Problem 14 - Calculus and its applications

Question

Find the area under the given curve over the indicated interval. $$ y=x^{2}-4 x ; \quad[-4,-2] $$

Step-by-Step Solution

Answer

The area under the curve from \(-4\) to \(-2\) is \(\frac{128}{3}\).

Step 1: Set up the definite integral

To find the area under the curve \(y = x^2 - 4x\) over the interval \([-4, -2]\), we need to calculate the definite integral of the function from \(-4\) to \(-2\). We express this as \[ \int_{-4}^{-2} (x^2 - 4x) \, dx. \]

Step 2: Find the antiderivative

Next, find the antiderivative of the integrand \(x^2 - 4x\). The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\), and the antiderivative of \(-4x\) is \(-2x^2\). Therefore, the antiderivative of \(x^2 - 4x\) is \[ F(x) = \frac{1}{3}x^3 - 2x^2. \]

Step 3: Evaluate the definite integral

Now, evaluate \(F(x)\) at the upper and lower limits of the integral. This means we need to find \(F(-2)\) and \(F(-4)\).- \(F(-2) = \frac{1}{3}(-2)^3 - 2(-2)^2 = \frac{1}{3}(-8) - 8 = -\frac{8}{3} - 8 = -\frac{32}{3}\).- \(F(-4) = \frac{1}{3}(-4)^3 - 2(-4)^2 = \frac{1}{3}(-64) - 32 = -\frac{64}{3} - 32 = -\frac{160}{3}\).

Step 4: Calculate the area

The area under the curve is the difference \(F(-2) - F(-4)\).\[-\frac{32}{3} - (-\frac{160}{3}) = -\frac{32}{3} + \frac{160}{3} = \frac{128}{3}.\]

Step 5: Final answer

After performing the calculations, the area under the curve \(y = x^2 - 4x\) from \(-4\) to \(-2\) is \(\frac{128}{3}\).

Key Concepts

Antiderivative

Understanding the concept of an antiderivative is crucial when working with integrals. In simple terms, an antiderivative is a function whose derivative matches the function you started with. An easier way to think about it is the 'reverse process' of differentiation. In the context of our problem, when we find the antiderivative of the function \(y = x^2 - 4x\), we look for a function \(F(x)\) that, when differentiated, gives us back \(x^2 - 4x\).
Here's how it works:

• The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\) because if you differentiate \(\frac{1}{3}x^3\), you get back \(x^2\).
• The antiderivative of \(-4x\) is \(-2x^2\) because if you differentiate \(-2x^2\), you get back \(-4x\).

Hence, the combined antiderivative of \(x^2 - 4x\) is \[ F(x) = \frac{1}{3}x^3 - 2x^2. \] This is a foundational step in evaluating the area under a curve using definite integrals.

Area under the curve

The concept of finding the "area under the curve" involves calculating the space enclosed by the curve of a function, the x-axis, and vertical lines at two points on the x-axis. To visualize: imagine shading this area under the curve from \(x = -4\) to \(x = -2\). In our scenario, this gives you the section of space we're interested in, confined by the boundaries of the curve \(y = x^2 - 4x\).
Using integration helps us find this area precisely. It's important to remember that if the entire area is above the x-axis, integration gives the exact positive area value. If parts lie below, like in many real-world applications, you'll receive negative values. In any such instance, only the absolute value contributes to physical area measurements.

Definite Integral Evaluation

Evaluating a definite integral takes us through using the antiderivative to find the exact area under a curve over a specified interval. Think of it as calculating the net area between the curve and the x-axis, spanning two specific bounds.
For our problem:

• We calculate the antiderivative at the upper limit \(x = -2\), resulting in \(-\frac{32}{3}\).
• At the lower limit \(x = -4\), we evaluate to find \(-\frac{160}{3}\).
• The difference between these two outputs gives us the area: \[-\frac{32}{3} - (-\frac{160}{3}) = \frac{128}{3}.\]

The calculation results in \(\frac{128}{3}\), representing the area under the curve from \(-4\) to \(-2\). This neat outcome showcases how integrating functions over specific intervals can effectively measure such regions under curves.