logo

Ask Questions, Get Answers

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

Calculate the area under the curve $y=2\sqrt x$ included between the lines $x=0$ and $x=1$.

$\begin{array}{1 1} \frac{4}{3}\;sq.units \\ \frac{2}{3}\;sq.units \\ \frac{1}{3}\;sq.units \\ None\;of\;the\;above \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c$.
Step 1:
$y=2\sqrt x$ between $x=0$ and $x=1$.
$A=\int_0^1ydx.$
$\;\;=\int_0^22\sqrt xdx.$
$\;\;=2\int_0^2\sqrt xdx.$
On integrating we get,
$\;\;=2\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^1$
$\;\;2\times \large\frac{2}{3}\begin{bmatrix}x^{\Large\frac{3}{2}}\end{bmatrix}_0^1$
Step 2:
On applying limits we get,
$\;\;=\large\frac{4}{3}$$[1-0]=\large\frac{4}{3}$.
Hence the required area is $\large\frac{4}{3}$sq.units.
answered Apr 28, 2013 by sreemathi.v
 

Related questions

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
...