Problem 10 - Active Calculus

Question

Let \(f(x)=3-x^{2}\) and \(g(x)=2 x^{2}\) a. On the interval [-1,1] , sketch a labeled graph of \(y=f(x)\) and write a definite integral whose value is the exact area bounded by \(y=f(x)\) on [-1,1] . b. On the interval \([-1,1],\) sketch a labeled graph of \(y=g(x)\) and write a definite integral whose value is the exact area bounded by \(y=g(x)\) on [-1,1] . c. Write an expression involving a difference of definite integrals whose value is the exact area that lies between \(y=f(x)\) and \(y=g(x)\) on [-1,1] . d. Explain why your expression in (c) has the same value as the single integral \(\int_{-1}^{1}[f(x)-\) \(g(x)] d x\) e. Explain why, in general, if \(p(x) \geq q(x)\) for all \(x\) in \([a, b],\) the exact area between \(y=p(x)\) and \(y=q(x)\) is given by $$ \int_{a}^{b}[p(x)-q(x)] d x $$

Step-by-Step Solution

Answer

The area bounded by \(f(x)\) is \(\int_{-1}^{1} (3 - x^2) \, dx\), bounded by \(g(x)\) is \(\int_{-1}^{1} 2x^2 \, dx\), and the area in between is \(\int_{-1}^{1} (3 - 3x^2) \, dx\).

Step 1: Graph of y = f(x)

The function is given as \(f(x) = 3 - x^2\). To sketch the graph, note that it is a downward parabola with vertex at \((0, 3)\) and it intersects the x-axis at \(x = \pm \sqrt{3}\). On the interval \([-1, 1]\), evaluate f(x) at the endpoints: \(f(-1) = 3 - (-1)^2 = 2\) and \(f(1) = 3 - 1^2 = 2\). Plot these points and join them to form the parabola from -1 to 1.

Step 2: Integral for the Area Bounded by f(x)

The area bounded by \(y = f(x)\) on \([-1, 1]\) is given by the definite integral \(\int_{-1}^{1} (3 - x^2) \, dx\).

Step 3: Graph of y = g(x)

The function \(g(x) = 2x^2\) is an upward parabola. On the interval \([-1, 1]\), evaluate g(x) at the endpoints: \(g(-1) = 2(-1)^2 = 2\) and \(g(1) = 2(1)^2 = 2\). Plot these points along with the vertex at \((0, 0)\) to form the parabola from -1 to 1.

Step 4: Integral for the Area Bounded by g(x)

The area bounded by \(y = g(x)\) on \([-1, 1]\) is given by the definite integral \(\int_{-1}^{1} 2x^2 \, dx\).

Step 5: Area Between f(x) and g(x)

To find the area between \(y = f(x)\) and \(y = g(x)\) on \([-1, 1]\), write the expression involving a difference of definite integrals: \(\int_{-1}^{1} (3 - x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx\).

Step 6: Simplifying the Area Expression

The expression mentioned in (c) can be simplified to a single integral: \(\int_{-1}^{1} (3 - x^2 - 2x^2) \, dx = \int_{-1}^{1} (3 - 3x^2) \, dx\).

Step 7: General Case Explanation

If \(p(x) \geq q(x)\) for all \(x\) in \([a, b]\), the area between \(y=p(x)\) and \(y=q(x)\) is given by \(\int_{a}^{b} [p(x) - q(x)] \, dx\) because the integral subtracts the area under \(q(x)\) from the area under \(p(x)\), leaving the area in between.

Key Concepts

Area between curves

When finding the area between two curves, such as between the functions \(y = f(x)\) and \(y = g(x)\), on a given interval, you essentially subtract the area under one curve from the area under another.
This concept is crucial because it helps isolate the region that lies between the two curves.
For example, consider the interval \([-1, 1]\) and the functions \(f(x) = 3 - x^2\) and \(g(x) = 2x^2\).
To find the area between these curves, it involves:

• First, computing the area under each curve separately.
• Then, taking the difference of these areas.

This can be mathematically represented as: \ \ \ \ \[ \int_{-1}^{1} (3 - x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx = \int_{-1}^{1} [f(x) - g(x)] \, dx \] \ \ \ This integral gives the exact area between the curves \(y = f(x)\) and \(y = g(x)\) within the given interval.

Parabolas in calculus

Understanding parabolas in calculus involves recognizing their standard forms and properties.
The function \(f(x) = 3 - x^2\) represents a downward parabola, while \(g(x) = 2x^2\) represents an upward parabola.
Key characteristics for parabolas include:

• The vertex, or highest/lowest point, depending on whether the parabola opens up or down.
• Points where the parabola intersects the x-axis (known as roots or zeros).

For instance, the parabola **\(f(x) = 3 - x^2\)**:

• Has a vertex at \((0,3)\).
• Intersects the x-axis at \(x = \pm \sqrt{3}\).

Meanwhile, the parabola **\(g(x) = 2x^2\)**:

• Has a vertex at \((0,0)\).
• Passes through the points \((1,2)\) and \((-1, 2)\).

These properties can be used in calculus to sketch the graphs and compute areas under or between these curves.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus that connects differentiation and integration.
It consists of two parts:

• The first part states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral of \(f(x)\) from \(a\) to \(b\) is given by \(F(b) - F(a)\).
• The second part states that if \(f\) is continuous on \([a, b]\), then the function \(F\) defined by \(F(x) = \int_a^x f(t) dt\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(F'(x) = f(x)\).

Applying this theorem to our example, consider the integral: \ \ \ \ \[ \int_{-1}^{1} (3 - x^2 - 2x^2) \, dx \] \ \ \ This simplifies to: \ \ \ \ \ \ \ \[ \int_{-1}^{1} (3 - 3x^2) \, dx \] \ \ \ Using antiderivatives, we find:

• The antiderivative of \(3 - 3x^2\)
\ \
• Evaluate it at the endpoints \(1\) and \(-1\)
\ \
• Subtract the values to obtain the total area.

By linking the areas we compute back to their respective functions' antiderivatives, we're harnessing the power of the Fundamental Theorem of Calculus to provide exact values.