$$\text{Area} = \int_c^d x \, dy = \int_c^d g(y) \, dy$$

Crucial Rule (Sign of Area): If a curve lies entirely below the x-axis, the definite integral will evaluate to a negative number. Since geometric area cannot be negative, you MUST take the absolute value (modulus): $\text{Area} = \left| \int_a^b y \, dx \right|$.
If a curve crosses the x-axis within the interval $[a, b]$, you must split the integral at the root and add the absolute values of the separate regions.

Practice Problem 2: Simple Curve Question: Find the area of the region bounded by the curve $y = x^2$, the x-axis, and the lines $x = 1$ and $x = 2$.

Solution:
1. Trace: The curve $y = x^2$ is an upward-opening parabola. We need the area between $x=1$ and $x=2$ above the x-axis. It is entirely positive in this region.
2. Formula: We use vertical strips. Area $= \int_a^b y \, dx$.
3. Substitute: Area $= \int_1^2 x^2 \, dx$.
4. Integrate: $= \left[ \frac{x^3}{3} \right]_1^2$
5. Evaluate: $= \left( \frac{2^3}{3} \right) - \left( \frac{1^3}{3} \right) = \frac{8}{3} - \frac{1}{3} = \mathbf{\frac{7}{3} \text{ sq. units}}$.

3. Area of the Region Bounded by a Curve and a Line

Standard Algorithm:
1. Sketch the curve and the straight line roughly on the same axes.
2. Intersect: Find their exact points of intersection by solving their equations simultaneously.
3. Limits: The x-coordinates of these intersection points will become your limits of integration (if using vertical strips).
4. Setup: Identify if the required region drops all the way to the axis, or if you need to subtract a geometric shape (like a triangle) from the curve's area.

Practice Problem 3: Curve and Line Question: Find the area of the region bounded by the parabola $y^2 = x$ and the straight line $x + y = 2$.

Solution:
1. Trace: $y^2 = x$ is a right-opening parabola. The line $x + y = 2$ crosses axes at $(2,0)$ and $(0,2)$.
2. Intersect: Substitute $x = 2 - y$ into the parabola's equation.
$y^2 = 2 - y \implies y^2 + y - 2 = 0 \implies (y+2)(y-1) = 0$.
So, $y = -2$ and $y = 1$.
Corresponding $x$ values (using $x=y^2$): $x = 4$ and $x = 1$.
Points are $(4,-2)$ and $(1,1)$.
3. Setup: Look at the bounded region. It's much easier to use horizontal strips (integrating w.r.t $y$) because the right boundary is ALWAYS the line ($x_{line} = 2-y$) and the left boundary is ALWAYS the parabola ($x_{curve} = y^2$) throughout the interval $y=-2$ to $y=1$.
4. Integral: Area $= \int_{y_1}^{y_2} (x_{line} - x_{curve}) \, dy$
Area $= \int_{-2}^1 [(2 - y) - (y^2)] \, dy$
5. Integrate: $= \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{-2}^1$
6. Evaluate Upper limit (1): $2(1) - \frac{1}{2} - \frac{1}{3} = \frac{12 - 3 - 2}{6} = \frac{7}{6}$.
7. Evaluate Lower limit (-2): $2(-2) - \frac{4}{2} - \frac{-8}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = \frac{-10}{3}$.
8. Total Area: $\frac{7}{6} - \left(\frac{-10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \mathbf{\frac{27}{6} = \frac{9}{2} \text{ sq. units}}$.

4. Area Between Two Curves (High Board/JEE Priority)

This is the most frequently asked 5/6 mark question in standard examinations.

Area Between Two Intersecting Curves

Area = $\int_a^b [\text{Upper Curve } f(x) - \text{Lower Curve } g(x)] dx$

Algorithm: Area Between Two Curves 1. Sketch both curves $y_1 = f(x)$ and $y_2 = g(x)$.
2. Intersect: Set $f(x) = g(x)$ to find the points of intersection. Let the roots be $x = a$ and $x = b$.
3. Identify Top/Bottom: Look at the sketch to determine which curve lies "above" the other in the interval $[a, b]$.
4. Formula: $$ \text{Area} = \int_a^b [\text{Upper Curve} - \text{Lower Curve}] \, dx $$
5. Splitting (If they cross): If the curves cross each other at $x=c$ inside $(a,b)$ such that $f(x)$ is on top in $(a,c)$ but $g(x)$ is on top in $(c,b)$, you MUST split the integral:
$\text{Area} = \int_a^c [f(x) - g(x)] \, dx + \int_c^b [g(x) - f(x)] \, dx$.

Practice Problem 4: Two Parabolas Question: Find the area of the region bounded by the two parabolas $y = x^2$ and $y^2 = x$.

