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Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is

Q: Is there a shortcut formula for the area between two parabolas?

Yes, for two parabolas $y = a_1x^2 + b_1x + c_1$ and $y = a_2x^2 + b_2x + c_2$, the area is $\frac{|a_1 - a_2|}{6} (\beta - \alpha)^3$, where $\alpha$ and $\beta$ are the roots. While this is a fantastic tool for competitive exams, I encourage you to master the integration method first to build your conceptual foundation!

Enter Numerical Value:

Visualized Solution

Visualizing the Parabolas

Given parabolas: $y = x^{2} - 5 x$ (Upward opening)
And $y = 7 x - x^{2}$ (Downward opening)
Objective: Find the area of the region enclosed by these two curves.

Finding Intersection Points

To find the boundaries of our enclosed region, we need the intersection points.
We do this by equating the two functions.

Equating the Functions

Set $y_{1} = y_{2}$
$x^{2} - 5 x = 7 x - x^{2}$

Simplifying the Equation

Bring all terms to one side:
$x^{2} + x^{2} - 5 x - 7 x = 0$
$2 x^{2} - 12 x = 0$

Solving for $x$

Factor out the common term $2 x$ :
$2 x (x - 6) = 0$
Roots are $x = 0$ and $x = 6$

Marking the Intersections

The parabolas intersect at $x = 0$ and $x = 6$ .
Corresponding points are $(0, 0)$ and $(6, 6)$ .

Setting up the Integral

The area between curves is given by:
$A = \int_{a}^{b} (y_{upper} - y_{lower}) d x$
Here, $y_{upper} = 7 x - x^{2}$ and $y_{lower} = x^{2} - 5 x$

Substituting into the Integral

Substitute the limits and functions:
$A = \int_{0}^{6} ((7 x - x^{2}) - (x^{2} - 5 x)) d x$

Simplifying the Integrand

Expand the brackets:
$A = \int_{0}^{6} (7 x - x^{2} - x^{2} + 5 x) d x$
Combine like terms:
$A = \int_{0}^{6} (12 x - 2 x^{2}) d x$

Performing Integration

Integrate term by term:
$\int 12 x d x = 12 \cdot \frac{x ^{2}}{2} = 6 x^{2}$
$\int 2 x^{2} d x = \frac{2 x ^{3}}{3}$
$A = [6 x^{2} - \frac{2 x ^{3}}{3}]_{0}^{6}$

Applying the Limits

Substitute the upper limit $x = 6$ :
$A = (6 (6)^{2} - \frac{2 ( 6 ) ^{3}}{3}) - (0)$
$A = (6 \cdot 36 - \frac{2 \cdot 216}{3})$

Final Calculation

Calculate the numerical values:
$A = 216 - \frac{432}{3}$
$A = 216 - 144$
$A = 72$

Summary and Key Takeaway

Key Takeaway:
1. Find intersection points by equating functions.
2. Identify the upper and lower curves.
3. Integrate $(y_{upper} - y_{lower})$ over the interval.
Final Answer: $72$ sq. units

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

The problem involves two parabolas: the upward-opening curve

y = x^{2} - 5 x

and the downward-opening curve

y = 7 x - x^{2}

. These curves intersect to enclose a finite region in the Cartesian plane.

To determine the boundaries of this region, we set the two equations equal to each other:

x^{2} - 5 x = 7 x - x^{2}

Finding the Intersection Points

Rearranging the terms to one side, we obtain the following quadratic equation:

2 x^{2} - 12 x = 0

Factoring this expression yields:

2 x (x - 6) = 0

Thus, the curves intersect at the points

x = 0

and

x = 6

. These values serve as the lower and upper limits of our integration.

The Master Equation

The area

A

trapped between the two curves is defined by the integral of the upper curve minus the lower curve over the interval

[0, 6]

A = \int_{0}^{6} ((7 x - x^{2}) - (x^{2} - 5 x)) d x

Simplifying the integrand, we arrive at:

A = \int_{0}^{6} (12 x - 2 x^{2}) d x

Final Calculation

We now integrate the expression term by term:

A = [6 x^{2} - \frac{2 x ^{3}}{3}]_{0}^{6}

Substituting the limits of integration into the expression:

A = (6 (6)^{2} - \frac{2 ( 6 ) ^{3}}{3}) - (0)

A = 6 (36) - \frac{2 ( 216 )}{3}

A = 216 - 144 = 72

The area of the region enclosed by the two parabolas is 72 square units.

Common Traps & FAQs

How do I know which parabola is the upper curve without graphing it perfectly?

Great question! You can test a point inside the interval. Pick an

x

value between your intersection points, say

x = 1

. Plug it into both equations. The one yielding the higher

y

-value is your upper curve. In this case,

7 (1) - (1)^{2} = 6

(1)^{2} - 5 (1) = - 4

. Clearly, the downward parabola is on top.

What happens if I accidentally subtract the upper curve from the lower curve?

You will get a negative area! Remember, area is a physical quantity and must be positive. If you get a negative result, it simply means you swapped the functions. Just take the absolute value, or better yet, re-evaluate your setup to ensure you are subtracting the lower function from the upper one.

Is there a shortcut formula for the area between two parabolas?

Yes, for two parabolas

y = a_{1} x^{2} + b_{1} x + c_{1}

and

y = a_{2} x^{2} + b_{2} x + c_{2}

, the area is

\frac{∣ a _{1} - a _{2} ∣}{6} (β - α)^{3}

, where

α

and

β

are the roots. While this is a fantastic tool for competitive exams, I encourage you to master the integration method first to build your conceptual foundation!

Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Parabolas

Finding Intersection Points

Equating the Functions

Simplifying the Equation

Solving for $x$

Marking the Intersections

Setting up the Integral

Substituting into the Integral

Simplifying the Integrand

Performing Integration

Applying the Limits

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Intersection Points

The Master Equation

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region enclosed by the parabolas y=x2−5x and y=7x−x2 is

Enter Numerical Value:

Visualized Solution

Visualizing the Parabolas

Finding Intersection Points

Equating the Functions

Simplifying the Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Marking the Intersections

Setting up the Integral

Substituting into the Integral

Simplifying the Integrand

Performing Integration

Applying the Limits

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Intersection Points

The Master Equation

Final Calculation

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Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the parabolas $y = x^{2} - 5 x$ and $y = 7 x - x^{2}$ is

Solving for $x$

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