
The slope of the tangent to a curve $y = f\left( x \right)$ at $\left( {x,f\left( x \right)} \right)$ is $2x + 1$ . If the curve passes through the point $\left( {1,2} \right)$ , then what is the area of the region bounded by the curve, the x-axis and the line $x = 1$ ?
A. $\dfrac{5}{6}$ sq. unit
B. $\dfrac{6}{5}$ sq. unit
C. $\dfrac{1}{6}$ sq. unit
D. $6$ sq. unit
Answer
162.9k+ views
Hint:In the above question, we are provided with the equation of the tangent to the curve. We are also provided with a point through which the curve is passing. Use this to evaluate the equation of the curve. Then, proceed further to calculate the area enclosed within the curve, the line, and the x-axis using integration.
Formula used: Area under a curve $y = f\left( x \right)$ , above x-axis, between two points $x = a$ and $x = b$ is calculated as $A = \int_a^b {f\left( x \right)dx} $ .
Complete step by step Solution:
The equation of the tangent to a curve $y = f\left( x \right)$ at a point is $y = 2x + 1$ .
Therefore,
$f'\left( x \right) = 2x + 1$
Integrating both sides with respect to $x$,
$f\left( x \right) = \int {f'\left( x \right)dx = \int {\left( {2x + 1} \right)dx} } $
Integrating,
$f\left( x \right) = {x^2} + x + c$
It is also given that the curve passes through $\left( {1,2} \right)$ , therefore,
$2 = f\left( 1 \right) = {\left( 1 \right)^2} + 1 + c$
Thus, $c = 0$ .
Hence, the equation of the curve is $f\left( x \right) = {x^2} + x$ .
The line $x = 1$ will intersect this curve at $\left( {1,2} \right)$ .
Let us plot the curves and the line and shade the enclosed area.

Clearly, the area enclosed in the area under the curve $y = {x^2} + x$ from $x = 0$ to $x = 1$ .
Hence, the enclosed area is calculated as:
$\int_0^1 {\left( {{x^2} + x} \right)dx} $
Integrating,
$\int_0^1 {\left( {{x^2} + x} \right)dx} = \left[ {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2}} \right]_0^1$
Calculating the definite integral,
$\int_0^1 {\left( {{x^2} + x} \right)dx} = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6}$
Thus, enclosed within the curve, the line $x = 1$ and the x-axis is $\dfrac{5}{6}$ sq. units.
Therefore, the correct option is (A).
Note:The slope of the tangent of a curve at a particular point gives the value of its derivative at that point. Thus, the equation of the tangent gives the equation of the first derivative of that curve.
Formula used: Area under a curve $y = f\left( x \right)$ , above x-axis, between two points $x = a$ and $x = b$ is calculated as $A = \int_a^b {f\left( x \right)dx} $ .
Complete step by step Solution:
The equation of the tangent to a curve $y = f\left( x \right)$ at a point is $y = 2x + 1$ .
Therefore,
$f'\left( x \right) = 2x + 1$
Integrating both sides with respect to $x$,
$f\left( x \right) = \int {f'\left( x \right)dx = \int {\left( {2x + 1} \right)dx} } $
Integrating,
$f\left( x \right) = {x^2} + x + c$
It is also given that the curve passes through $\left( {1,2} \right)$ , therefore,
$2 = f\left( 1 \right) = {\left( 1 \right)^2} + 1 + c$
Thus, $c = 0$ .
Hence, the equation of the curve is $f\left( x \right) = {x^2} + x$ .
The line $x = 1$ will intersect this curve at $\left( {1,2} \right)$ .
Let us plot the curves and the line and shade the enclosed area.

Clearly, the area enclosed in the area under the curve $y = {x^2} + x$ from $x = 0$ to $x = 1$ .
Hence, the enclosed area is calculated as:
$\int_0^1 {\left( {{x^2} + x} \right)dx} $
Integrating,
$\int_0^1 {\left( {{x^2} + x} \right)dx} = \left[ {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2}} \right]_0^1$
Calculating the definite integral,
$\int_0^1 {\left( {{x^2} + x} \right)dx} = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6}$
Thus, enclosed within the curve, the line $x = 1$ and the x-axis is $\dfrac{5}{6}$ sq. units.
Therefore, the correct option is (A).
Note:The slope of the tangent of a curve at a particular point gives the value of its derivative at that point. Thus, the equation of the tangent gives the equation of the first derivative of that curve.
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