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Chapter 4: Problem 14

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

Short Answer

Expert verified
\( \frac{128}{3} \)

Step by step solution

01

- Set up the Integral

To find the area under the curve, the integral of the function over the given interval needs to be set up. The function is given by \[ y = x^{2} - 4x \] and the interval is \[ [-4, -2] \]. Thus, the integral to evaluate is \[ \int_{-4}^{-2} (x^{2} - 4x) \,dx \].
02

- Integrate the Function

Integrate the function \[ x^{2} - 4x \] with respect to x. The antiderivative of \[ x^{2} \] is \[ \frac{x^{3}}{3} \] and the antiderivative of \[ 4x \] is \[ 2x^{2} \]. Therefore, the integral of \[ x^{2} - 4x \] is: \[ \int (x^{2} - 4x) \,dx = \frac{x^{3}}{3} - 2x^{2} + C \], where C is the constant of integration.
03

- Evaluate the Definite Integral

To find the area under the curve between \[ x = -4 \] and \[ x = -2 \], substitute these limits into the antiderivative: \[ \left[ \frac{x^{3}}{3} - 2x^{2} \right]_{-4}^{-2} \]. First, evaluate at \( x = -2 \): \[ \frac{(-2)^{3}}{3} - 2(-2)^{2} = \frac{-8}{3} - 2(4) = \frac{-8}{3} - 8 = \frac{-8 - 24}{3} = \frac{-32}{3} \]. Then, evaluate at \( x = -4 \): \[ \frac{(-4)^{3}}{3} - 2(-4)^{2} = \frac{-64}{3} - 2(16) = \frac{-64}{3} - 32 = \frac{-64 - 96}{3} = \frac{-160}{3} \].
04

- Subtract the Values

Subtract the value of the antiderivative at \[ -4 \] from the value at \[ -2 \]: \[ \frac{-32}{3} - \frac{-160}{3} = \frac{-32 + 160}{3} = \frac{128}{3} \].

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Key Concepts

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

Definite Integral
A definite integral helps us find the area under a curve between two points. Think of it as summing up tiny rectangles under the curve.
In our exercise, we used the boundaries \([-4, -2]\) to calculate the area under \( y = x^2 - 4x \).
This process involves a few steps:
  • Set up the integral with the given function and interval.
  • Find the antiderivative (indefinite integral) of the function.
  • Evaluate the antiderivative at both endpoints.
  • Subtract the values to get the area.
The notation for a definite integral, \(\int_{a}^{b} f(x)\,dx\), tells us to compute the area from \(x = a\) to \(x = b\). It's essential for precisely finding areas in calculus.
Antiderivative
The antiderivative, or indefinite integral, is a function that reverses what differentiation does. In other words, if we take the derivative of the antiderivative, we get our original function back.
For our exercise, the antiderivative of \(x^2 - 4x\) is \[\frac{x^3}{3} - 2x^2 + C\].
Here's how it works:
  • The antiderivative of \(x^2\) is \[\frac{x^3}{3}\]
  • The antiderivative of \(4x\) is \[2x^2\]
The constant \(C\) appears because any constant added to an antiderivative will disappear when differentiated. However, when dealing with definite integrals, the constant cancels out, so we don't include it in our final calculation.
Calculus
Calculus, the branch of mathematics that explores change, focuses mainly on differentiation and integration.
To see why it鈥檚 important:
  • Differentiation finds out the rate of change or slope of a function at any given point.
  • Integration, on the other hand, accumulates quantities, calculating areas under curves as we did in our exercise.
The concepts of limits form the backbone of both differentiation and integration. They allow us to handle infinities or small quantities in a controlled way.
Understanding calculus is crucial because it applies to various real-world problems:
  • Physics (motion, forces)
  • Economics (cost, profit maximization)
  • Engineering (design, optimization)
  • Biology (population modeling)
This example of finding the area under the curve using a definite integral showcases just one of the many applications of calculus.

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Most popular questions from this chapter

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