Basic Applications of CalculusPYQ May 25Question 4036 of 28
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Evaluate: ∫24(3x−2)2dx\displaystyle \int_2^4 (3x - 2)^2 dx

Options

A104
B100
C10
D52
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Correct Answer

✅ Option a — 104

All Options:

  • A104
  • B100
  • C10
  • D52

Detailed Solution & Explanation

We want to evaluate the definite integral:
∫24(3x−2)2dx\int_2^4 (3x - 2)^2 dx
Let\'s expand the integrand first:
(3x−2)2=9x2−12x+4(3x - 2)^2 = 9x^2 - 12x + 4
Now, integrate each term individually using the power rule ∫xndx=xn+1n+1\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1}:
∫(9x2−12x+4)dx=9(x33)−12(x22)+4x=3x3−6x2+4x\int (9x^2 - 12x + 4) dx = 9\left(\frac{x^3}{3}\right) - 12\left(\frac{x^2}{2}\right) + 4x = 3x^3 - 6x^2 + 4x
Now apply the limits of integration from 2\displaystyle 2 to 4\displaystyle 4:
Let F(x)=3x3−6x2+4x\displaystyle F(x) = 3x^3 - 6x^2 + 4x.
Evaluate F(x)\displaystyle F(x) at the upper limit x=4\displaystyle x = 4:
F(4)=3(4)3−6(4)2+4(4)=3(64)−6(16)+16=192−96+16=112F(4) = 3(4)^3 - 6(4)^2 + 4(4) = 3(64) - 6(16) + 16 = 192 - 96 + 16 = 112
Evaluate F(x)\displaystyle F(x) at the lower limit x=2\displaystyle x = 2:
F(2)=3(2)3−6(2)2+4(2)=3(8)−6(4)+8=24−24+8=8F(2) = 3(2)^3 - 6(2)^2 + 4(2) = 3(8) - 6(4) + 8 = 24 - 24 + 8 = 8
Now calculate the difference:
∫24(3x−2)2dx=F(4)−F(2)=112−8=104\int_2^4 (3x - 2)^2 dx = F(4) - F(2) = 112 - 8 = 104
Hence, **Option A** is the correct answer.

About This Chapter: Basic Applications of Calculus

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Limits, Continuity, Derivatives, Integrals

This chapter covers Limits, Continuity, Derivatives, Integrals and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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