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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 = ax^{2} + bx + c (which represents a parabola).

On integrating the function y = ax^{2} + bx + c within the limit x = x_{1} to x = x_{2 }(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 = 2x ^{2} + 3x + 1 between x = 0 and x = 2.**

Solution:

Given, y = 2x^{2} + 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, x_{1} = 0 and x_{2} = 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 = 3x ^{2} + 2x + 1 between x = 1 and x = 4 using the integration method.**

Solution:

Given, y = 3x^{2} + 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, x_{1} = 1 and x_{2} = 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.**

Solution:

**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.**

Solution:

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.**

Solution:

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