Question

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

Short answer

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

Step-by-step solution

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 \(x = 3\).

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.

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 \(9\).

Final Answer

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