What is the area of the region bounded by y = x^3 + x, x = 2 and y = 0?

I got 10. Is this correct?

1 Answer
Feb 2, 2018

Area #= 6 " square units"#

Explanation:

#y=x^3+x#

Equate to #0#:

#x^3+x=0=>x(x^2+1)=0=>x=0#, Only real root.

For interval #[0,2]#

#y=x^3+x=(1)^3+(1)=2#

So we are above the x axis:

We need the integral:

#Area=int_(0)^(2)(x^3+x) dx=1/4x^4+1/2x^2=[1/4x^4+1/2x^2]_(0)^(2)#

Area#=[1/4x^4+1/2x^2]^(2)-[1/4x^4+1/2x^2]_(0)#

Plugging in upper and lower bounds:

Area#=[1/4(2)^4+1/2(2)^2]^(2)-[1/4(0)^4+1/2(0)^2]_(0)#

Area#=[4+2]-[0]=6#

Graph:

enter image source here

You must have made an error somewhere. It looks like you did not integrate #x^3+x#. This would give you #10# .i.e.:

#[x^3+x]^(2)-[x^3+x]_(0)#

#[(2)^3+(2)]^(2)-[(0)^3+(0)]_(0)#

#[8+2]-[0]=10#