Problem 27 - Calculus

Question

Find \(c\) of the Mean Value Theorem for Integrals for \(f(x)=3 x^{2}\) on [-4,-1].

Step-by-Step Solution

Answer

The value of \( c \) is \(-\sqrt{7}\).

Step 1: Recall the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f(x) \) is continuous on \([a, b]\), then there exists a point \( c \) in the interval \([a, b]\) such that \( f(c) \) times the length of the interval \([a, b]\) equals the integral of \( f(x) \) from \( a \) to \( b \). Mathematically, \( \int_{a}^{b} f(x) \, dx = f(c) \cdot (b-a) \).

Step 2: Determine the Length of the Interval

The interval given is \([-4, -1]\). The length \((b-a)\) of this interval is: \((-1) - (-4) = 3\).

Step 3: Calculate the Definite Integral of the Function

The function given is \( f(x) = 3x^2 \). Calculate the definite integral from \(-4\) to \(-1\): \[ \int_{-4}^{-1} 3x^2 \, dx. \]To solve this, find the antiderivative of \( 3x^2 \), which is \( x^3 \). Thus,\[ \int 3x^2 \, dx = x^3 + C. \]Now evaluate from \(-4\) to \(-1\):\[ x^3 \bigg|_{-4}^{-1} = (-1)^3 - (-4)^3 = -1 - (-64) = 63. \]

Step 4: Solve for \( c \)

Using the integral result and the length of the interval, set up the equation for the Mean Value Theorem:\[ 63 = f(c) \cdot 3. \]We have \( f(c) = 3c^2 \). Substitute and solve:\[ 63 = 3c^2 \cdot 3. \]\[ 63 = 9c^2. \]\[ c^2 = 7. \]Solve for \( c \):\[ c = \pm \sqrt{7}. \]

Step 5: Select the Appropriate Value of \( c \) in the Interval

Since \( c \) must lie in the interval \([-4, -1]\), the appropriate value is the negative one. Thus, \( c = -\sqrt{7}. \)

Key Concepts

Definite Integral

The concept of the definite integral is central to calculus and helps in calculating the area under a curve from one point to another. Consider the function\( f(x) = 3x^2 \) on the interval \([a, b] = [-4, -1]\). To find the area under the curve, we calculate the definite integral \( \int_{-4}^{-1} 3x^2 \, dx \). This integral represents the sum of infinitesimal rectangles' areas under the curve, providing the total area between the function and the x-axis.
The evaluation of the integral involves finding the antiderivative (or indefinite integral) of \( 3x^2 \). An antiderivative for \( 3x^2 \) is \( x^3 \), since differentiating \( x^3 \) gives us \( 3x^2 \) again. After obtaining the antiderivative, we apply the limits of integration: \(-4\) to \(-1\).

• We compute the antiderivative at the upper limit \(-1\): \( (-1)^3 = -1 \).
• Next, evaluate it at the lower limit \(-4\): \( (-4)^3 = -64 \).
• Subtract these values: \(-1 - (-64) = 63 \).

This result, 63, is the area under \( f(x) \) from \(-4\) to \(-1\), and it is used in finding \( c \) in the Mean Value Theorem for Integrals.

Continuous Function

A continuous function is one without breaks, jumps, or holes in its graph within a specified interval. This property is crucial when applying the Mean Value Theorem for Integrals, ensuring the existence of a real number \( c \) in the interval where the theorem holds.
For our function \( f(x) = 3x^2 \), continuity on the interval \([-4, -1]\) means there are no abrupt changes as \( x \) moves from \(-4\) to \(-1\). In simpler terms, imagine tracing the graph of \( f(x) \) with a pencil from left to right without lifting it from the paper.
Why is this important? Because the Mean Value Theorem for Integrals states that for any continuous function on a closed interval, there must be at least one point \( c \) within this interval. At \( c \), the function's value equals the average value of the function over that interval. Ensuring \( f(x) \) is continuous allows us to trust that such a \( c \) genuinely exists.

Antiderivative

An antiderivative of a function is another function whose derivative results in the original function. For instance, consider our function \( f(x) = 3x^2 \). We need to find a function \( F(x) \) such that \( F'(x) = 3x^2 \).
In this case, \( F(x) = x^3 \) serves as an antiderivative, because differentiating \( x^3 \) with respect to \( x \) provides \( 3x^2 \).
Finding the antiderivative is an essential step in computing definite integrals, as it allows us to transform the problem of calculating the area under a curve into evaluating a couple of values and doing a simple subtraction.

• The general form of an antiderivative includes a constant \( C \), so formally, it is \( F(x) = x^3 + C \).
• However, when dealing with definite integrals, this constant cancels itself out since we are interested in a specific interval.
• This process of using the antiderivative to find the value at the bounds of an interval leads us to the integral's value.

Thus, antiderivatives are integral (pun intended!) for solving any problems involving areas under curves.