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

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

## 1 Answer

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

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