Find the area of the region bounded by the curve $y={{x}^{2}}+2$ , and the lines $y=x$ , $x=0$ and $x=3$ .
VerifiedHint: For solving this question first we will plot the given curves on the same $x-y$ plane. Then, we will find the desired region whose area is asked in the question. After that, we will take an elementary vertical strip of width $dx$ and try to write its height in terms of the variable $x$ . Then, we will write the area of the elementary in terms of $x$ and $dx$ , by multiplying its height and width. And finally, we will integrate the area of the elementary strip with suitable limits to get the total area of the desired region.
Complete step-by-step answer:
Given:
We have to find the area of the region bounded by the curve $y={{x}^{2}}+2$ , and the lines $y=x$ , $x=0$ and $x=3$ .
Now, before we proceed we should plot the curve $y={{x}^{2}}+2$ , and the lines $y=x$ , $x=0$ and $x=3$ on the same $x-y$ plane. For more clarity look at the figure given below:
In the above figure, we have to find the area of the region OBAC.
Now, the coordinates of points $B\equiv \left( 0,0 \right)$ , $C\equiv \left( 0,2 \right)$ and for coordinates of the point A and B we should put $x=3$ in the equation $y={{x}^{2}}+2$ and $y=x$ respectively. Then,
$\begin{align}
& {{y}_{A}}={{x}^{2}}+2 \\
& \Rightarrow {{y}_{A}}={{3}^{2}}+2 \\
& \Rightarrow {{y}_{A}}=9+2 \\
& \Rightarrow {{y}_{A}}=11 \\
& {{y}_{B}}=x \\
& \Rightarrow {{y}_{B=3}} \\
\end{align}$
Now, we take an elementary vertical strip at $x$ of width $dx$ . For more clarity look at the figure given below:
Now, to find the height of the elementary strip, we should subtract the ${{y}_{above}}={{x}^{2}}+2$ and ${{y}_{below}}=x$ . Then,
Height of the elementary strip $={{y}_{above}}-{{y}_{below}}={{x}^{2}}+2-x$ .
Now, as we know, the width of the elementary strip is $dx$ . So, the area of the elementary strip will be height multiplied by width. Then,
Area of the elementary strip $=dA=\left( {{x}^{2}}+2-x \right)dx$ .
Now, to get the total area of the region OBAC we should add the area of such elementary strips from $x=0$ to $x=3$ so, to get the desired area we should integrate the expression $\left( {{x}^{2}}+2-x \right)dx$ from $x=0$ to $x=3$ . Then,
Area of the desired region $=\int\limits_{0}^{3}{\left( {{x}^{2}}+2-x \right)dx}$ .
Now, we will use the formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}$ to integrate the above integral. Then,
$\begin{align}
& \int\limits_{0}^{3}{\left( {{x}^{2}}+2-x \right)dx} \\
& \Rightarrow \left[ \dfrac{{{x}^{3}}}{3}+2x-\dfrac{{{x}^{2}}}{2} \right]_{0}^{3} \\
& \Rightarrow \left[ \dfrac{{{3}^{3}}}{3}+2\times 3-\dfrac{{{3}^{2}}}{2}-0 \right] \\
& \Rightarrow \left[ \dfrac{27}{3}+6-\dfrac{9}{2} \right] \\
& \Rightarrow 9+6-4.5 \\
& \Rightarrow 15-4.5 \\
& \Rightarrow 10.5 \\
\end{align}$
Now, from the above result, we conclude that the area of the desired region will be $10.5\text{ sq}\text{.units}$ .
Thus, the area of the region bounded by the curve $y={{x}^{2}}+2$ , and the lines $y=x$ , $x=0$ and $x=3$ will be equal to $10.5\text{ sq}\text{.units}$ .
Note: Here, the student should first plot the given curves carefully and then find the desired region whose area is asked in the question and proceed in a stepwise manner. Then, we should be careful while writing the dimensions of the elementary strip and for that, we should take help from the plot of the given curves. Moreover, though the integration part is very easy, we should take upper and lower limits correctly, to get the correct answer and whenever we get stuck at some point we should see the plot of the given curves and use the basic concepts of integral calculus.
