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ECET 2026 Preparation

Day 45 – Morning Session: C Integration – Area Under Curves – ECET 2026

In ECET 2026 Mathematics, Definite Integrals and Area under Curves are highly important. Almost every year, questions appear from this topic. With proper understanding and practice, you can score easy marks.


📘 Concept Notes – Area Under Curves

🔹 Definition

The area under a curve  y = f(x) between limits  x = a and  x = b is given by the definite integral:

 A = \int_{a}^{b} f(x) , dx


🔹 Basic Cases

  1. Area under x-axis:
    • If  f(x) < 0 in the interval, the integral gives negative value.
    • Actual area =  | \int_{a}^{b} f(x) , dx |
  2. Area between two curves:
    If  y_1 = f(x) and  y_2 = g(x) , then:  A = \int_{a}^{b} \big( f(x) - g(x) \big) dx
    (where  f(x) \ge g(x) )

🔹 Important Formulas

  • Area under curve:

 A = \int_{a}^{b} f(x) , dx

Area between curve and x-axis:

 A = \int_{a}^{b} |f(x)| , dx

Area between two curves:

 A = \int_{a}^{b} (f(x) - g(x)) , dx

Area in polar coordinates:
For  r = f(\theta) from  \theta = \alpha to  \theta = \beta :

 A = \dfrac{1}{2} \int_{\alpha}^{\beta} r^2 , d\theta


📐 Example 1

Find the area under the curve  y = x^2 between  x = 0 and  x = 2 .

 A = \int_{0}^{2} x^2 , dx

 A = \left[ \dfrac{x^3}{3} \right]_{0}^{2} = \dfrac{8}{3}

So area =  \dfrac{8}{3} , \text{sq. units} .


📐 Example 2

Find the area between the curves  y = x+1 and  y = x^2 from  x=0 to  x=1 .

 A = \int_{0}^{1} \big( (x+1) - (x^2) \big) dx

 A = \int_{0}^{1} (x + 1 - x^2) dx

 A = \left[ \dfrac{x^2}{2} + x - \dfrac{x^3}{3} \right]_{0}^{1}

 A = \dfrac{1}{2} + 1 - \dfrac{1}{3} = \dfrac{7}{6}

So area =  \dfrac{7}{6} , \text{sq. units} .


🔟 10 Expected MCQs – ECET 2026

Q1. The area under the curve  y = f(x) from  x = a to  x = b is given by:
A)  f(b) - f(a)
B)  \int_{a}^{b} f(x) , dx
C)  \int f(x) dx
D) None

Q2. If  f(x) \leq 0 in [a, b], then actual area is:
A) Negative integral
B) Zero
C)  |\int_{a}^{b} f(x) , dx|
D) None

Q3. Area between  y = f(x) and  y = g(x) is:
A)  \int_{a}^{b} (f(x)+g(x)) dx
B)  \int_{a}^{b} (f(x)-g(x)) dx
C)  f(b)-f(a)
D) None

Q4. Area under  y = x^2 from 0 to 1 is:
A)  \dfrac{1}{3}
B)  \dfrac{1}{2}
C) 1
D)  \dfrac{2}{3}

Q5. Area in polar coordinates is given by:
A)  \int r^2 d\theta
B)  \dfrac{1}{2} \int r^2 d\theta
C)  \int r d\theta
D) None

Q6. Area under  y = \sin x from  0 to  \pi is:
A) 1
B) 0
C) 2
D) -2

Q7. If  A = \int_{0}^{2} (2x - x^2) dx , then A = ?
A) 0
B) 4/3
C) 2
D) 8/3

Q8. Area between y=x and y=x² from 0 to 1 is:
A) 1/3
B) 2/3
C) 1/6
D) 5/6

Q9. For  y = x between 0 and 2, area = ?
A) 1
B) 2
C) 3
D) 2

Q10. Area between  y = \cos x and x-axis from 0 to  \pi/2 is:
A) 0
B) 1
C) 2
D) None


✅ Answer Key

Q.NoAnswer
Q1B
Q2C
Q3B
Q4A
Q5B
Q6C
Q7C
Q8C
Q9C
Q10B

🧠 Explanations

  • Q1 → B: By definition of area.
  • Q2 → C: Actual area is absolute value.
  • Q3 → B: Formula for area between two curves.
  • Q4 → A:  \int_{0}^{1} x^2 dx = 1/3 .
  • Q5 → B: Formula in polar coordinates is half integral.
  • Q6 → C:  \int_{0}^{\pi} \sin x dx = 2 .
  • Q7 → C: Evaluate integral = 2.
  • Q8 → C:  \int_{0}^{1} (x - x^2) dx = 1/6 .
  • Q9 → C:  \int_{0}^{2} x dx = (x^2/2)|_0^2 = 2 → actually 2, correction: Answer = B.
  • Q10 → B:  \int_{0}^{\pi/2} \cos x dx = 1 .

🎯 Why Practice Matters

  • Direct formula-based problems are asked every year.
  • Most answers are simple definite integrals.
  • With practice, you can score full marks quickly.

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