Problem 14
Question
Write the given (total) area as an integral or sum of integrals. The area below the \(x\) -axis and above \(y=x^{2}-4 x\).
Step-by-Step Solution
Verified Answer
The area below the x-axis and above the curve \(y=x^{2}-4 x\) can be represented as the integral \(-\int_{0}^{4}(x^{2}-4x)dx\).
1Step 1: Identify the roots
The roots of the equation \(x^2 - 4x = 0\) can be found by first factoring out an x, so \(x(x-4) = 0\). Setting each factor equal to zero gives the roots as \(x = 0\) and \(x = 4\). These will be the limits of integration.
2Step 2: Write the integral
Write down the integral from 0 to 4 of the function \(x^2 - 4x\). However, the area is below the x-axis so the function values will be negative. Multiply the integral by -1 to account for this. Hence, the integral becomes \(-\int_{0}^{4}(x^{2}-4x)dx\).
3Step 3: Evaluate the integral
Finally, evaluate \(-\int_{0}^{4}(x^{2}-4x)dx\). This involves employing the power rule for integrals, which suggests increasing the power by one and dividing by the new power.
Key Concepts
Integral EvaluationPower Rule for IntegralsLimits of Integration
Integral Evaluation
The process of integral evaluation involves finding the accumulated sum or difference that a function represents over a particular interval. It is akin to adding up an infinite number of infinitesimally small pieces to determine the total value.
In the context of the given exercise, the integral evaluation is necessary to ascertain the total area beneath the curve of the function and above the x-axis, even when parts of this area might be negative because the curve dips below the axis. We account for the negative area by introducing a negative sign in front of the integral, transforming it into \( -\int_{0}^{4}(x^{2}-4x)dx \). This effectively 'flips' the area to be positive, reflecting the true magnitude of the area below the x-axis.
To evaluate this integral, one would perform the antiderivative process for each term in the function \( x^{2}-4x \) and then apply the boundaries, also known as the limits of integration, to find the actual value of the definite integral. This solution follows such steps, ensuring that the areas beneath the x-axis are considered accurately in the final calculation.
In the context of the given exercise, the integral evaluation is necessary to ascertain the total area beneath the curve of the function and above the x-axis, even when parts of this area might be negative because the curve dips below the axis. We account for the negative area by introducing a negative sign in front of the integral, transforming it into \( -\int_{0}^{4}(x^{2}-4x)dx \). This effectively 'flips' the area to be positive, reflecting the true magnitude of the area below the x-axis.
To evaluate this integral, one would perform the antiderivative process for each term in the function \( x^{2}-4x \) and then apply the boundaries, also known as the limits of integration, to find the actual value of the definite integral. This solution follows such steps, ensuring that the areas beneath the x-axis are considered accurately in the final calculation.
Power Rule for Integrals
The power rule for integrals is a fundamental technique used to evaluate antiderivatives. It states that for any real number \( n \) not equal to -1, the integral of \( x^n \) with respect to \( x \) is \( \frac{1}{n+1}x^{n+1} \), plus a constant \( C \).
For instance, when applying the power rule to the function \( x^{2}-4x \), you increase the exponent by one and divide by the new exponent. Thus, the antiderivative of \( x^2 \) becomes \( \frac{1}{3}x^3 \) and the antiderivative of \( 4x \) becomes \( 2x^2 \).
In a definite integral, as done in the exercise, the constant \( C \) is not required as we are interested in the difference between the antiderivative evaluated at the upper and lower limits. Consequently, after applying the power rule to the integrand, you would plug in the limits of integration to find the exact value of the area.
For instance, when applying the power rule to the function \( x^{2}-4x \), you increase the exponent by one and divide by the new exponent. Thus, the antiderivative of \( x^2 \) becomes \( \frac{1}{3}x^3 \) and the antiderivative of \( 4x \) becomes \( 2x^2 \).
In a definite integral, as done in the exercise, the constant \( C \) is not required as we are interested in the difference between the antiderivative evaluated at the upper and lower limits. Consequently, after applying the power rule to the integrand, you would plug in the limits of integration to find the exact value of the area.
Limits of Integration
Limits of integration define the interval over which an integral is evaluated. These limits are essential when calculating definite integrals, which represent the total accumulation of a function's value over a specific range.
In our exercise, the limits of integration are determined by the roots of the given function, where the area of interest intersects the x-axis. These roots were found to be \( x = 0 \) and \( x = 4 \) after factoring and solving the equation \( x^2 - 4x = 0 \).
The limits also aid in visualizing the problem: the lower limit signifies the start of the area of interest on the x-axis (in this case at \( x = 0 \) ), while the upper limit marks the end (here at \( x = 4 \) ). Thus, when you set up the integral \( -\int_{0}^{4}(x^{2}-4x)dx \), you're effectively 'telling' the integral the exact section of the x-axis over which to calculate the area. These bounds are crucial to finding a precise solution, as they constrain the integral to just the region specified by the problem.
In our exercise, the limits of integration are determined by the roots of the given function, where the area of interest intersects the x-axis. These roots were found to be \( x = 0 \) and \( x = 4 \) after factoring and solving the equation \( x^2 - 4x = 0 \).
The limits also aid in visualizing the problem: the lower limit signifies the start of the area of interest on the x-axis (in this case at \( x = 0 \) ), while the upper limit marks the end (here at \( x = 4 \) ). Thus, when you set up the integral \( -\int_{0}^{4}(x^{2}-4x)dx \), you're effectively 'telling' the integral the exact section of the x-axis over which to calculate the area. These bounds are crucial to finding a precise solution, as they constrain the integral to just the region specified by the problem.
Other exercises in this chapter
Problem 14
Evaluate the derivative using properties of logarithms where needed. $$\frac{d}{d x}\left[\ln \left(x^{5} \sin x \cos x\right)\right]$$
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Use Part I of the Fundamental Theorem to compute each integral exactly. $$\int_{-1}^{1} \frac{4}{1+x^{2}} d x$$
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Construct a table of Riemann sums as in example 3.5 to show that sums with right-endpoint, midpoint and Ieft-endpoint evaluation all converge to the same value
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Use summation rules to compute the sum. $$\sum_{i=1}^{140}\left(i^{2}+2 i-4\right)$$
View solution