Sketch the graph of y = |x + 3| and evaluate ∫ −6 0 |x + 3| dx .

V-vertex lies at x = −3 . |x + 3| = −(x + 3) on [−6, −3] and (x + 3) on [−3, 0] . I = ∫ −6 −3 −(x + 3) dx + ∫ −3 0 (x + 3) dx = −[ (x+3)² 2 ] −6 −3 + [ (x+3)² 2 ] −3 0 = 9 2 + 9 2 = 9 sq. units .

NCERT Plus Two (Class 12) Mathematics • Chapter 8

Application of Integrals

Q: Using integration, find the area enclosed by the circle x² + y² = a² .

By symmetry about both axes, Area = 4 × (Area in First Quadrant) . In Q-I, y = √(a² − x²) from x = 0 to x = a . Area = 4 ∫ 0 a √(a² − x²) dx = 4 [ x 2 √(a²−x²) + a² 2 sin⁻¹( x a ) ] 0 a = 4 [ 0 + a² 2 · π 2 − 0 ] = πa² sq. units .

Q: Find the area enclosed by the ellipse x² 16 + y² 9 = 1 using integration.

Here a = 4, b = 3 . In Q-I, y = 3 4 √(16 − x²) . By 4-quadrant symmetry: Area = 4 · 3 4 ∫ 0 4 √(16 − x²) dx = 3 [ x 2 √(16−x²) + 8 sin⁻¹( x 4 ) ] 0 4 = 3 [ 0 + 8 · π 2 ] = 12π sq. units .

Q: Find the area bounded by the curve y² = 4x , the y-axis, and the line y = 3 .

Since the region is bounded with the y-axis from y = 0 to y = 3 , use horizontal strips x dy : Area = ∫ 0 3 x dy = ∫ 0 3 y² 4 dy = [ y³ 12 ] 0 3 = 27 12 = 9 4 sq. units .

Q: Find the area of the region bounded by the parabola y² = 4ax and its latus rectum.

The latus rectum is the vertical line x = a . By symmetry about the x-axis: Area = 2 ∫ 0 a 2√(ax) dx = 4√a [ 2 3 x^(3/2) ] 0 a = 8 3 √a · a^(3/2) = 8 3 a² sq. units .

Q: Find the area bounded by the line y = 3x + 2 , the x-axis, and the ordinates x = −1 and x = 1 .

The line crosses the x-axis at 3x + 2 = 0 ⇒ x = − 2 3 (below x-axis on [−1, −2/3] , above on [−2/3, 1] ). Area = |∫ −1 −2/3 (3x + 2) dx| + ∫ −2/3 1 (3x + 2) dx = |[ 3x² 2 + 2x] −1 −2/3 | + [ 3x² 2 + 2x] −2/3 1 = 1 6 + 25 6 = 13 3 sq. units .

Q: Find the area bounded by the curve y = sin x between x = 0 and x = 2π .

Q: Find the area lying in the first quadrant bounded by the circle x² + y² = 4 and the lines x = 0 , x = 2 .

Radius a = 2 . First-quadrant area = ∫ 0 2 √(4 − x²) dx = [ x 2 √(4−x²) + 2 sin⁻¹( x 2 )] 0 2 = 2( π 2 ) = π sq. units .

Area under simple curves using vertical and horizontal strips, symmetry of circles & ellipses, parabolas with lines, absolute-value areas below the x-axis & curve-sketching checklist.

📄 Official NCERT Textbook PDF

Topic 8.2: Area Under y = f(x) Using Vertical Strips (∫ y dx)

Imagine measuring an uneven Kuttanad paddy plot bordering a straight canal along the x-axis! A standard rectangle formula fails because the outer bund curves smoothly. Instead, slice the plot vertically into paper-thin green ribbon strips of height y = f(x) and tiny width dx. One strip has area dA = y dx — sweep and add every strip from x = a to x = b using a definite integral!

VERTICAL STRIP RULEdA = y dx
Area = ∫_a^b y dx = ∫_a^b f(x) dx

BOUNDARIES NEEDEDTop: curve y = f(x) | Bottom: x-axis
Sides: ordinates x = a to x = b

UNITS & SIGNAlways write sq. units at the end. No + C in definite area!

