Find Area Between Curves: Y=x-x^2 And Y=-6

Understanding the Problem: Finding the Area Between Curves

When we talk about finding the area of a region bounded by the graphs of equations, we're essentially looking to quantify the space enclosed between two or more functions. This is a fundamental concept in calculus, specifically within the realm of integration. The core idea is to slice the region into infinitesimally thin vertical or horizontal strips, calculate the area of each strip, and then sum them all up using an integral. This process allows us to find the precise area of irregular shapes defined by mathematical functions. In this particular problem, we're given two specific functions: $y = x - x^2$ and $y = -6$. Our goal is to determine the exact area that lies between the curve of the parabola $y = x - x^2$ and the horizontal line $y = -6$. This involves identifying where these two graphs intersect, which will define the limits of our integration, and then setting up the integral to calculate the enclosed area. It's a bit like finding the volume of a loaf of bread by slicing it and measuring each slice, but in two dimensions and with mathematical precision.

Visualizing the Graphs: The Parabola and the Horizontal Line

To effectively find the area between $y=x-x^2$ and $y=-6$, it's crucial to visualize these graphs. The equation $y = x - x^2$ represents a parabola. Since the coefficient of the $x^2$ term is negative (-1), this parabola opens downwards. We can find its vertex by completing the square or by using the formula $x = -b/(2a)$. For $y = -x^2 + x$, $a=-1$ and $b=1$, so the x-coordinate of the vertex is $x = -1/(2(-1)) = 1/2$. Plugging this back into the equation, we get $y = (1/2) - (1/2)^2 = 1/2 - 1/4 = 1/4$. So, the vertex of the parabola is at $(1/2, 1/4)$. The parabola also intersects the x-axis when $y=0$, so $x - x^2 = 0$, which means $x(1-x) = 0$, giving us x-intercepts at $x=0$ and $x=1$. The other equation, $y = -6$, is much simpler. It represents a horizontal line that passes through the y-axis at -6. When we sketch these two graphs, we'll see the parabola opening downwards, with its highest point at $(1/2, 1/4)$, and the horizontal line $y=-6$ positioned well below the vertex. The region whose area we need to find is the space enclosed between the upper curve (the parabola) and the lower curve (the horizontal line). Understanding this visual relationship is the first step in setting up the correct integral for calculating the area.

Finding the Intersection Points: Defining the Limits of Integration

To calculate the area between two curves, we first need to determine the points where they intersect. These intersection points will serve as the limits of integration, defining the boundaries along the x-axis over which we will integrate. To find these points for $y=x-x^2$ and $y=-6$, we set the two equations equal to each other: $x - x^2 = -6$. Now, we need to solve this quadratic equation for $x$. Rearranging the terms to get a standard quadratic form ($ax^2 + bx + c = 0$), we have: $x^2 - x - 6 = 0$. This is a quadratic equation that we can solve by factoring, using the quadratic formula, or completing the square. Let's try factoring. We are looking for two numbers that multiply to -6 and add up to -1. These numbers are -3 and +2. So, we can factor the equation as $(x - 3)(x + 2) = 0$. Setting each factor to zero gives us the solutions for $x$: $x - 3 = 0 ightarrow x = 3$ and $x + 2 = 0 ightarrow x = -2$. Therefore, the two graphs intersect at $x = -2$ and $x = 3$. These values, -2 and 3, are our limits of integration. They tell us that the region bounded by the two curves extends from $x = -2$ to $x = 3$. Knowing these points is absolutely critical because it defines the exact horizontal span of the area we are trying to calculate. Without these intersection points, we wouldn't know where to start and stop our integration.

