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Section 7.1 #6: Set up the definite integral that gives the area in the region bounded by y1=(x1)3 and y2=x1:
Solution: First draw the region:

We will use a dx integral. It will require two such integrals. For the left region, the function on top is y1 and the function on the bottom is y2 and in the right region, the function on top is y2 and the function on the bottom is y1. Therefore we may compute the area with Area=LeftArea+RightArea=10(x1)3(x1)dx+21(x1)(x1)3dx. Section 7.1 #16: Find the area of the region bounded by y=x2 and y=6x by integrating...
a.) with respect to x, and
b.) with respect to y.
c.) Compare your results. Which was simpler?
Solution: First draw the region:

For a.), the top function is always the linear function and the bottom function is the quadratic. Therefore the area is given by Area=23(6x)x2dx=6xx22x33|23=(124283)(1892+273)=1256. For b.), the right and left functions change at (2,4). First rewrite y=x2 as x=±y and rewrite y=6x as x=6y. Between y=0 and y=4, the function x=y is on the right and the function x=y is on the left. Between y=4 and y=9, the function x=6y is on the right and the function x=y is on the left. Therefore, Area=40y(y)dy+94(6y)(y)dy=240y12dy+946y+y12dy=2[y3232|40+[6yy22+y3232|94=43[4320]+[(54812+23932)(24162+23432)]=323+[632643]=1256. For c.), we got the same result both ways and the dx integral was much simpler!

Section 7.1 #17: Sketch the region bounded by the graphs of the equations and find the area of region: y=x21, y=x+2, x=0, and x=1.
Solution: First find intersection points of the first two curves by setting them equal and solving:
x21=x+2, which is algebraically equivalent to x2+x3=0. Solving this equation (using the quadratic formula) yields the roots x=1±124(1)(3)2=12±132. Now sketch the region:

It is easier to do this as a dx integral: compute Area=10(x+2)(x21)dx=10x2x+3dx=x33x22+3x|10=(1312+3)0=136. Section 7.1 #21: Sketch the region bounded by the graphs of the equations and find the area of region: y=x, y=2x, y=0.
Solution: First find the intersection points between y=x and y=2x by setting them equal: x=2x, hence x=1. Therefore (1,1) is the intersection point of the two curves. Now sketch the region:

It is easier to do this integral as a dy (notice that if we did a dx, then it would require two integrals). The "right" function is x=2y and the "left" function is x=y. Therefore we compute Area=10(2y)ydy=1022ydy=2yy2|10=(21)0=1.

Section 7.2 #22: Sketch the region bounded by the graphs of the equations and find the area of region: y=4x3, y=0, x=1, and x=4.
Solution: Sketch the region:
This is easier to do as a dx integral, so compute 414x3dx=441x3dx=42x2|41=42142+4211=218=158.