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Chapter 4: Problem 16

Find all values of c that satisfy the Mean Value Theorem for Integrals on the given interval. $$ f(x)=x^{2} ; \quad[-1,1] $$

Short Answer

Expert verified
The values of \(c\) are \(c = \pm\frac{1}{\sqrt{3}}\).

Step by step solution

01

Understand the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function is continuous on a closed interval \([a, b]\), then there exists at least one point \(c\) in the interval where \([a, b]\) such that \(f(c)\) equals the average value of \(f\) on \([a, b]\). This is mathematically expressed as: \[ f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \]
02

Set up the Integral

For the function \(f(x) = x^2\) on the interval \([-1, 1]\), we need to compute \[ \int_{-1}^{1} x^2 \, dx \]
03

Calculate the Integral

Compute the integral: \[ \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} \]Evaluate it by plugging in the bounds:\[ \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \]
04

Calculate the Average Value of the Function

The average value of the function on the interval \([-1, 1]\) is\[ \text{Average} = \frac{1}{1 - (-1)} \times \int_{-1}^{1} x^2 \, dx = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \]
05

Find All Values of c

According to the theorem, we need to find \(c\) such that \[ f(c) = \frac{1}{3} \]Substitute \(f(c)\):\[ c^2 = \frac{1}{3} \]Solving for \(c\), we get:\[ c = \pm \frac{1}{\sqrt{3}} \] Both these values lie within the interval \([-1, 1]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
Continuity is a crucial concept when applying the Mean Value Theorem (MVT) for Integrals. It assures us that a function behaves nicely, without any break, throughout a given interval. If you visualize a continuous function, you can draw its curve on a graph without lifting your pen. In mathematical terms, a function is continuous on a closed interval \[a, b\] if it is continuous at every point in \[a, b\].

For the Mean Value Theorem to apply, the function must meet this requirement. In the problem provided, \(f(x) = x^2\) is a polynomial. Since all polynomials are continuous everywhere on the real number line, \(f(x)\) is continuous on the interval \([-1, 1]\).

  • This means there are no jumps or breaks in the graph of \(f(x)\) from \(x = -1\) to \(x = 1\).
  • Ensures that the MVT can be applied to find a value \(c\) within this interval where the function's value equals its average value.
Definite Integral
A definite integral is a fundamental concept in calculus used to find the accumulated area under a curve within a given interval. When you calculate a definite integral, you're summing up infinitely small slices of the area between the curve of a function and the x-axis over a specific range.

For the function \(f(x) = x^2\) on the interval \([-1, 1]\), we calculate the definite integral:\[ \int_{-1}^{1} x^2 \, dx \]
This integral gives us the total area under the curve from \(x = -1\) to \(x = 1\). Solving this, as shown in the steps, yields:\[ \frac{2}{3} \]

  • Essential for finding the average value of the function on this interval.
  • Involves evaluating an antiderivative and applying the limits of integration.
Average Value of a Function
The average value of a function over an interval gives us a sense of the function's typical behavior between two points. For a function \(f\) continuous on \[a, b\], the average value is given by:\[ \text{Average} = \frac{1}{b-a} \int_a^b f(x) \, dx \]

In essence, this formula calculates the mean height of the curve \(f(x)\) from \(x = a\) to \(x = b\). Applying this to \(f(x) = x^2\) over \([-1, 1]\), we find:\[ \text{Average} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \]

  • This average value represents a flat line across the interval that encloses the same area under \(f(x)\).
  • According to the Mean Value Theorem for Integrals, there are points within the interval where \(f(x)\) matches this height.

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Most popular questions from this chapter

Show that if \(f(x)=\int_{2 x}^{5 x} \frac{1}{t} d t,\) then \(f\) is a constant function on \((0, \infty)\)

In Problems \(49-52,\) sketch the graph of the given function over the interval \([a, b] ;\) then divide \([a, b]\) into \(n\) equal subintervals. Finally, calculate the area of the corresponding circumscribed polygon. \(f(x)=x+1 ; a=-1, b=2, n=3\)

Approximate \(\int_{1}^{2} \frac{1}{1+x^{4}} d x\) using left, right, and midpoint Riemann sums with \(n=8\).

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Find the average value of the function on the given interval. $$ f(x)=\cos x ; \quad[0, \pi] $$
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