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Chapter 11: Problem 35

Use a graphing utility to graph the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=x^{2}-4 x, g(x)=0 $$

Short Answer

Expert verified
The area of the region bounded by the graphs of the functions \( f(x)=x^{2} - 4x \) and \( g(x)=0 \) is \(-16/3\) square units

Step by step solution

01

Finding x-intercepts

To find the x-intercepts of the function \( f(x)=x^{2} - 4x \), set \( f(x) = 0 \) and solve for \( x \). This gives us the quadratic equation \( x^{2} - 4x = 0 \). Factoring out an \( x \) gives \( x(x - 4) = 0 \). From this we can see that the x-intercepts are \( x = 0 \) and \( x = 4 \).
02

Graphing the bounding curves

Graphing the function \( f(x)=x^{2} - 4x \) and the x-axis, \( g(x)=0 \), will give an understanding of the region covered. Since f(x) is a positively oriented parabola with roots at \( x = 0 \) and \( x = 4 \), the region of interest is the area enclosed by the parabola and the x-axis, between \( x = 0 \) and \( x = 4 \).
03

Calculating the Area

The area of the region between the x-axis and the curve \( f(x)=x^{2} - 4x \), between \( x = 0 \) and \( x = 4 \), is calculated by evaluating the definite integral of \( f(x) =x^{2} - 4x \) from 0 to 4. That is, the area A is given by the formula \( A = \int_{0}^{4}(x^{2} - 4x)dx \). By applying the Fundamental theorem of calculus, this gives \( A = [x^{3}/3 - 2x^{2}]_{0}^{4} = 16/3 - 32 = -16/3 \) square units.

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