🧠 Memory Trick: Stand your strip UP vertically like a sliding bamboo pole of height y and thickness dx, marching left-to-right from x = a to x = b!

💡 Simple Kid-Check: A thin vertical rectangle has Area = Height × Width = y × dx. Integration symbol ∫ literally stretches an S for "Sum" — gluing millions of skinny rectangles together from x = a to x = b!

1
Write y purely in terms of x: If given x² = 4y, first solve y = x²4.
2
Pick vertical limits on the x-axis: Set lower limit x = a and upper limit x = b (a < b).
3
Integrate and evaluate: Compute [F(x)]_a^b = F(b) − F(a) and label sq. units.

🌾 Classic Strip Example (NCERT Misc Q1): Find the area bounded by y = x², the x-axis, and lines x = 1, x = 2.
• Vertical strip: height y = x², width dx, sliding from x = 1 to x = 2.
• Area = ∫₁² x² dx = [x³3]₁² = 83 − 13 = 73 sq. units!

Watch TWO sneaky mistakes! (1) Trying to integrate ∫₁² y dx while leaving the letter y untouched instead of substituting f(x)! (2) Swapping limits upside down (∫₂¹ gives a negative sign — ALWAYS put smaller x = a at the bottom and larger x = b on top!).

Topic 8.2: Area Bounded with the Y-Axis Using Horizontal Strips (∫ x dy)

Sometimes a curve hugs the y-axis cleanly while its height runs from y = c to y = d — like a curved Kerala house roof sloping sideways from a vertical wall! Standing strips vertically here would need awkward square roots or split bounds, whereas laying horizontal strips flat (like stacked appams from bottom to top!) solves it in two clean lines!

HORIZONTAL STRIPdA = x dy
Area = ∫_c^d x dy = ∫_c^d g(y) dy

WHEN TO CHOOSE dyRegion rests against the y-axis between horizontal lines y = c and y = d

REWRITE RULESolve for x = g(y) FIRST, and plug y-limits (NOT x-limits!)

🧠 Memory Trick: A sleeping horizontal strip has length x and thickness dy — stack it from bottom floor y = c up to ceiling y = d!

💡 Simple Kid-Check: Tilt your head 90°! Horizontal strips measure width x out from the y-axis, multiplied by skinny vertical thickness dy, stacked cleanly from bottom line y = c to top line y = d!

1
Express x in terms of y: E.g., from y² = 4x, write x = y²4 (zero messy square roots!).
2
Read limits on the y-axis: Find bottom bound y = c and top bound y = d.
3
Integrate wrt y: Evaluate ∫_c^d g(y) dy and write sq. units.

🥞 NCERT Ex 8.1 Q4 Classic: Find the area bounded by y² = 4x, the y-axis, and the line y = 3.
• Curve hits y-axis at y = 0 and reaches top line y = 3. Solve x = y²4.
• Area = ∫₀³ y²4 dy = [y³12]₀³ = 2712 = 94 sq. units!

Watch the Limit-Swap Trap! For y² = 4x, y-axis, and y = 3, students accidentally convert to y = 2√x and integrate from x = 0 to 3 — WRONG! At y = 3, x = 9/4 (not 3!), AND the region sits between the curve and the y-axis (left side!), NOT the x-axis! Always use ∫ x dy whenever the region touches the y-axis!

Topic 8.2: Area of Circle (x² + y² = a²) & Ellipse Using Quadrant Symmetry

When you cut a round Unniyappam or an oval coconut shell neatly down the centre horizontally AND vertically, all four quarters are exact mirror twins! Why integrate across negative numbers and fight sign flips when you can find ONE clean slice in the First Quadrant (x ≥ 0, y ≥ 0) and multiply by 4?

CIRCLE (x² + y² = a²)A = 4 ∫₀^a √(a²−x²) dx = πa²
(First quadrant only: πa²4)

ELLIPSE (x²a² + y²b² = 1)y = ba√(a²−x²) (in Q-I)
A = 4 · ba ∫₀^a √(a²−x²) dx = πab

KEY ANTIDERIVATIVE∫ √(a²−x²) dx = x2√(a²−x²) + a²2 sin⁻¹(xa)

🧠 Memory Trick: Remember Ellipse Area = πab! Set b = a and it shrinks straight back to Circle Area πa²! At limits 0 → a, the x√(a²−x²) part vanishes at BOTH ends, leaving only a²2 · π2 = πa²4!