Setting Up the Integral: The Formula for Area Between Curves

With the intersection points found, we can now set up the definite integral to calculate the area between the curves. The general formula for the area $A$ between two continuous functions $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) gtrsim g(x)$ on the interval $[a, b]$, is given by: $A = \int_{a}^{b} [f(x) - g(x)] dx$. In our case, the upper curve is the parabola $y = x - x^2$, and the lower curve is the horizontal line $y = -6$. So, $f(x) = x - x^2$ and $g(x) = -6$. Our limits of integration, $a$ and $b$, are the intersection points we found: $a = -2$ and $b = 3$. Substituting these into the formula, we get: $A = \int_{-2}^{3} [(x - x^2) - (-6)] dx$. Simplifying the integrand, we have: $A = \int_{-2}^{3} (x - x^2 + 6) dx$. This integral represents the sum of the areas of infinitesimally thin vertical strips, where the height of each strip is the difference between the y-value of the upper curve and the y-value of the lower curve at that particular x. The width of each strip is $dx$. Integrating from -2 to 3 sums up all these infinitesimal areas to give us the total enclosed area. This is the mathematical expression that will lead us to our final answer for the area of the region bounded by the graphs.

Evaluating the Integral: Calculating the Exact Area

Now that we have our definite integral set up, $A = \int_{-2}^{3} (x - x^2 + 6) dx$, we need to evaluate it to find the exact area. First, we find the antiderivative of the integrand $(x - x^2 + 6)$. The antiderivative of $x$ is $\frac{1}{2}x^2$. The antiderivative of $-x^2$ is $-\frac{1}{3}x^3$. And the antiderivative of $6$ is $6x$. So, the antiderivative of $(x - x^2 + 6)$ is $\frac{1}{2}x^2 - \frac{1}{3}x^3 + 6x$. Now, we apply the Fundamental Theorem of Calculus, which states that $\int_{a}^{b} F'(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $F'(x)$. In our case, $F(x) = \frac{1}{2}x^2 - \frac{1}{3}x^3 + 6x$, $a = -2$, and $b = 3$. So, we need to calculate $F(3) - F(-2)$.

Let's calculate $F(3)$: $F(3) = \frac{1}{2}(3)^2 - \frac{1}{3}(3)^3 + 6(3) = \frac{1}{2}(9) - \frac{1}{3}(27) + 18 = \frac{9}{2} - 9 + 18 = \frac{9}{2} + 9 = \frac{9}{2} + \frac{18}{2} = \frac{27}{2}$.

Now, let's calculate $F(-2)$: $F(-2) = \frac{1}{2}(-2)^2 - \frac{1}{3}(-2)^3 + 6(-2) = \frac{1}{2}(4) - \frac{1}{3}(-8) - 12 = 2 + \frac{8}{3} - 12 = -10 + \frac{8}{3} = -\frac{30}{3} + \frac{8}{3} = -\frac{22}{3}$.

Finally, we subtract $F(-2)$ from $F(3)$: $A = F(3) - F(-2) = \frac{27}{2} - (-\frac{22}{3}) = \frac{27}{2} + \frac{22}{3}$.

To add these fractions, we find a common denominator, which is 6: $A = \frac{27 \times 3}{2 \times 3} + \frac{22 \times 2}{3 \times 2} = \frac{81}{6} + \frac{44}{6} = \frac{81 + 44}{6} = \frac{125}{6}$.

Therefore, the exact area of the region bounded by the graphs of $y=x-x^2$ and $y=-6$ is $\frac{125}{6}$ square units.

Conclusion: The Power of Calculus in Finding Areas

In conclusion, we have successfully determined the area of the region bounded by the graphs of $y=x-x^2$ and $y=-6$. This process involved several key steps: first, visualizing the functions to understand the shape of the region; second, finding the intersection points of the curves to establish the limits of integration; third, setting up the definite integral using the formula for the area between curves, ensuring we correctly identified the upper and lower functions; and finally, evaluating the integral using the Fundamental Theorem of Calculus. The result, $\frac{125}{6}$ square units, represents the precise amount of space enclosed between the downward-opening parabola and the horizontal line. This problem beautifully illustrates the power and utility of calculus, particularly integration, in solving geometric problems that would be incredibly difficult, if not impossible, to solve using basic algebra or geometry alone. The ability to sum up infinitesimal quantities allows us to find exact areas of complex shapes. Understanding these concepts is fundamental for further studies in mathematics, physics, engineering, and many other fields where quantifying areas and volumes is essential.

For more in-depth information on calculating areas between curves and other calculus topics, you can explore resources like Khan Academy's calculus section, which offers excellent tutorials and practice problems.