Problem 39
Question
Evaluate the integral by interpreting it in terms of areas. \( \displaystyle \int^3_{-4} \biggl| \frac{1}{2}x \biggr| \, dx \)
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{25}{4} \).
1Step 1: Understand the Function Inside the Integral
The integral to be evaluated is \( \int_{-4}^{3} \left| \frac{1}{2}x \right| \, dx \). The function \( f(x) = \left| \frac{1}{2}x \right| \) represents the absolute value of the function \( \frac{1}{2}x \), which forms a V-shape graph that is symmetric about the y-axis. The point of symmetry is the origin (0,0).
2Step 2: Identify Points Where Function Changes
Since we have an absolute value, the function changes from negative to positive at \( x = 0 \). Therefore, we will break the integral at this point and analyze it as two separate regions for \( x < 0 \) and \( x > 0 \).
3Step 3: Evaluate Negative Part of the Function
For \( x \in [-4, 0] \), the absolute value function can be rewritten as \( -\frac{1}{2}x \). Thus, the integral from \(-4\) to \(0\) is \( \int_{-4}^{0} \left(-\frac{1}{2}x\right) \, dx \). Find the area under this line: \[\int_{-4}^0 \left(-\frac{1}{2}x\right) \, dx = \left[ -\frac{1}{4}x^2 \right]_{-4}^{0}\]Calculate the result: \(-\frac{1}{4}(0)^2 + \frac{1}{4}(-4)^2 = 0 + 4 = 4\).
4Step 4: Evaluate Positive Part of the Function
For \( x \in [0, 3] \), the function remains as \( \frac{1}{2}x \). Thus, the integral from \(0\) to \(3\) is \( \int_{0}^{3} \left(\frac{1}{2}x\right) \, dx \). Evaluate this integral to find the area:\[\int_{0}^{3} \frac{1}{2}x \, dx = \left[ \frac{1}{4}x^2 \right]_{0}^{3}\]Calculate the result: \(\frac{1}{4}(3)^2 - \frac{1}{4}(0)^2 = \frac{9}{4}\).
5Step 5: Sum of Areas
Add the two calculated areas to find the total area under the curve from \( x = -4 \) to \( x = 3 \), which represents the value of the integral. The sum is \( 4 + \frac{9}{4} = \frac{16}{4} + \frac{9}{4} = \frac{25}{4}\).
Key Concepts
Absolute Value FunctionDefinite IntegralArea Under Curve
Absolute Value Function
The absolute value function is a mathematical operation that takes any real number and returns its non-negative value. This means it converts negative inputs into positive ones, while positive inputs and zero remain unchanged. In the function \(f(x) = \left| \frac{1}{2}x \right|\), we deal with the absolute value of a linear function, specifically \(\frac{1}{2}x\). This results in a V-shaped graph that is symmetric about the y-axis. It meets at the origin, where the value is zero.
The nature of this function is such that for \(x < 0\), the graph is a mirror image of the graph when \(x > 0\). This happens because the absolute value of a negative number is the same as the absolute value of its positive counterpart. Hence, the absolute value function shifts the graph upwards, forming two linear segments.
In calculus, understanding the behavior of the absolute value function is crucial for properly evaluating integrals, especially when the function crosses the x-axis, such as in the integral \(\int^3_{-4} \left| \frac{1}{2}x \right| \, dx\). This is why the integral is split into two different parts at the point \(x = 0\), where the function's behavior changes.
The nature of this function is such that for \(x < 0\), the graph is a mirror image of the graph when \(x > 0\). This happens because the absolute value of a negative number is the same as the absolute value of its positive counterpart. Hence, the absolute value function shifts the graph upwards, forming two linear segments.
In calculus, understanding the behavior of the absolute value function is crucial for properly evaluating integrals, especially when the function crosses the x-axis, such as in the integral \(\int^3_{-4} \left| \frac{1}{2}x \right| \, dx\). This is why the integral is split into two different parts at the point \(x = 0\), where the function's behavior changes.
