Problem 16 - Essential Calculus

Question

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(x)=-x^{2}+4 x+1, g(x)=x+1 $$

Step-by-Step Solution

Answer

The area of the region bounded by the two given functions is 9 square units.

Step 1: Finding Intersection Points

First, set the two functions equal to each other and solve for x. So, solve \(f(x) = g(x)\), or \(-x^2 + 4x + 1 = x + 1\). This simplifies to \(x^2 - 3x = 0\), which factoring x out gives \(x(x - 3) = 0\). From this, it's deduced that the two functions intersect at \(x = 0\) and \(x = 3\).

Step 2: Drawing the Graph

Next, the region of interest is between \(x=0\) and \(x=3\). By plotting both functions, it can be seen that \(f(x)\) is above \(g(x)\) between these two points.

Step 3: Calculating the Area

The area of the region between these two functions is given by the definite integral \(\int_{a}^{b} (f(x) - g(x)) dx\), where [a,b] is the interval on which the point of intersection lie. Here, calculate \(\int_{0}^{3} ((-x^2 + 4x + 1) - (x + 1)) dx\), which simplifies to \(\int_{0}^{3} (-x^2 + 3x) dx\). The integral of \(-x^2\) with respect to x is \(-\frac{1}{3}x^3\) and the integral of \(3x\) is \(\frac{3}{2}x^2\). Now, substitute the upper and lower limits of integration: \(-\frac{1}{3}(3)^3 + \frac{3}{2}(3)^2 - (-\frac{1}{3}(0)^3 + \frac{3}{2}(0)^2)\). This simplifies to \(9\).

Step 4: Final Answer

So, the area of the region bounded by the two functions is 9 square units.