Okay Update,

1.

Find the area enclosed by the curves:

y² = 4 + x and x + 2y = 4

I worked it out myself, so dont worry

2.

Fine the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by the y-axis and the curves y = 2x^2 - 1 and y =

√(x)

V = integral 1 to 0 of ( pi (√(x) - 2x^2 - 1)^2 )

So i expanded (√(x) - 2x^2 - 1)^2

to get (simplified): 4x^4 - 4x^2.5 - 4x^2 + 2√(x) + x + 1

then got the integral is:

pi((4/5)x^5 - (4/3.5)x^3.5 - (4/3)x^3 + (2/1.5)x^1.5 + (x^2)/2 + x) between 0.42 and 0

so subbing in those values: pi((81/70) - 0)

= 3.64 units^3

Correct?

3.

Use the method of cylindrical shells to find the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by y = √(x+1), y = 0, x=0 and x=4

V = integral_{4 to 0} of [ 2pi x ( √(x)+1 ) ] dx

= (2pi)integral_{4 to 0} of (x^1.5 + x) dx

= (2pi)[(x^2.5)/2.5 + (x^2)/2] between 4 and 0

= (2pi)(12.8 + 8)

= 41.6 pi units^3

Correct?

4.

Solve

dy/dx = (x(y²+3)/y)

(y/(y^2+3)) dy/dx = x

so: integral (y/(y^2+3)) dy = integral x dx + C

so the integral is: (1/2)ln(y^2+3) = x^2/2 + C

then, we say ln(y^2+3) = x^2 + A

where A = 2C

so, y^2 = = e^(x^2 + A) - 3

y = √(e^(x^2 + A) - 3)

Correct?

5.

Solve

dy/dx = e^(x-2y), y(1) = 0

ok.. so im pretty much stumped after this step

ln(dy) - ln(dx) = x - 2y

ln(dy) + 2y = ln(dx) + x

Help?

6.

dy/dx = x²y-2e^(x), y(1) = -1

Use Euler method with 4 steps to find and approx value for y(2)

The formula for calculating succcessive values of x and y is :

y_n+1 = y_n + 0.25[(x_n^2)(y_n)-2e^(x_n)]

(_n+1 means subscript n+1 etc)

y_1 = -1 + 0.25[(1^2)(-1)-2e^(1)] = -2.6091

y_2 = -2.6091 + 0.25[(1.25^2)(-2.6091)-2e^(1.25)] = -5.3735

y_3 = -5.3735 + 0.25[(1.50^2)(-5.3735)-2e^(1.50)] = -10.6369

y_4 = -10.6369 + 0.25[(1.75^2)(-10.6369)-2e^(1.75)] = -21.6581

So the value at y(2) = -21.6581

Yes? Correct?