💡 Simple Kid-Check: Both x² + y² = a² and x²/a² + y²/b² = 1 only contain even powers (x² and y²), meaning swapping x → −x or y → −y changes nothing — perfect 4-way mirror symmetry!

4-Mark Board Derivation Skeleton (Ellipse):
1. In First Quadrant: y = ba √(a² − x²) from x = 0 to x = a.
2. Whole Area = 4 × ba [ x2√(a²−x²) + a²2 sin⁻¹(xa) ]₀^a.
3. Plug x = a (root = 0, sin⁻¹(1) = π/2) and x = 0 (both 0): 4 · ba · a²2 · π2 = πab!

🥚 NCERT Ex 8.1 Q1: Find the area enclosed by the ellipse x²16 + y²9 = 1.
• Identify a² = 16 ⇒ a = 4, and b² = 9 ⇒ b = 3.
•

Area = 4 · 34 ∫₀⁴ √(16 − x²) dx = 3 · [x2√(16−x²) + 8 sin⁻¹(x4)]₀⁴ = 3(0 + 8·π2) = 12π sq. units

Watch THREE deadly DHSE traps! (1) In x²/16 + y²/9 = 1, using a = 16, b = 9 instead of taking square roots a = 4, b = 3! (2) Reading a question asking for "area in the first quadrant" (like x² + y² = 4, x = 0, x = 2) and mistakenly multiplying by 4 (First quadrant answer is π, NOT 4π!). (3) Dropping the a²/2 inside ∫ √(a² − x²) dx!

Topic 8.2: Area Bounded by Parabola & Latus Rectum / Straight Line

Look at a parabolic car headlight mirror or dish antenna sliced across its focal point — that vertical slice through the focus (a, 0) is the Latus Rectum (x = a)! Because the parabola y² = 4ax mirrors itself evenly above and below the x-axis, finding the upper bowl area and doubling it gives the whole cross-section in seconds!

PARABOLA + LATUS RECTUMy² = 4ax and x = a:
A = 2 ∫₀^a 2√(ax) dx = 83 a²

ORIENTATION QUICK CHECKy² = 4ax → symmetric about x-axis
x² = 4ay → symmetric about y-axis

KEY INTEGRAL∫ √x dx = 23 x^(3/2)
(At x = a: a^(3/2) = a√a)

🧠 Memory Trick: The squared letter points out the symmetry trap — y² means +y and −y look identical (symmetric about x-axis → multiply upper half by 2); x² means symmetric about y-axis!

💡 Simple Kid-Check: A parabola opening right (y² = 4ax) is folded symmetrically along the x-axis like an open crocodile mouth! If a vertical line x = k closes the mouth, the enclosed region has BOTH an upper half and a matching lower half — so double the first-quadrant integral!

Standard Curve + Line	Symmetry	Strip & Integral Setup
`y² = 4ax`, latus rectum `x = a`	About x-axis (`2×`)	`2 ∫₀^a 2√a · x^(1/2) dx = 83a²`
`y² = 4ax`, line `x = k` (whole)	About x-axis (`2×`)	`2 ∫₀^k 2√a · x^(1/2) dx = 83√a k^(3/2)`
`x² = 4y`, `y = 2`, `y = 4`, y-axis (Q-I)	Q-I only (`1×`)	`∫₂⁴ 2√y dy = 43(8 − 2√2)`

📡 Classic Latus Rectum Problem: Find the area bounded by the parabola y² = 4x and its latus rectum.
• Compare with y² = 4ax ⇒ 4a = 4 ⇒ a = 1, so the latus rectum is the vertical line x = 1.
• By symmetry about x-axis: Area = 2 ∫₀¹ 2√x dx = 4 [23 x^(3/2)]₀¹ = 83 sq. units!

Watch TWO common exam traps! (1) Missing the factor 2 when finding the area between y² = 4ax and x = a (giving 4a²/3 instead of 8a²/3) — only drop the 2 if the problem explicitly says "in the first quadrant"! (2) Remembering that for y² = 4x, the latus rectum is x = 1 (NOT x = 4 — solve 4a = 4 ⇒ a = 1 first!).

