Applications of Integration (simple problem, but not getting right answer)

[deadsupra]

Homework Statement

Applications of Integration: y=x^2+1 and y=sq root(1-x^2)
the graph (shown in the book) shows two curves from the two given equations (one obivously positive and the other negative) the shaded area is in between the two curves

the bounds are x=1 and the shaded area is only in quadrant I, so i assume x=0

Homework Equations

∫[a,b] f(x)-g(x)dx

The Attempt at a Solution

the bounds i chose are [0,1]
∫x^2+1 -(sqrt(1-x^2) dx
1/3*x^3 + x -(2/3*(1-x^2)^(3/2) * (-1/2x)) |1 to 0
simplified:
1/3*x^3 + x (1/(3x))(1-x^2)^(3/2)) |1 to 0

1/3*(1)^3 + (1) +(1/(3(1)))(1-(1)^2)^(3/2)) -[1/3*(0)^3 + (0) +(1/(3(0)))(1-(0)^2)^(3/2))]

simply 1/3 + 1 + (1/3)(0) - [0 + 0+ (1/0)]

the back of the book speakeths: 4/3 - pi/4

 

[Mark44]

Homework Statement

Applications of Integration: y=x^2+1 and y=sq root(1-x^2)
the graph (shown in the book) shows two curves from the two given equations (one obivously positive and the other negative) the shaded area is in between the two curves

the bounds are x=1 and the shaded area is only in quadrant I, so i assume x=0

Homework Equations

∫[a,b] f(x)-g(x)dx

The Attempt at a Solution

the bounds i chose are [0,1]
∫x^2+1 -(sqrt(1-x^2) dx
1/3*x^3 + x -(2/3*(1-x^2)^(3/2) * (-1/2x)) |1 to 0

Your work for the integral of the radical is incorrect. The standard way of doing this type of integral is using a trig substitution.

simplified:
1/3*x^3 + x (1/(3x))(1-x^2)^(3/2)) |1 to 0

1/3*(1)^3 + (1) +(1/(3(1)))(1-(1)^2)^(3/2)) -[1/3*(0)^3 + (0) +(1/(3(0)))(1-(0)^2)^(3/2))]

simply 1/3 + 1 + (1/3)(0) - [0 + 0+ (1/0)]

the back of the book speakeths: 4/3 - pi/4

 

[deadsupra]

duh dude... I knew that...

lol! thanks. That helped answer my next few questions. Thanks!