logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the definite integral\[\int\limits_0^\frac{\large\pi}{4}(2\sec^2x+x^3+2)dx\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \sec^2xdx=\tan x$
Given $I=\int\limits_0^\frac{\large\pi}{4}(2\sec^2x+x^3+2)dx$
 
On integrating we get,
 
$I=2[\tan x]_0^{\frac{\pi}{4}}+[\large\frac{x^4}{4}]_0^{\frac{\pi}{4}}+[2x]_0^{\frac{\pi}{4}}$
 
On applying limits,
 
$2[\tan \large\frac{\pi}{4}-\tan 0]+[(\frac{\pi}{4})^4-0]+[2.\frac{\pi}{4}-0]$
 
But $\tan \frac{\pi}{4}=1\; and\; \tan 0=0$
 
Hence $I=(2 \times 1)+\large \frac{\pi^4}{1024}+\frac{\pi}{2}$
 
$\int\limits_0^\frac{\large\pi}{4}(2\sec^2x+x^3+2)dx=2 +\large \frac{\pi^4}{1024}+\frac{\pi}{2}$

 

answered Feb 11, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...