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Wagner Horton
06/22/2023 · Middle School
a. \(\frac { 32} { 3} \)
UpStudy Free Solution:
To find the area enclosed by the curves \(y = 2x - x^ 2\) and \(y = 2x - 4\), we need to determine the points of intersection and then integrate the difference of the functions between these points.
First, set the equations equal to each other to find the points of intersection:
\[2x - x^ 2 = 2x - 4\]
Simplify the equation:
\[- x^ 2 = - 4\]
\[x^ 2 = 4\]
\[x = \pm 2\]
Thus, the curves intersect at \(x = - 2\) and \(x = 2\).
Next, we find the area between the curves by integrating the difference of the functions from \(x = - 2\) to \(x = 2\):
\[\text { Area} = \int _ { - 2} ^ { 2} [ ( 2x - 4) - ( 2x - x^ 2) ] \space dx\]
Simplify the integrand:
\[\int _ { - 2} ^ { 2} [ ( 2x - 4) - ( 2x - x^ 2) ] \space dx = \int _ { - 2} ^ { 2} ( - 4 + x^ 2) \space dx\]
Now, integrate term by term:
\[\int _ { - 2} ^ { 2} ( - 4 + x^ 2) \space dx = \int _ { - 2} ^ { 2} - 4 \space dx + \int _ { - 2} ^ { 2} x^ 2 \space dx\]
Calculate each integral separately:
\[\int _ { - 2} ^ { 2} - 4 \space dx = - 4 \int _ { - 2} ^ { 2} 1 \space dx = - 4 [ x] _ { - 2} ^ { 2} = - 4 ( 2 - ( - 2) ) = - 4 \cdot 4 = - 16\]
\[\int _ { - 2} ^ { 2} x^ 2 \space dx = \left [ \frac { x^ 3} { 3} \right ] _ { - 2} ^ { 2} = \frac { 2^ 3} { 3} - \frac { ( - 2) ^ 3} { 3} = \frac { 8} { 3} - \left ( - \frac { 8} { 3} \right ) = \frac { 8} { 3} + \frac { 8} { 3} = \frac { 16} { 3} \]
Combine the results:
\[\text { Area} = - 16 + \frac { 16} { 3} \]
To combine these terms, convert \(- 16\) to a fraction with the same denominator:
\[- 16 = - \frac { 48} { 3} \]
So,
\[\text { Area} = - \frac { 48} { 3} + \frac { 16} { 3} = - \frac { 32} { 3} \]
Since we are looking for the absolute area, we take the absolute value:
\[\text { Area} = \frac { 32} { 3} \]
Therefore, the correct answer is:
a. \(\frac { 32} { 3} \)
Supplemental Knowledge
To find the area enclosed between two curves, we use the integral of the difference between the functions over the interval where they intersect. Specifically, if \(y = f( x) \) is the upper curve and \(y = g( x) \) is the lower curve, then the area \(A\) between them from \(x = a\) to \(x = b\) is given by:
\[A = \int _ { a} ^ { b} [ f( x) - g( x) ] \space dx\]
First, we need to determine the points of intersection by setting \(f( x) = g( x) \). Then we integrate the difference of these functions over this interval.
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