Mathematics

$$\displaystyle \overset{e^2}{\underset{1}{\int}} [log_e \,x]dx, x > 0$$ and $$[\cdot]$$ is greatest integer function, is equal to


ANSWER

$$e^2 - e$$


SOLUTION
Breaking limits in 1 to e and e to $$e^2$$

$$\displaystyle \int_1^{e^2} [\log_ex] . \Rightarrow \int_1^e [\log_ex ] d_2t \int_e^{e^2} [\log_ex] dx$$

Now, $$\log_ex $$ for $$x = 1 \Rightarrow \log = 0$$

$$\log_e^x $$ for $$x = e \Rightarrow \log_e^e  = 1 \,\, \log_e^x $$ for $$x = e^2 \Rightarrow \log_e^{e^2} = 2 \log_e^{e = 2}$$

$$\therefore$$ for $$(1 to e) \log_ex $$ will be between $$(0, 1)$$

$$\therefore [\log_ex] = 0$$

and for $$(e \, to \, e^2), \, \log_ex$$ will be between $$(1, 2)$$

$$\therefore [\log_e x] = 1$$

$$\therefore \displaystyle \int_1^{e^2} [\log_ex] = 0 + \int_e^{e^2} . 1 dx = x|^{e^2}_e$$

$$\therefore \displaystyle \int_1^{e^2} [\log_ex] = e^2 - e$$
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Evaluate: $$\displaystyle\int_{4}^{9}\dfrac{\sqrt{x}}{(30-x^{3/2})^{2}}\, dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
If $$\int \dfrac {1}{(x^{2} + 4)(x^{2} + 9)}dx = A\tan^{-1}\dfrac {x}{2} + B\tan^{-1}\left (\dfrac {x}{3}\right ) + C$$ then $$A - B =$$
  • A. $$\dfrac {1}{30}$$
  • B. $$-\dfrac {1}{30}$$
  • C. $$-\dfrac {1}{6}$$
  • D. $$\dfrac {1}{6}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Solve $$\displaystyle\int {x{{\tan }^{ - 1}}x\,dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate the following : $$\displaystyle\int \sqrt{\dfrac{9+x}{9-x}}.dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
$$\underset {n\rightarrow \infty}{lim}\dfrac{1^2+2^2+3^2+.....+n^2}{n^3}=.................$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer