Problem 28 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves. \(y=x^{2}, y=3 x-2\)

Step-by-Step Solution

Answer

The area between the curves is \(\frac{11}{6}\) square units.

Step 1: Sketch the Graphs

Before doing any calculations, it is helpful to sketch the functions to visually identify the region of interest. The graph of `y = x^2` is a parabola opening upwards with a vertex at the origin (0,0). The line `y = 3x - 2` is a straight line with a y-intercept at -2 and a slope of 3.

Step 2: Find Points of Intersection

To find the points where the two curves intersect, set the equations equal to each other: \(x^2 = 3x - 2\). Rearrange the equation to \(x^2 - 3x + 2 = 0\) and solve for \(x\). Factoring gives \((x - 1)(x - 2) = 0\), so \(x = 1\) and \(x = 2\). Thus, the intersection points are (1,1) and (2,4).

Step 3: Establish Limits of Integration

The area to be found is between \(x = 1\) and \(x = 2\), where the curves intersect, hence the limits for integration will be these x-values.

Step 4: Set Up the Integral for Area

To calculate the area between the curves, we integrate the difference of the functions. The line \(y = 3x - 2\) is above the parabola \(y = x^2\) between x = 1 and x = 2. The integral for the area is: \ \[ \int_{1}^{2} ((3x - 2) - x^2) \, dx \]

Step 5: Evaluate the Integral

First, simplify the integrand: \[ (3x - 2) - x^2 = -x^2 + 3x - 2 \]Integrate term by term: \[ \int_{1}^{2} (-x^2 + 3x - 2) \, dx = \left[ \frac{-x^3}{3} + \frac{3x^2}{2} - 2x \right]_{1}^{2} \]Calculate the definite integral:For \(x = 2\): \[ \frac{-(2)^3}{3} + \frac{3(2)^2}{2} - 2(2) = \frac{-8}{3} + 6 - 4 = \frac{2}{3} \]For \(x = 1\): \[ \frac{-(1)^3}{3} + \frac{3(1)^2}{2} - 2(1) = \frac{-1}{3} + \frac{3}{2} - 2 = \frac{-7}{6} \]Subtract the results: \[ \frac{2}{3} - \left(\frac{-7}{6}\right) = \frac{2}{3} + \frac{7}{6} = \frac{4}{6} + \frac{7}{6} = \frac{11}{6} \]

Step 6: Conclude the Area

The absolute value of the result from the integration provides the area between the two curves. Therefore, the area of the region bounded by \(y=x^2\) and \(y=3x-2\) is \(\frac{11}{6}\) square units.

Key Concepts

Definite Integral

In calculus, a definite integral is used to compute the area under a curve, between a set of bounds or limits. It's represented generally as \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function being integrated between the limits \( a \) and \( b \). This integral gives us the net area between the curve of \( f(x) \) and the x-axis in that interval.
For the problem at hand, the definite integral was used to find the area between two curves, \( y = x^2 \) and \( y = 3x - 2 \), over the interval \([1, 2]\). When integrating to find the area between curves, it’s important to integrate the difference between the higher curve \( y = 3x - 2 \) and the lower curve \( y = x^2 \), ensuring you are calculating the area of the region enclosed by the curves.

Points of Intersection

Finding the points of intersection is a crucial step when solving problems involving areas between curves. These points help us establish the limits for integration, which is where the two curves cross or intersect each other.
To find these points, we set the equations of the curves equal to each other, which in this case are \( x^2 = 3x - 2 \). Solving the resulting equation allows us to identify the x-values where both graphs cross.
By rearranging and factoring \( x^2 - 3x + 2 = 0 \), we obtain the factorized form \((x - 1)(x - 2) = 0\). Consequently, the intersection points for \( x \) are 1 and 2, meaning the curves intersect at points \((1,1)\) and \((2,4)\). These intersection points mark where the curves overlap in our integration process.

Limits of Integration

Limits of integration define the bounds within which we apply our integral calculation. When determining the area between curves, it's vital to know where the region of interest starts and ends along the x-axis.
For this problem, the limits of integration are based on the x-values of the points of intersection we calculated earlier, x = 1 and x = 2. These limits ensure that the integral covers the entire segment between the intersections where our area calculation is focused.

• The start of the interval is at x = 1.
• The end of the interval is at x = 2.

This range \([1, 2]\) forms the boundaries of our definite integral, \( \int_{1}^{2} ((3x - 2) - x^2) \, dx \), ensuring we are only evaluating the area within the specified region.

Integration Techniques

Integrating functions involves applying techniques that simplify the function into a form which is easier to integrate term by term. In our example, once we set the integrand as \( (3x - 2) - x^2 \), it requires simplifying to \( -x^2 + 3x - 2 \) before integrating.
Using the power rule of integration for each term separately is the key technique here:

• \( \int -x^2 \, dx = \frac{-x^3}{3} \)
• \( \int 3x \, dx = \frac{3x^2}{2} \)
• \( \int -2 \, dx = -2x \)

After integrating term by term from x = 1 to x = 2, we substitute these boundaries into the integrated expression and compute the difference to find the final area. This approach of breaking down integral components into manageable parts is fundamental to evaluating definite integrals effectively.