Table of Contents
Last modified on October 8th, 2024
The area under a parabola refers to the space enclosed between a parabola and the x-axis, typically over a given interval.
Since a parabola extends indefinitely, it can be obtained by integrating the general quadratic function y = ax2 + bx + c (which represents a parabola).
On integrating the function y = ax2 + bx + c within the limit x = x1 to x = x2 (definite integral), we get
${A=\int ^{x_{2}}_{x_{1}}\left( ax^{2}+bx+c\right) dx}$
⇒ ${A=\left[ \dfrac{a}{3}x^{3}+\dfrac{b}{2}x^{2}+cx\right] _{x_{1}}^{x_{2}}}$
Calculate the area under the parabola y = 2x2 + 3x + 1 between x = 0 and x = 2.
Given, y = 2x2 + 3x + 1
As we know, the area of the parabola is ${A=\int ^{x_{2}}_{x_{1}}\left( ax^{2}+bx+c\right) dx}$
Here, x1 = 0 and x2 = 2
Now, ${A=\int ^{2}_{0}\left( 2x^{2}+3x+1\right) dx}$
= ${\left[ \dfrac{2x^{3}}{3}+\dfrac{3}{2}x^{2}+x\right] _{0}^{2}}$
= ${\left( \dfrac{2\left( 2\right) ^{3}}{3}+\dfrac{3}{2}\left( 2\right) ^{2}+2-\dfrac{2\left( 0\right) ^{3}}{3}-\dfrac{3\left( 0\right) ^{2}}{2}-0\right)}$
= ${\left( \dfrac{16}{3}+6+2\right)}$
= ${\dfrac{40}{3}}$
Thus, the area of the parabola is ${\dfrac{40}{3}}$ square units.
Find the area under the parabola described by the function y = 3x2 + 2x + 1 between x = 1 and x = 4 using the integration method.
Given, y = 3x2 + 2x + 1
As we know, the area of the parabola is ${A=\int ^{x_{2}}_{x_{1}}\left( ax^{2}+bx+c\right) dx}$
Here, x1 = 1 and x2 = 4
Now, ${A=\int ^{4}_{1}\left( 3x^{2}+2x+1\right) dx}$
= ${\left[ \dfrac{3x^{3}}{3}+\dfrac{2}{2}x^{2}+x\right] _{1}^{4}}$
= ${\left[ x^{3}+x^{2}+x\right] _{1}^{4}}$
= (4)3 + (4)2 + 4 – (1)3 – (1)2 – 1
= 81
Thus, the area of the parabola is 81 square units.
Problem – Finding the area of a parabola when the CHORD LENGTH and the LENGTH OF AXIS OF SYMMETRY are given.
The formula for the area of a parabola is:
${A=\dfrac{2}{3}bh}$
Here,
b = length of chord,
h = length along the axis of symmetry
It is applied in specific parabolas, which are symmetric about the x-axis or the y-axis.
A parabola has a chord with a length of 12 cm, and the height from the chord to the vertex along the axis of symmetry is 9 cm. Calculate the area of the region enclosed by the parabola and the chord.
A parabolic arch has a chord that is 15 meters long, with a height of 10 meters from the midpoint of the chord to the vertex along the axis of symmetry. Calculate the area enclosed by the parabola.
As we know, the area of a parabola is ${A=\dfrac{2}{3}bh}$
Here,
b = 15 meters
h = 10 meters
Thus, the area enclosed by the parabola is
${A=\dfrac{2}{3}\times 15\times 10}$
= ${2\times 5\times 10}$
= ${100}$ square meters
Thus, the area enclosed by the parabola is 100 square meters.
If the area of a parabola is 72 square units, and the height along the axis of symmetry is 6 units, find the length of the chord.
As we know, the area of a parabola is ${A=\dfrac{2}{3}bh}$
Here,
A = 72 square units
h = 6 units
Thus, the length of the chord is
${72=\dfrac{2}{3}\times b\times 6}$
⇒ 72 = 4b
⇒ b = 18
Thus, the length of the chord is 18 units.
Last modified on October 8th, 2024