
What is the area bounded by the curve $y = 4x - {x^2} - 3$ and the x-axis?
A. $\dfrac{2}{3}sq.\,unit$
B. $\dfrac{4}{3}sq.\,unit$
C. $\dfrac{5}{3}sq.\,unit$
D. $\dfrac{4}{5}sq.\,unit$
Answer
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Hint: To find the area of the required region we first need to calculate the points of intersection of the two curves (here, between the curve and the x-axis). We will then shade the required region and use the concept of integration to find the area of the region bounded by the given curve and the x-axis.
Formula Used: Area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Complete step by step Solution:
Given curve: $y = 4x - {x^2} - 3$ ...(1)
We have to find the area between the curve in equation (1) and the x-axis, that is, $y = 0$.
Putting $y = 0$ in equation (1), we get, $4x - {x^2} - 3 = 0$
That is , ${x^2} - 4x + 3 = 0$
This implies that, ${x^2} - 3x - x + 3 = 0$
That gives, $x\left( {x - 3} \right) - 1\left( {x - 3} \right) = 0$
Solving this, $\left( {x - 1} \right)\left( {x - 3} \right) = 0$
Thus, $x = 1$ and $x = 3$ .
We know that, if \[f\left( x \right)\] and \[g\left( x \right)\] are continuous on \[\left[ {a,{\text{ }}b} \right]\] and \[g\left( x \right){\text{ }} < {\text{ }}f\left( x \right)\] for all $x$ in \[\left[ {a,{\text{ }}b} \right]\] , then we have the following formula.
Area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Therefore, the required area is, $A = \int_1^3 {\left( {4x - {x^2} - 3} \right)dx} $
Integrating this, we get, \[A = \left( {4\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - 3x} \right)_1^3\]
This implies that, $A = \left( {2{x^2} - \dfrac{{{x^3}}}{3} - 3x} \right)_1^3$
Substituting the limits, we get, $A = \left( {2 \times {3^2} - \dfrac{{{3^3}}}{3} - 3 \times 3} \right) - \left( {2 \times {1^2} - \dfrac{{{1^3}}}{3} - 3 \times 1} \right)$
Solving this, we get, $A = \left( {18 - 9 - 9} \right) - \left( {2 - \dfrac{1}{3} - 3} \right)$
Solving further, $A = 0 - \left( { - \dfrac{4}{3}} \right)$
Thus, $A = \dfrac{4}{3}sq.\,unit$
Therefore, the correct option is (B).
Note: In this question the points of intersection between the given curve and the x-axis gives the upper and lower limits of the integration for $x$. Also, the limits of integration should be substituted carefully. One must know the rules of integration for all such questions.
Formula Used: Area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Complete step by step Solution:
Given curve: $y = 4x - {x^2} - 3$ ...(1)
We have to find the area between the curve in equation (1) and the x-axis, that is, $y = 0$.
Putting $y = 0$ in equation (1), we get, $4x - {x^2} - 3 = 0$
That is , ${x^2} - 4x + 3 = 0$
This implies that, ${x^2} - 3x - x + 3 = 0$
That gives, $x\left( {x - 3} \right) - 1\left( {x - 3} \right) = 0$
Solving this, $\left( {x - 1} \right)\left( {x - 3} \right) = 0$
Thus, $x = 1$ and $x = 3$ .
We know that, if \[f\left( x \right)\] and \[g\left( x \right)\] are continuous on \[\left[ {a,{\text{ }}b} \right]\] and \[g\left( x \right){\text{ }} < {\text{ }}f\left( x \right)\] for all $x$ in \[\left[ {a,{\text{ }}b} \right]\] , then we have the following formula.
Area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Therefore, the required area is, $A = \int_1^3 {\left( {4x - {x^2} - 3} \right)dx} $
Integrating this, we get, \[A = \left( {4\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - 3x} \right)_1^3\]
This implies that, $A = \left( {2{x^2} - \dfrac{{{x^3}}}{3} - 3x} \right)_1^3$
Substituting the limits, we get, $A = \left( {2 \times {3^2} - \dfrac{{{3^3}}}{3} - 3 \times 3} \right) - \left( {2 \times {1^2} - \dfrac{{{1^3}}}{3} - 3 \times 1} \right)$
Solving this, we get, $A = \left( {18 - 9 - 9} \right) - \left( {2 - \dfrac{1}{3} - 3} \right)$
Solving further, $A = 0 - \left( { - \dfrac{4}{3}} \right)$
Thus, $A = \dfrac{4}{3}sq.\,unit$
Therefore, the correct option is (B).
Note: In this question the points of intersection between the given curve and the x-axis gives the upper and lower limits of the integration for $x$. Also, the limits of integration should be substituted carefully. One must know the rules of integration for all such questions.
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