Definite Integral
In calculus, a definite integral calculates the net area under a curve over a specified interval. The term 'definite' specifies that the integration is bound by two values, a lower and an upper limit. In our exercise, we're asked to evaluate the definite integral \(\int_{-4}^{3} \left| \frac{1}{2}x \right| \, dx\).
To solve this definite integral, we observe that the absolute value function \(\left| \frac{1}{2}x \right|\) behaves differently on the intervals \(x < 0\) and \(x > 0\). This requires breaking the integral into two parts: one from \(-4\) to \(0\) and another from \(0\) to \(3\).
For \(x\) in the first interval negative \([-4, 0]\), the function is \(-\frac{1}{2}x\). The integral over this interval results in 4, as calculated from substituting the limits into the expression \(-\frac{1}{4}x^2\).
For the second interval \([0, 3]\) where \(x\) is positive, the function remains \(\frac{1}{2}x\). Solving the integral yields \(\frac{9}{4}\) from substituting the limits into \(\frac{1}{4}x^2\). The combination of these integrals allows for a more comprehensive analysis of the behavior of piecewise functions like the absolute value function.
To solve this definite integral, we observe that the absolute value function \(\left| \frac{1}{2}x \right|\) behaves differently on the intervals \(x < 0\) and \(x > 0\). This requires breaking the integral into two parts: one from \(-4\) to \(0\) and another from \(0\) to \(3\).
For \(x\) in the first interval negative \([-4, 0]\), the function is \(-\frac{1}{2}x\). The integral over this interval results in 4, as calculated from substituting the limits into the expression \(-\frac{1}{4}x^2\).
For the second interval \([0, 3]\) where \(x\) is positive, the function remains \(\frac{1}{2}x\). Solving the integral yields \(\frac{9}{4}\) from substituting the limits into \(\frac{1}{4}x^2\). The combination of these integrals allows for a more comprehensive analysis of the behavior of piecewise functions like the absolute value function.
Area Under Curve
The area under a curve, typically calculated using definite integrals, reflects the sum of values represented by the curve over a specific interval. It represents a key concept in integral calculus for understanding the accumulated change over time or distance.
In the integral \(\int_{-4}^{3} \left| \frac{1}{2}x \right| \, dx\), we calculate the total area between the graph of the function and the x-axis from \(x = -4\) to \(x = 3\). This involves considering the sections of the graph where the curve dips below the x-axis and then rises above it. Because the function involves an absolute value, the curve essentially reflects any negative values into positive ones, facilitating a straightforward calculation of the total area.
For this exercise, we computed the area as follows:
In the integral \(\int_{-4}^{3} \left| \frac{1}{2}x \right| \, dx\), we calculate the total area between the graph of the function and the x-axis from \(x = -4\) to \(x = 3\). This involves considering the sections of the graph where the curve dips below the x-axis and then rises above it. Because the function involves an absolute value, the curve essentially reflects any negative values into positive ones, facilitating a straightforward calculation of the total area.
For this exercise, we computed the area as follows:
- From \(-4\) to \(0\), the area under the curve represented by \(-\frac{1}{2}x\) results in an area of 4.
- From \(0\) to \(3\), the area under the curve \(\frac{1}{2}x\) results in \(\frac{9}{4}\).
- Summing these areas gives a total area under the curve of \(\frac{25}{4}\).
Other exercises in this chapter
Problem 39
Evaluate the integral. \( \displaystyle \int^{8}_{1} \frac{2 + t}{\sqrt[3]{t^2}} \,dt \)
View solution Problem 39
Evaluate the integral. \( \displaystyle \int^{\sqrt{3}}_{1/\sqrt{3}} \frac{8}{1 + x^2} \,dx \)
View solution Problem 40
Evaluate the indefinite integral. \( \displaystyle \int \frac{\sin x}{1 + \cos^2 x} \, dx \)
View solution Problem 40
Evaluate the integral. \( \displaystyle \int^{10}_{-10} \frac{2e^x}{\sinh x + \cosh x} \,dx \)
View solution