Problem 23 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves. \(y=x^{2}-1, y=x+1\)

Step-by-Step Solution

Answer

The area of the region is \(\frac{25}{6}\) square units.

Step 1: Set the equations equal to find points of intersection

To find the points where the curves intersect, equate the two expressions: \[ x^2 - 1 = x + 1 \]Rearranging gives: \[ x^2 - x - 2 = 0 \]This equation can be factored as: \[ (x - 2)(x + 1) = 0 \]Therefore, the solutions are: \[ x = 2 \text{ and } x = -1 \]

Step 2: Verify intersection points

Substitute the x-values into either function to find the y-values.For \(x = 2\):\[ y = 2 + 1 = 3 \]For \(x = -1\):\[ y = -1 + 1 = 0 \]Thus, the points of intersection are \((2, 3)\) and \((-1, 0)\).

Step 3: Determine which function is above the other

For values between \(-1\) and \(2\), determine which curve lies above by comparing the functions.Choose a point in this interval, like \(x = 0\).For \(y = x^2 - 1\), when \(x = 0\), \( y = 0^2 - 1 = -1 \).For \(y = x + 1\), when \(x = 0\), \( y = 0 + 1 = 1 \).Thus, \(y = x + 1\) is above \(y = x^2 - 1\) on this interval.

Step 4: Set up the integral to find the area

The area between the curves from \(x = -1\) to \(x = 2\) is given by the integral:\[ \int_{-1}^{2} ((x + 1) - (x^2 - 1)) \, dx \]Simplify the expression inside the integral:\[ \int_{-1}^{2} (x + 1 - x^2 + 1) \, dx \]\[ \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

Step 5: Evaluate the integral

Integrate the simplified expression:\[ \int (-x^2 + x + 2) \, dx = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^{2} \]Compute the definite integral by substituting the bounds:\[ \left(-\frac{2^3}{3} + \frac{2^2}{2} + 2 \times 2 \right) - \left(-\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2 \times (-1) \right) \]\[ = \left(-\frac{8}{3} + 2 + 4 \right) - \left(\frac{1}{3} + \frac{1}{2} - 2 \right) \]\[ = \left( -\frac{8}{3} + 6 \right) - \left(\frac{1}{3} + \frac{1}{2} - 2 \right) \]\[ = \frac{10}{3} - \left( \frac{1}{3} + \frac{3}{6} - 2 \right) \]\[ = \frac{10}{3} - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) \]\[ = \frac{10}{3} - \frac{-5}{6} \]\[ = \frac{20}{6} + \frac{5}{6} \]\[ = \frac{25}{6} \]

Step 6: Conclusion

The area of the region bounded by the curves \(y = x^2 -1\) and \(y = x + 1\) is \(\frac{25}{6}\) square units.

Key Concepts

Intersection Points

Intersection points are where two curves meet on a graph. In this problem, the task is to identify the points at which the parabola, given by \( y = x^2 - 1 \), intersects with the line, expressed as \( y = x + 1 \).
To find these intersection points, set the two equations equal to each other:
\( x^2 - 1 = x + 1 \).
This simplifies to:

• \( x^2 - x - 2 = 0 \)

This is a quadratic equation, which can be factored as follows:

• \( (x - 2)(x + 1) = 0 \)

This gives us two solutions or intersection points at:

• \( x = 2 \)
• \( x = -1 \)

To get the corresponding \( y \) values, substitute these \( x \) values back into either original equation. This verification process will ensure the points are indeed where the curves intersect. Thus, the points are \( (2, 3) \) and \( (-1, 0) \).
Knowing these points is crucial when solving area problems, as they define the bounds for integration and help identify regions of interest in calculus. Let's delve further into these concepts in the next sections.

Definite Integral

The definite integral is an essential concept in calculus, especially when dealing with areas. It allows us to calculate the exact area under a curve within specific boundaries. Once we have the intersection points, which in this problem are \( x = -1 \) and \( x = 2 \), we can set up a definite integral. This integral will calculate the area of the region between the curves from one intersection point to the other.
For the problem at hand, the definite integral is used to find the area between \( y = x^2 - 1 \) and \( y = x + 1 \) on the interval from \( x = -1 \) to \( x = 2 \). The integral is structured as:

• \[ \int_{-1}^{2} ((x + 1) - (x^2 - 1)) \, dx \]

After simplification, this becomes:

• \[ \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

The definite integral calculates the net area between the curves from the lower limit to the upper limit. This process involves finding the anti-derivative,and then evaluating it at the upper and lower limits.
In this context, the definite integral determines the real-world quantity of area between the two curves, effectively giving us the final solution.

Area Between Curves

Finding the area between curves is a common topic in calculus. It involves determining how much space lies between two functions over an interval. Once we have identified the intersection points and set up the definite integral, we proceed to solve for the area.
For the curves \( y = x^2 - 1 \) and \( y = x + 1 \), once we've determined which function is on top (in this case \( y = x + 1 \) is above \( y = x^2 - 1 \) for \(-1 \leq x \leq 2\)), the problem simplifies to subtracting the lower curve from the upper one within the integral:

• \[ \int_{-1}^{2} ((x + 1) - (x^2 - 1)) \, dx \]
• Simplifies to: \[ \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

The need to set up a correct integral by establishing what curve is on top helps ensures the answer correctly reflects the desired area.
Upon evaluating this integral using fundamental calculus techniques, the area between the curves is found to be \( \frac{25}{6} \) square units.
Understanding how to calculate the area between curves using integration not only helps in mathematics but also in real-world applications such as economics, engineering, and even biology, where understanding these spatial relationships can be critical.