Question

The area bounded by $y={{x}^{2}}+3$and $y=2x+3$ is ( in sq units)
A . $\dfrac{12}{7}$
B. $\dfrac{4}{3}$
C. $\dfrac{3}{4}$
D. $\dfrac{8}{3}$
E. $\dfrac{3}{8}$

Answer 1

Hint: In this question, we are given two curves $y={{x}^{2}}+3$ and $y=2x+3$ and we have to find the area bounded by the two curves. For this, we first draw the curves and find the bounded area. Then we use the integration to find the area.

Formula Used: We use the power rule of integration, which is
$\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$

Complete step by step Solution:
Here we are given the curves $y={{x}^{2}}+3$ and $y=2x+3$
First, we draw the curves on the plane to get the bounded region.
Therefore, we get

We have to find the area bound between these two curves. We can see that the region of the curved line is the required region and we have to find out the area of this region. By using the integration we are able to find the area of the shaded region.
Therefore, the
Area = $\int_{a}^{b}{|f(x)-g(x)|dx}$
 Then the Required area = $\int_{0}^{2}{(2x+3)-({{x}^{2}}}+3)dx$
By solving the above equation with integration, we get
$\int_{0}^{2}{(2x+3)-({{x}^{2}}}+3)$= $\int_{0}^{2}{(2x-{{x}^{2}})dx}$
Now we will apply the power rule of integration, which is
$\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$
Integrate the above equation from 2 to 0, and we get
$\int_{0}^{2}{(2x-{{x}^{2}})dx}$ = ${{\left[ \dfrac{2{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3} \right]}_{0}}^{2}$
Now we expand the limits,
That is $\left[ \dfrac{2{{(2)}^{2}}}{2}-\dfrac{{{(2)}^{3}}}{3}-\dfrac{2{{(0)}^{2}}}{2}+\dfrac{{{(0)}^{3}}}{3} \right]$
Solving further, we get
$\left[ 4-\dfrac{8}{3} \right]$ = $\dfrac{4}{3}$ square units
Hence the area of region enclosed by curves $y={{x}^{2}}+3$ and $y=2x+3$ is $\dfrac{4}{3}$ square units.

Therefore, the correct option is (B).

Note: We should take care while writing the coordinates of the intersection of curves as it is used to solve the whole question. When there are intersection points, we should break the intervals into sub-intervals and find out which curve is greater on each sub-interval. Then we find the area of the region by integrating the difference between larger and smaller functions.