Topic 8.2: Handling Curves Below the X-Axis (Absolute Value Rule)

Suppose you walk 4 metres forward and then 4 metres backward — your net displacement is 0, yet your step-counter still logs 8 metres of real walking! Definite integrals naturally subtract regions hanging below the x-axis as negative numbers, so whenever you compute actual geometric Area, you must split at x-intercepts and flip negative dips positive using absolute value bars |...|!

BELOW X-AXIS RULEIf f(x) < 0 on [a, b]:
Area = |∫_a^b f(x) dx| = −∫_a^b f(x) dx

MIXED SIGN CURVESSolve f(x) = 0 inside (a, b) to find sign-flip root c, then add:
|∫_a^c f dx| + |∫_c^b f dx|

FAMOUS TRIG AREASy = sin x on [0, 2π] → 2 + 2 = 4
y = cos x on [0, 2π] → 1 + 2 + 1 = 4

🧠 Memory Trick: Area is paint on the floor — ALWAYS POSITIVE! Spot where the curve crosses the x-axis (y = 0), chop the integral at those zero-crossings, and take |...| of every piece!

💡 Simple Kid-Check: Integrating ∫₀^2π sin x dx blindly gives 0 because the top hill (+2) cancels the bottom valley (−2)! Measuring total painted area needs |+2| + |−2| = 4 sq. units!

1
Find all x-intercepts inside (a, b): Solve f(x) = 0 to catch every crossing point.
2
Split the integral at each crossing: Never integrate straight across a sign change!
3
Wrap every negative sub-integral in |...| (or negate it) before adding pieces together.

📉 Two NCERT Miscellaneous Gems:
1. Line y = 3x + 2, x-axis, x = −1 to 1 (Example 3): Crosses x-axis at 3x + 2 = 0 ⇒ x = −23.
Area = |∫₋₁^−2/3 (3x+2) dx| + ∫_−2/3¹ (3x+2) dx = |−16| + 256 = 133 sq. units!
2. Cubic y = x³, x-axis, x = −2 to 1 (Misc Q4): Crosses at x = 0.
Area = |∫₋₂⁰ x³ dx| + ∫₀¹ x³ dx = |−4| + 14 = 174 sq. units!

#1 BOARD EXAM TRAP IN CHAPTER 8: Computing ∫₋₂¹ x³ dx = −15/4 or ∫₋₁¹ x|x| dx = 0 and writing negative or zero area! Whenever a curve dips below the x-axis, you MUST split at y = 0 and take absolute values — otherwise y = sin x on [0, 2π] wrongly gives 0 instead of 4!

Topic 8.2: Step-by-Step Curve Sketching & Limit Extraction Checklist

A skilled Ernakulam tailor NEVER cuts expensive Kasavu cloth without first chalking the shape! In the same way, spending 20 seconds drawing a tidy rough sketch before touching ∫ instantly reveals your strip orientation, symmetry factor, crossing roots, and exact limits — turning every 3- or 4-mark area question into guaranteed marks!

4-STEP BLUEPRINT1. Sketch curve & lines
2. Shade target enclosed zone
3. Strip (y dx vs x dy)
4. Limits & Integrate

MODULUS V-GRAPHy = |x + c| has sharp V-corner at x = −c!
Always split integral at the V-corner!

STRIP CHOICEBounded with x-axis → ∫ y dx
Bounded with y-axis → ∫ x dy

🧠 Memory Trick: S-S-S-I Checklist: Sketch rough axes & shape → Shade exact region → Slice thin strip → Integrate with clean limits!

💡 Simple Kid-Check: Your rough sketch doesn't need to be an art masterpiece! Just mark the centre/vertex, axis intercepts (put x = 0 and y = 0), and shade ONLY the region touched by ALL lines named in the question!

Equation You See	Instant Shape & Key Landmark	Smart Strip & Symmetry
`x² + y² = a²`	Circle centred `(0,0)`, radius `a`	`4 ∫₀^a √(a²−x²) dx = πa²`
`x²a² + y²b² = 1`	Ellipse intercepts `(±a, 0)`, `(0, ±b)`	`4 ba ∫₀^a √(a²−x²) dx = πab`
`y² = 4ax`	Right parabola, vertex `(0,0)`	Symm x-axis (`2×`) with `y dx` or `x dy`
`y = \|x − c\|`	V-shape bouncing off `(c, 0)`	Split at `x = c` (two right triangles!)