Solution:
1. Trace: $y=x^2$ is an upward parabola. $y^2=x \implies y = \sqrt{x}$ (taking positive branch for 1st quadrant) is a rightward half-parabola.
2. Intersect: Substitute $y=x^2$ into $y^2=x$:
$(x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0$.
Roots are $x = 0$ and $x = 1$.
3. Identify Top/Bottom: In the interval $[0, 1]$, the curve $y = \sqrt{x}$ lies ABOVE the curve $y = x^2$. (e.g., at $x=0.5$, $\sqrt{0.5} \approx 0.707$ while $0.5^2 = 0.25$).
4. Formula: Area $= \int_0^1 (\text{Upper} - \text{Lower}) dx = \int_0^1 (\sqrt{x} - x^2) dx$.
5. Integrate: $= \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_0^1 = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 \right]_0^1$
6. Evaluate: $= \left( \frac{2}{3}(1) - \frac{1}{3}(1) \right) - (0) = \frac{2}{3} - \frac{1}{3} = \mathbf{\frac{1}{3} \text{ sq. units}}$.

5. Standard Short-cut Formulas (For JEE Verification)

Warning: In CBSE Board exams, you MUST show the full integration steps. However, these formulas are absolute lifesavers in JEE Mains and great for double-checking your long board calculations.

Must-Memorize Shortcuts 1. Area of an Ellipse:
For $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \mathbf{\text{Area} = \pi a b}$

2. Parabola and its Latus Rectum:
Area bounded by $y^2 = 4ax$ and the vertical line $x = a$ (Latus Rectum) $\implies \mathbf{\text{Area} = \frac{8a^2}{3}}$

3. Parabola and a Line passing through origin:
Area bounded by $y^2 = 4ax$ and line $y = mx \implies \mathbf{\text{Area} = \frac{8a^2}{3m^3}}$

4. Area between Two Standard Parabolas:
Area bounded between $y^2 = 4ax$ and $x^2 = 4by \implies \mathbf{\text{Area} = \frac{16ab}{3}}$

Practice Problem 5: Shortcut Verification Question: Use the shortcut formula to find the area bounded between the parabolas $y^2 = 12x$ and $x^2 = 16y$.

Solution:
1. Compare $y^2 = 12x$ with $y^2 = 4ax \implies 4a = 12 \implies \mathbf{a = 3}$.
2. Compare $x^2 = 16y$ with $x^2 = 4by \implies 4b = 16 \implies \mathbf{b = 4}$.
3. Apply shortcut formula: Area $= \frac{16ab}{3}$.
4. Area $= \frac{16(3)(4)}{3} = 16 \times 4 = \mathbf{64 \text{ sq. units}}$.

6. Symmetry (Crucial to Minimize Calculation)

If the region you have sketched is completely symmetrical across the x-axis, y-axis, or both (like a circle or an ellipse), do not integrate the whole thing. Calculate the area of the region lying in the First Quadrant ONLY, and then multiply by the appropriate symmetry factor (usually 2 or 4).

Always state clearly in your board exam: "Because the curve is symmetrical about both axes, Total Area = 4 $\times$ Area in First Quadrant".

Practice Problem 6: Symmetry & Full Integration Question: Find the area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ using integration.

Solution:
1. Trace: It is an ellipse centered at origin with x-intercepts $\pm 4$ ($a=4$) and y-intercepts $\pm 3$ ($b=3$). It is fully symmetrical in all 4 quadrants.
2. Isolate $y$: $\frac{y^2}{9} = 1 - \frac{x^2}{16} \implies y^2 = 9 \left( \frac{16 - x^2}{16} \right) \implies y = \frac{3}{4} \sqrt{16 - x^2}$. (Taking +ve branch for 1st quadrant).
3. Setup Integral: We find area in 1st quadrant (from $x=0$ to $x=4$) and multiply by 4.
Total Area $= 4 \times \int_0^4 \frac{3}{4} \sqrt{4^2 - x^2} \, dx = 3 \int_0^4 \sqrt{4^2 - x^2} \, dx$.
4. Use Standard Formula: $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.
$= 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_0^4$
5. Evaluate Upper limit (4): $3 \left[ 0 + 8\sin^{-1}(1) \right] = 3 [8(\pi/2)] = 12\pi$.
6. Evaluate Lower limit (0): $3 \left[ 0 + 8\sin^{-1}(0) \right] = 0$.
7. Final Area: $12\pi - 0 = \mathbf{12\pi \text{ sq. units}}$.
(Verification check using shortcut: Area $= \pi a b = \pi(4)(3) = 12\pi$. It perfectly matches!).