Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
96 views
in Calculus by (113k points)
closed by
If \(\mathop \smallint \nolimits_1^2 \ln x\;dx = ln(\frac{A}{e})\) then the value of A is equal to?
1. 2
2. 3
3. 4
4. 5
5. 1

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 3 : 4

Concept:

Integration by parts: Integration by parts is a method to find integrals of products

  • The formula for integrating by parts is given by,
  • ∫u v dx = u∫v dx −∫u' (∫v dx) dx

Where u is the function u(x) and v is the function v(x)

ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation: 

Let , I = \(\mathop \smallint \ln x\cdot1\;dx\)

Using integration by parts, 

 \(\begin{array}{l} = \quad \ln x \int d x-\int\left(\frac{1}{x} \cdot \int d x\right) d x \\= \quad x \cdot \ln x-\int d x \\ \end{array}\)

= x ln x - x

= x(ln x - 1)

 Now,

 \(\mathop \smallint \nolimits_1^2 \ln x\;dx\)

\(\begin{aligned} &=[x(\ln x-1)]_{1}^{2}\\ &=[2 \ln 2-2-(\ln 1-1)]\\ &=\left[\operatorname{ln}\left(2)^{2}-2-\ln 1+1\right]\right.\\ &=[\ln 4-1]\\ &=[\ln 4-\ln e]\\ &=\ln\frac{4}{e} \end{aligned}\)

= [2 ln 2 - 2 - (ln 1 - 1)]

= [ln (2)2 - 2 - ln 1 + 1]

= ln 4 - 1

= ln 4 - ln e    (∵ ln e = 1)

= ln (4/e) = ln (A/e)

∴ The value of A is 4

Hence, option (3) is correct. 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...