ANSWERS
a) 3
b) -6.28
c) -1.78
EXPLANATION
The value of the integral of a function is the area "under" the graph of the function. It is usually called the area under, but it is the area between the curve and the x-axis.
a) The graph of the function between x = 0 and x = 1 is a straight line, and it forms a rectangle with a triangle on top,
The area under the curve is the sum of the areas of each shape.
The rectangle's base is 1 unit and its height is 2 units. The triangle's base is the same as the rectangle, 1 unit, and its height is also 2 units.
[tex]\int ^1_0g(x)dx=A_{rec\tan le}+A_{triangle}=(1\cdot2)+\frac{1\cdot2}{2}=2+1=3[/tex]
b) Between x = 2 and x = 6, the graph of the function is a semicircle with radius 2, so the integral of g(x) between 2 and 6 is the area of the semicircle, but since it is below the x-axis, we have a negative area,
[tex]\int ^6_2g(x)dx=-A_{semicircle}=-\frac{1}{2}\pi r^2=-\frac{1}{2}\pi\cdot2^2=-\frac{1}{2}\pi\cdot4=-2\pi\approx-6.28[/tex]
c) Now we have to find the area under the curve for the domain. Since the graph has different shapes between 0 and 7, we can split this integral with the intervals of each shape,
[tex]\int ^7_0g(x)dx=\int ^2_0g(x)dx+\int ^6_2g(x)dx+\int ^7_6g(x)dx[/tex]
The second integral is the one we found in part b. The first integral is the area of the first triangle in the graph. Its base is 2 units and its height is 4 units,
[tex]\int ^2_0g(x)dx=\frac{2\cdot4}{2}=4[/tex]
And the last integral is also the area of a triangle, but its base is 1 unit and its height is 1 unit,
[tex]\int ^7_6g(x)dx=\frac{1\cdot1}{2}=\frac{1}{2}[/tex]
The result of the integral of g(x) between 0 and 7 is,
[tex]\int ^7_0g(x)dx=4-2\pi+\frac{1}{2}=\frac{9}{2}-2\pi\approx-1.78[/tex]
