Application of Integrals – Class 12 Maths Chapter 8 | Area Under Curves & Between Curves

Here’s the complete guide for Chapter 8: Application of Integrals — Class 12 CBSE Maths:

---

🧠 A. Key Concepts

1. Area Under a Curve

   $$\text{Area} = \int_a^b f(x)\,dx$$

   Valid when $f(x) \ge 0$ on $[a,b]$.

2. Area Between Two Curves
   If $f(x) \ge g(x)$, then

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

3. Area with Respect to Y-axis

   $$\text{Area} = \int_c^d [x_{\text{right}}(y) - x_{\text{left}}(y)]\,dy$$

4. Combined Regions
   Split the integration region when functions intersect; sum individual definite integrals.

---

📘 B. NCERT Exercises & Solutions

✅ Ex 8.1 – Area under curve

• Identity: $\int x^2, \int \sqrt{x}$, etc.

• Check correct application of limits.

✅ Ex 8.2 – Area between curves

• Determine intersection points

• Use $\int (f - g) \,dx$ with proper limits

✅ Ex 8.3 – Area w.r.t y-axis

• Rearrange curves to $x = f(y)$

• Integrate between y-limits

Each solution includes:

1. Determine region or limits

2. Set up integral

3. Solve with clarity

---

🏆 C. CBSE PYQs with Best Answers

• 2023 (4 marks):
  Find area between $y=x^2$ and $y=x+2$.
  Solution: Intersection at $x^2 = x+2 ⇒ x=2, -1$.
  Use $\int_{-1}^2 [(x+2) - x^2]\,dx$ → evaluate and simplify.

• 2022 (5 marks):
  Area enclosed by $y^2 = x$ and $x=4y$.
  Solution: Solve intersection $y^2 = 4y ⇒ y=0,4$. Express $x$ in both forms, integrate w.r.t $y$:

  $$\int_0^4 [4y - y^2]\,dy$$

• 2020 (3 marks):
  Find area under $y = 1/(1+x^2)$ from 0 to 1:
  Recognize derivative of $\tan^{-1}x$; area = $\tan^{-1}(1) - \tan^{-1}(0) = \pi/4$.

#Tags:
#Class12Maths, #Integrals, #AreaUnderCurve, #AreaBetweenCurves, #CBSEPYQs, #NCERTSolutions, #Chapter8Maths, #Calculus, #BoardPrep