Math Problem Statement

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.) y=x^2- x, y=4 x+6 Then find the area S of the region.

Solution

To solve this problem, let's break it down step by step.

Step 1: Sketch the region enclosed by the curves y=x2xy = x^2 - x and y=4x+6y = 4x + 6.

  1. Set up the equations:

    • The first curve is a parabola: y=x2xy = x^2 - x.
    • The second curve is a line: y=4x+6y = 4x + 6.
  2. Find the points of intersection between the curves by setting the two equations equal: x2x=4x+6x^2 - x = 4x + 6 Simplify and solve the quadratic equation: x25x6=0x^2 - 5x - 6 = 0 Factor the quadratic: (x6)(x+1)=0(x - 6)(x + 1) = 0 So, x=6x = 6 and x=1x = -1.

    These are the points of intersection: x=1x = -1 and x=6x = 6.

  3. Plot the curves and shade the region between x=1x = -1 and x=6x = 6. The region is bounded between these two points along the x-axis, with the parabola below the line.

Step 2: Decide whether to integrate with respect to xx or yy.

Since we are integrating between two vertical lines (from x=1x = -1 to x=6x = 6), we will integrate with respect to xx. The height of the approximating rectangle will be the vertical distance between the two curves, which is the difference in their yy-values at each xx.

Step 3: Draw a typical approximating rectangle.

  • The width of the rectangle is a small change in xx, denoted by dxdx.
  • The height of the rectangle is the difference between the two functions at any point xx: Height=(4x+6)(x2x)=4x+6x2+x=x2+5x+6\text{Height} = (4x + 6) - (x^2 - x) = 4x + 6 - x^2 + x = -x^2 + 5x + 6

Step 4: Set up the integral.

To find the area, we integrate the difference between the two functions over the interval x=1x = -1 to x=6x = 6: A=16((4x+6)(x2x))dxA = \int_{-1}^{6} \left( (4x + 6) - (x^2 - x) \right) dx Simplifying the integrand: A=16(x2+5x+6)dxA = \int_{-1}^{6} (-x^2 + 5x + 6) dx

Step 5: Solve the integral.

First, find the indefinite integral: (x2+5x+6)dx=x33+5x22+6x\int (-x^2 + 5x + 6) dx = -\frac{x^3}{3} + \frac{5x^2}{2} + 6x

Now, evaluate the definite integral: A=[x33+5x22+6x]16A = \left[ -\frac{x^3}{3} + \frac{5x^2}{2} + 6x \right]_{-1}^{6}

Substitute the limits:

  • At x=6x = 6: 633+5(6)22+6(6)=2163+5(36)2+36=72+90+36=54-\frac{6^3}{3} + \frac{5(6)^2}{2} + 6(6) = -\frac{216}{3} + \frac{5(36)}{2} + 36 = -72 + 90 + 36 = 54
  • At x=1x = -1: (1)33+5(1)22+6(1)=13+5(1)26=13+526=13+2.56=3.1667-\frac{(-1)^3}{3} + \frac{5(-1)^2}{2} + 6(-1) = -\frac{-1}{3} + \frac{5(1)}{2} - 6 = \frac{1}{3} + \frac{5}{2} - 6 = \frac{1}{3} + 2.5 - 6 = -3.1667

Now, subtract the values: A=54(3.1667)=54+3.1667=57.1667A = 54 - (-3.1667) = 54 + 3.1667 = 57.1667

Thus, the area of the region is approximately 57.17 square units.

Step 6: Conclusion

  • The area of the region enclosed by the curves is approximately 57.17 square units.
  • The typical rectangle is vertical, with width dxdx and height x2+5x+6-x^2 + 5x + 6.

Would you like further details or clarifications on any of the steps?

Here are some related questions to deepen your understanding:

  1. How would the process change if the curves were given in terms of y=y = instead of x=x =?
  2. What would the integral look like if you had to integrate with respect to yy instead of xx?
  3. How can you use the symmetry of the curves to simplify the integral?
  4. What happens to the area if the limits of integration are changed?
  5. How do the properties of parabolas and linear functions affect the shape of the enclosed region?

Tip: Always visualize the region before setting up the integral—it helps in deciding the limits of integration and choosing whether to integrate with respect to xx or yy.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Area under Curves
Quadratic Equations
Linear Functions

Formulas

Definite Integral: A = ∫[a, b] f(x) dx
Difference of functions: Height = f(x) - g(x)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12