Question

Find the total area enclosed by the functions \(f\) and \(g\). \(f(x)=-x^{3}+5 x^{2}+2 x+1, g(x)=3 x^{2}+x+3\)

Short answer

The total area is \(\frac{16}{3}\).

Step-by-step solution

Find Points of Intersection

To find the points of intersection of the two curves, we must solve the equation \( f(x) = g(x) \). This means setting \(-x^{3}+5x^{2}+2x+1 = 3x^{2}+x+3\). After rearranging, we get a cubic equation: \(-x^{3}+2x^{2}+x-2 = 0\). Solving this gives us the intersection points \(x = -1, \ x = 1, \ x = 2\).

Set Up the Area Integral

The area between the two curves from \(a\) to \(b\) is given by the integral \(\int_a^b |f(x)-g(x)| \, dx\). Since the region is divided by the intersection points, we need to set up separate integrals. This results in:1. \( \int_{-1}^{1} (g(x) - f(x)) \, dx\)2. \( \int_{1}^{2} (f(x) - g(x)) \, dx\)

Solve the First Integral

Calculate the integral \(\int_{-1}^{1} (g(x) - f(x)) \, dx\):\[\int_{-1}^{1} [(3x^2 + x + 3) - (-x^3 + 5x^2 + 2x + 1)] \, dx= \int_{-1}^{1} (x^3 - 2x^2 - x + 2) \, dx\]Now, integrate term by term:- Integral of \(x^3\) is \(\frac{x^4}{4}\)- Integral of \(-2x^2\) is \(-\frac{2x^3}{3}\)- Integral of \(-x\) is \(-\frac{x^2}{2}\)- Integral of \(2\) is \(2x\)Evaluate from \(x = -1\) to \(x = 1\):\[\left[\frac{x^4}{4} - \frac{2x^3}{3} - \frac{x^2}{2} + 2x\right]_{-1}^{1} = \left(\frac{1}{4} - \frac{2}{3} - \frac{1}{2} + 2\right) - \left(\frac{-1}{4} + \frac{2}{3} - \frac{1}{2} - 2\right)\]Calculating this gives us an area of \(\frac{37}{12}\) between the curves from \(x = -1\) to \(x = 1\).

Solve the Second Integral

Calculate the integral \(\int_{1}^{2} (f(x) - g(x)) \, dx\):\[\int_{1}^{2} [(-x^3 + 5x^2 + 2x + 1) - (3x^2 + x + 3)] \, dx= \int_{1}^{2} (-x^3 + 2x^2 + x - 2) \, dx\]Now, integrate term by term:- Integral of \(-x^3\) is \(-\frac{x^4}{4}\)- Integral of \(2x^2\) is \(\frac{2x^3}{3}\)- Integral of \(x\) is \(\frac{x^2}{2}\)- Integral of \(-2\) is \(-2x\)Evaluate from \(x = 1\) to \(x = 2\):\[\left[-\frac{x^4}{4} + \frac{2x^3}{3} + \frac{x^2}{2} - 2x\right]_{1}^{2} = \left(-\frac{16}{4} + \frac{16}{3} + 2 - 4\right) - \left(-\frac{1}{4} + \frac{2}{3} + \frac{1}{2} - 2\right)\]Calculating this gives us an area of \(\frac{9}{4}\) between the curves from \(x = 1\) to \(x = 2\).

Find Total Area

Add the areas calculated in Step 3 and Step 4:Total Area = \(\frac{37}{12} + \frac{9}{4}\)Convert \(\frac{9}{4}\) to \(\frac{27}{12}\) to have a common denominator:Total Area = \(\frac{37}{12} + \frac{27}{12} = \frac{64}{12} = \frac{16}{3}\)The total area enclosed by the curves is \(\frac{16}{3}\).

Key concepts

Integrals

Integrals are a fundamental concept in calculus, often defined as the area under a curve within a given interval. When we evaluate an integral, we are essentially summing up tiny slices of area to find a total. This process is called integration. There are two main types of integrals: definite and indefinite.

A definite integral, such as \( \int_a^b f(x) \, dx \), includes limits of integration \( a \) and \( b \), and gives a numerical result representing the area between the function \( f(x) \) and the x-axis over that interval. Conversely, an indefinite integral is more general, providing an antiderivative or a family of functions differing by a constant, indicated by \( \int f(x) \, dx \).

In our problem, using definite integrals is crucial since we are dealing with specific intervals to find the area between curves. We calculate the integral separately for each region where the two functions switch dominance.

Area Between Curves

When finding the area between two curves, we are interested in the space confined by the graphs of two functions over a particular interval. This area can be calculated using integrals by subtracting the lower function from the upper function within the desired limits.

For the functions \( f(x) \) and \( g(x) \), the formula to get the area between them from \( x = a \) to \( x = b \) can be expressed as:

• \( \int_a^b (f(x) - g(x)) \, dx \) when \( f(x) \) is above \( g(x) \)
• \( \int_a^b (g(x) - f(x)) \, dx \) when \( g(x) \) is above \( f(x) \)

In this exercise, we saw these two curves switching positions at the points of intersection, requiring us to evaluate separate integrals for each portion where the upper and lower roles of the functions switch to accurately find the total enclosed area.

Points of Intersection

Points of intersection are the x-values where two or more functions meet. Finding these points helps in determining the limits of integration when calculating the area between curves.

To find these points, set the equations for the two functions equal to each other: \( f(x) = g(x) \). For the given problem, solving \( -x^3 + 5x^2 + 2x + 1 = 3x^2 + x + 3 \) led us to a cubic equation,
\, \( -x^3 + 2x^2 + x - 2 = 0 \). Solving this, we obtained the points of intersection as \( x = -1 \), \( x = 1 \), and \( x = 2 \). These are critical to determining the boundaries for each integral used in the solution.

Cubic Equations

Cubic equations are polynomial equations of degree three, typically taking the form \( ax^3 + bx^2 + cx + d = 0 \). The challenge with these equations often lies in finding their roots, which are solutions for \( x \) that satisfy the equation.

Solving cubic equations can involve several methods, from graphical approaches, leveraging symmetry or patterns, to algebraic techniques, such as factoring or applying the cubic formula. In this exercise, solving the cubic equation \( -x^3 + 2x^2 + x - 2 = 0 \) helped us determine critical points of intersection. In doing so, we learned about where the functions \( f(x) \) and \( g(x) \) meet, allowing us to set the boundaries for our integrals and accurately calculate the area between these curves.