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# Find the area of the region bounded by the curve $y=x^3$ and $y=x+6$ and $x=0$.

$\begin{array}{1 1} \frac{10}{3}\;sq.units. \\ 15\;sq.units. \\ 10\;sq.units. \\ 12\;sq.units\end{array}$

Toolbox:
• Let $y=f(x)$ be a continiuous curve,then the area enclosed by the curve,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by $\int_a^b y dx.$
• $\int x^ndx=\large\frac{x^{n+1}}{n+1}$$+c$.
Step 1:
Given $y=x^3$ and $y=x+6$ and $x=0$
The sketch for $y=x^3$ is as shown in the fig and $y=x+6$ is a straight line.
The required area is bounded between the points (0,0),(0,6),(2,8) which is shaded portion.
The required area A=$\int_0^2(y_2-y_1)dx.$
Where $y_1=x^3$ and $y_2=x+6$.
$A=\int_0^2(x+6)dx-\int_0^2 x^3dx.$
Step 2:
On integrating we get,
$A=\begin{bmatrix}\large\frac{x^2}{2}\normalsize +6x\end{bmatrix}_0^2-\begin{bmatrix}\large\frac{x^4}{4}\end{bmatrix}_0^2$
On applying limits we get,
$A=\begin{bmatrix}\large\frac{2^2}{2}\normalsize -0+6\times 2-0\end{bmatrix}-\begin{bmatrix}\large\frac{2^4}{4}-\normalsize 0\end{bmatrix}$
$\;\;=[14-4]=10$
Hence the required area=10 sq. units.
edited Apr 28, 2013