Application of Integrals — Mathematics Class 12 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Application of Integrals (Class 12) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 2 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Application of Integrals (CBSE & HBSE)
Application of Integrals carries 5-6 marks in CBSE. One 5-mark question always asks for area under a curve (typically parabola, ellipse, or area between two curves). Drawing a rough sketch is mandatory.
Area Under Curves
Area Under a Curve
Area bounded by curve y=f(x), x-axis, and vertical lines x=a, x=b: A = ∫[a to b] f(x) dx (when f(x) ≥ 0)
If f(x) changes sign in [a,b], split the integral: A = ∫[a to c]|f(x)|dx + ∫[c to b]|f(x)|dx
Area bounded by curve x=g(y), y-axis, and y=c to y=d: A = ∫[c to d] g(y) dy
Standard Areas
| Shape | Formula |
|---|---|
| Circle x²+y²=r² | πr² (total area) |
| Parabola y²=4ax, x=a | (2/3) base × height |
| Ellipse x²/a²+y²/b²=1 | πab (total area) |
Area enclosed by parabola y²=4ax and its latus rectum (x=a): = 2∫[0 to a] 2√(ax) dx = (8/3)a²...
Detailed derivation for y=x² from x=0 to x=1: A = ∫[0 to 1] x² dx = [x³/3]₀¹ = 1/3
Steps for any area problem:
- Sketch the curve(s), shade the required region
- Identify the limits of integration (x or y)
- Set up the integral
- Evaluate using antiderivatives
CBSE Tip: ALWAYS draw the figure and shade the area. You get marks for the sketch. For circles and parabolas, exploit symmetry to halve or quarter the integral.
Area Between Two Curves
Area Between Two Curves
Area between y = f(x) (upper curve) and y = g(x) (lower curve) from x=a to x=b: A = ∫[a to b] [f(x) − g(x)] dx (when f(x) ≥ g(x) on [a,b])
General formula: A = ∫[a to b] |f(x) − g(x)| dx
Finding Intersection Points
- Set f(x) = g(x)
- Solve for x → these are the limits a and b
- Determine which curve is above in each sub-interval
Typical Problems
Area between parabola and line: y² = 4x and y = 2x intersect where (y/2)² = y/2 → y = 0 or y = 2 → x = 0 or x = 1 A = ∫[0 to 1] [2√x − 2x] dx = [4x^(3/2)/3 − x²]₀¹ = 4/3 − 1 = 1/3 sq units
Area between two parabolas: y = x² and y² = x intersect where (x²)² = x → x=0 or x=1 For 0≤x≤1: √x > x² (upper: y=√x, lower: y=x²) A = ∫0 to 1dx = [2x^(3/2)/3 − x³/3]₀¹ = 2/3 − 1/3 = 1/3 sq units
Area of Ellipse
Ellipse x²/a² + y²/b² = 1: A = 4∫[0 to a] b√(1−x²/a²) dx = 4·(b/a)·(πa²/4) = πab
CBSE Tip: For the 5-mark area question, always: (1) find intersection points, (2) draw the sketch showing both curves, (3) identify which is upper/lower, (4) integrate correctly. Losing the sketch means losing easy marks.
Frequently asked questions
Are these Application of Integrals notes free?
Yes — the Application of Integrals notes for Mathematics (Class 12) on Siksha Sarovar are completely free to read, with no account required.
Do these notes follow CBSE and HBSE?
Yes. The Application of Integrals notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Application of Integrals chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Application of Integrals.