📐 NCERT Misc Q2 Master Example: Sketch y = |x + 3| and evaluate ∫₋₆⁰ |x + 3| dx.
• Corner sits where x + 3 = 0 ⇒ x = −3. Left arm: y = −(x + 3); right arm: y = x + 3.
• Split at −3: ∫₋₆⁻³ −(x+3) dx + ∫₋₃⁰ (x+3) dx = 92 + 92 = 9 sq. units! (Geometry check: two triangles of base 3, height 3 → 2 × ½(3)(3) = 9!)

Watch out for Skipped Sketches on y = x|x| (Misc Q5) and y = |x + 3|! Remember y = x|x| equals −x² for x < 0 and +x² for x > 0 — so on [−1, 1], the area is |−1/3| + 1/3 = 2/3 (NOT 0!). Always double-check triangles/circles with elementary geometry when possible!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 4 Marks): Using integration, find the area enclosed by the circle x² + y² = a².

By symmetry about both axes, Area = 4 × (Area in First Quadrant).
In Q-I, y = √(a² − x²) from x = 0 to x = a.
Area = 4 ∫₀^a √(a² − x²) dx = 4 [ x2√(a²−x²) + a²2 sin⁻¹(xa) ]₀^a
= 4 [ 0 + a²2 · π2 − 0 ] = πa² sq. units.

Q2 (Kerala HSE Say • 4 Marks): Find the area enclosed by the ellipse x²16 + y²9 = 1 using integration.

Here a = 4, b = 3. In Q-I, y = 34√(16 − x²).
By 4-quadrant symmetry: Area = 4 · 34 ∫₀⁴ √(16 − x²) dx
= 3 [ x2√(16−x²) + 8 sin⁻¹(x4) ]₀⁴ = 3 [ 0 + 8 · π2 ] = 12π sq. units.

Q3 (Kerala HSE March • 2 Marks): Find the area bounded by the curve y² = 4x, the y-axis, and the line y = 3.

Since the region is bounded with the y-axis from y = 0 to y = 3, use horizontal strips x dy:
Area = ∫₀³ x dy = ∫₀³ y²4 dy = [y³12]₀³ = 2712 = 94 sq. units.

Q4 (Kerala HSE Model • 3 Marks): Find the area of the region bounded by the parabola y² = 4ax and its latus rectum.

The latus rectum is the vertical line x = a. By symmetry about the x-axis:
Area = 2 ∫₀^a 2√(ax) dx = 4√a [ 23 x^(3/2) ]₀^a = 83 √a · a^(3/2) = 83 a² sq. units.

Q5 (Kerala HSE March • 3 Marks): Find the area bounded by the line y = 3x + 2, the x-axis, and the ordinates x = −1 and x = 1.

The line crosses the x-axis at 3x + 2 = 0 ⇒ x = −23 (below x-axis on [−1, −2/3], above on [−2/3, 1]).
Area = |∫₋₁^−2/3 (3x + 2) dx| + ∫_−2/3¹ (3x + 2) dx
= |[3x²2 + 2x]₋₁^−2/3| + [3x²2 + 2x]_−2/3¹ = 16 + 256 = 133 sq. units.

Q6 (NCERT / DHSE Say • 3 Marks): Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Q7 (Kerala HSE Model • 3 Marks): Sketch the graph of y = |x + 3| and evaluate ∫₋₆⁰ |x + 3| dx.

V-vertex lies at x = −3. |x + 3| = −(x + 3) on [−6, −3] and (x + 3) on [−3, 0].
I = ∫₋₆⁻³ −(x + 3) dx + ∫₋₃⁰ (x + 3) dx
= −[(x+3)²2]₋₆⁻³ + [(x+3)²2]₋₃⁰ = 92 + 92 = 9 sq. units.

Q8 (Kerala HSE March • 1 Marks): Find the area lying in the first quadrant bounded by the circle x² + y² = 4 and the lines x = 0, x = 2.

Radius a = 2. First-quadrant area = ∫₀² √(4 − x²) dx = [x2√(4−x²) + 2 sin⁻¹(x2)]₀² = 2(π2) = π sq. units.