Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $$\int_{1}^{2}e^{x^2}~dx=\alpha$$ ,then show that $$\int_{e}^{e^4}\sqrt{\ln x}~dx=2e^4-e-\alpha$$

How I proceed:$$\int_{e}^{e^4}\sqrt{\ln x}~dx=2\int_{1}^{2}z^2e^{z^2}~dx$$ Then I can't proceed. Please help.

share|improve this question
    
The first integral has been edited to change $e^{x^2}$ to $e^2$. –  Daryl Dec 20 '12 at 10:56

2 Answers 2

up vote 4 down vote accepted

Integrating by parts we get $$ \int _1 ^2 z \cdot 2z e^{z^2} \, dz = \Big [ze^{z^2} \Big ]_1 ^2 - \int _1 ^2 e^{z^2}\, dz. $$

share|improve this answer
    
thanks for the step.Now I got this. –  Argha Dec 20 '12 at 10:54

You can also see it pictorially. In the figure below, the curve between the red and blue areas is $y=e^{x^2}$, so the blue area is $\alpha$, and the desired integral is the red area. The area of the large rectangle is clearly $2e^4$, and the area of the black rectangle is $e$, so $$\int_{e}^{e^4}\sqrt{\ln x}~dx=\text{red area}=2e^4-e-\text{blue area}=2e^4-e-\alpha\;$$

enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.