posted by Jessy152 .
1. Find the area of the region bounded by the curves and lines y=e^x sin e^x, x=0, y=0, and the curve's first positive intersection with the x-axis.
2. The area under the curve of y=1/x from x=a to x=5 is approximately 0.916 where 1<=a<5. Using your calculator, find a.
3. Let R be the region bounded by y=6e^(-.2x) and y=√x and the lines x=1 and x=4. Find the volume when R is rotated about the x-axis.
4. Find the value(s) of b if the vertical line x=b divides the region between y=16-2x and the x and y-axis into 2 equal areas.
Please show work.
1. y=e^x sin e^x
Naturally, the first place where y=0 is at e^x=π, or x = ln π.
So, the area is just
∫[0,π] e^x sin e^x dx
That's just sin u du, so the value is 1 + cos(1)
2. just solve ln5 - ln a = .916
3. using shells,
v = ∫[1,4] 2πrh dx
where r = 1+x and h = y = 6e^(-.2x)-√x
Pretty easy integration
4. the region is just a triangle, so you want the area of the trapezoid on 0<x<b to equal the triangle on b<x<8
b(16+y(b))/2 = 1/2 (8-b)*y(b)
evaluate y at x=b, plug it in, and solve for b